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Chapter
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Expressions and Equations ..................................................................... pages 31–72
2
INVESTIGATION 2A
2.01
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The long way would go as follows: 17 · 4 + 17 · 3 + 17 · 2 + 17 · 1 68 + 51 + 34 + 17 = 17 17 170 = 17 = 10 If you use the Distributive Property, you can do it a little faster: 17 · 4 + 17 · 3 + 17 · 2 + 17 · 1 17 17 · 4 17 · 3 17 · 2 17 · 1 + + = + 17 17 17 17 =4+3+2+1
EXPRESSIONS
Getting Started
For You to Explore 1. (a) Try different original numbers and see if you find a simple pattern. Spiro knows that all those different steps end up being the same as the single step “add 11 to the original.” (b) If you start with 2, you’ll get back 13. (c) If you start with −5, you’ll get back 6. 2. (a) The ending number is 2 every time. (b) To see why this trick works, follow these steps. • • • • • • •
Choose a number: x Multiply by 3: 3x Subtract 4: 3x − 4 Multiply by 2: 2(3x − 4) = 6x − 8 Add 20: 6x − 8 + 20 = 6x + 12 Divide by 6: 6x+12 =x+2 6 Subtract your original number: x + 2 − x = 2
So the ending number will always be 2 no matter what number you choose to start with. 3. Answers will vary, since students get to invent their own.
On Your Own 4. Yes it works and it will always work. If your starting number is called x, then after the first step you have x + 5. After the second step you have 2(x + 5) which equals 2x + 10. After the third step you have 2x + 10 − 7 which equals 2x + 3. Next, you add 1, which gives 2x + 4. Dividing by 2 next, you have x + 2, then subtracting 2 gives x. Aha! You start with x, which can be any number, and you get x back! So, the trick always works, unless a calculation error is made. 5. You get twice the starting number every time. The steps look like this, if you label your starting number x. x −→ x + 3 −→ 2x + 6 −→ 2x + 13 −→ 2x − 2 −→ 2x 6. Answers may vary. 7. You can evaluate this expression in two different ways: a long way, and a shorter way.
= 10 Maintain Your Skills 8. (a) 2 + 5 = 7, 7 − 3 = 4, 4 × 2 = 8, and 8 − 4 = 4. (b) 3 + 5 = 8, 8 − 3 = 5, 5 × 2 = 10, and 10 − 4 = 6. (c) 7 + 5 = 12, 12 − 3 = 9, 9 × 2 = 18, and 18 − 4 = 14. (d) (−2) + 5 = 3, 3 − 3 = 0, 0 × 2 = 0, and 0 − 4 = (−4). (e) The ending number is always double the starting number. 9. Take a number. Add three. Multiply by three. Subtract three. Divide by three. (a) 2 + 3 = 5, 3(5) = 15, 15 − 3 = 12, 12 ÷ 3 = 4. (b) 3 + 3 = 6, 3(6) = 18, 18 − 3 = 15, 15 ÷ 3 = 5. (c) 7 + 3 = 10, 3(10) = 30, 30 − 3 = 27, 27 ÷ 3 = 9. (d) −2 + 3 = 1, 3(1) = 3, 3 − 3 = 0, 0 ÷ 3 = 0. (e) The ending number is 2 more than the starting number.
2.02
Modeling General Situations— Writing Expressions
Check Your Understanding 1. (a) Barbara is 10, because she’s one year younger than Mary. (b) Barbara is 6, because she’s one year younger than Mary. (c) Barbara is 52, because she’s one year younger than Mary. (d) Barbara is 64. (e) Barbara is m − 1 years old. (f) Mary is n + 1 years old.
Mathematics I Solutions Manual
• Chapter 2, page 31
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2. (a) Doctors (one doctor for 100 adults means 1 adult for 1 doctors.) 100 (b) Gallons of water per week. (c) Medical kits (d) Pillows (e) Blankets 3. (a) II (b) IV (c) I (d) III (e) V 4. (a) Choose a number. Multiply by 2. Then add 1. (b) Choose a number. Subtract 1. Then multiply by −2. (c) Choose a number. Add 2. Multiply by 5. Then subtract 2. (d) Choose a number. Add 1. Multiply by 3. Subtract 2. Multiply by 7. Then subtract 9. 5. 2 · ((x + 5) − 11) + 3 On Your Own 6. Jeremy has 3 more apples than Ricardo. (a) Jeremy has 5 apples (three more than two). (b) Jeremy has 16 apples (three more than 13). (c) Jeremy has 110 apples (three more than 107). (d) Jeremy has r + 3 apples (three more than r). 7. (a) II (b) III (c) I (d) V 8. Travis forgot the parentheses. 10(x − 3) − 13 is the correct expression. 9. Since r is the number of quarters Rob has, the number of quarters that David has must be r + 24. Therefore, the total number of quarters that they both have together is r + (r + 24), or 2r + 24. The correct answer is D. 10. (a) The area is equal to lw. (b) The perimeter is equal to 2l + 2w. 11. (a) x2 (Half as many are in computer.) (b) 2x (Twice as many are in gym.) (c) x − 2 (Since two more are in math, that means two less are in history.) (d) x + 2 (Two more are in art class.) (e) The total number of students in the Computer, Gym, History, and Art classes is x2 + 2x + (x − 2) + (x + 2), or 29 x. Maintain Your Skills 12. (a) 38 in. (since 20 is double 10, you need to double 19, and that gives you 38). (b) 9.5 in. (since 5 is half of 10, you need to halve 19, which gives you 9.5). (c) 1.9 in. (since 1 is a tenth of 10, you need to divide 19 by 10, which is 1.9). (d) 1.9x inches (just check; every time the length is 1.9 times the width). 13. The long side should be four times the length of the short side.
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(a) 80 cm (that’s four times the short side, which is 20 cm) (b) 100 cm (that’s four times the short side, which is 25 cm) (c) 4x cm (four times the short side)
2.03
Evaluating Expressions
Check Your Understanding 1. For 500 adults and 100 children, there should be a + c + 10 = 500 + 100 + 10 = 610 beds. There should be 5a + 2c = 5(500) + 2(100) = 2700 pounds of food. There should be 9 · (500) + 5 · (100) = 5000 gallons of water. For 1000 adults and 300 children, there should be a + c + 10 = 1000 + 300 + 10 = 1310 beds. There should be 5a + 2c = 5(1000) + 2(300) = 5600 pounds of food. There should be 9 · (1000) + 5 · (300) = 9000 + 1500 = 10, 500 gallons of water. 2. (a) 10 (b) 10 (c) 10 (d) 10 3. (a) 3(n − 1) + 5 (b) 35 (c) 5 4. (a) 5 · (3n − 2) (b) 50 (c) 10 (b + t) 5. ·h 2 5 (2 + 3) (5 + 3) 8 ·7= ·7 (b) (a) ·6= ·6 2 2 2 2 =4·6 = 24 4 +2 7 7 2 (d) (c) · = 2 · 2 4 2 4 2 7 = · 2 4 7 =1· 4 7 3 = or 1 4 4 6. (a) 3 · (7 + 3) = 30 (b) (−3 + 3) · 5 + 11 = 11 (c) −3 + 3 · (5 + 11) = 45 (d) 25 − (5 + 4 · 5) = 0 (e) 25 − (5 + 4) · 5 = −20 3
1
On Your Own 7. (a)
Mathematics I Solutions Manual
1 2
(b)
1 3
(c)
1 4
(d)
• Chapter 2, page 32
1 5
=
35 or 17 21 2
14 1 (8 + 6) 1 · = · 2 4 2 4 1 =7· 4 7 3 = or 1 4 4
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8. Linda made an error with order of operations. When calculating 7 + 5 · (3), she should have multiplied first, then added.
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12. 14 − 6 + 32 ÷ 8 · 2 = 14 − 6 + 4 · 2 = 14 − 6 + 8 =8+8 = 16
7 + 5 · (3) = 7 + 15 = 22 She made the same mistake when x = 45 . She should have multiplied first. 7 + 5 · 45 = 7 + 4 = 11
The answer is C. Maintain Your Skills
9.
13. Evaluate the expression
5 + 3x (a) (b) (c) (d) (e) (f)
(x − 1) · (x − 2) · (x − 3) · (x − 4) · (x − 5)
5 + 3(3) = 5 + 9 = 14 5 + 3(−1) = 5 + (−3) = 2 5 + 3(−5) = 5 + (−15) = −10 5 + 3(0) = 5 + 0 = 5 5 + 3(−11) = 5 + (−33) = −28 5 + 3(−3) = 5 + (−9) = −4
for each value of x listed below: (a) (1 − 1) · (1 − 2) · (1 − 3) · (1 − 4) · (1 − 5) = 0 · (−1) · (−2) · (−3) · (−4) = 0 (b) (2 − 1) · (2 − 2) · (2 − 3) · (2 − 4) · (2 − 5) = 1 · 0 · (−1) · (−2) · (−3) = 0 (c) (3 − 1) · (3 − 2) · (3 − 3) · (3 − 4) · (3 − 5) = 2 · 1 · 0 · (−1) · (−2) = 0 (d) (4 − 1) · (4 − 2) · (4 − 3) · (4 − 4) · (4 − 5) = 3 · 2 · 1 · 0 · (−1) = 0 (e) (5 − 1) · (5 − 2) · (5 − 3) · (5 − 4) · (5 − 5) = 4·3·2·1·0= 0 (f) The answer every time is 0. There is not another similar case, because the only way for the product to be 0 is if one of the differences in parentheses is equal to 0, and all of those possibilities are covered.
10. (z + 3)2 − 4 (a) (b) (c) (d) (e) (f)
((3) + 3)2 − 4 = (6)2 − 4 = 36 − 4 = 32 ((−1) + 3)2 − 4 = (2)2 − 4 = 4 − 4 = 0 ((−5) + 3)2 − 4 = (−2)2 − 4 = 4 − 4 = 0 ((0) + 3)2 − 4 = (3)2 − 4 = 9 − 4 = 5 ((−11) + 3)2 − 4 = (−8)2 − 4 = 64 − 4 = 60 ((−3) + 3)2 − 4 = (0)2 − 4 = 0 − 4 = −4
11. (a) d t 100 m r= 40 sec r = 2.5 m per sec
r=
(b)
14. The first expression (the complicated one) is 3x plus the expression from Exercise 12. You can use your work from that exercise to simplify this problem. (a) (b) (c) (d)
d t 50 yds r= 40 sec r = 2 yd per sec r=
(c) If she can run 6 miles in 1 hour, how far would she run in a half hour? Well, you’d think that half the time she’d cover half the distance, and you’d be right. Assuming she runs at a steady pace, she’d run 3 miles in 30 minutes (which is a half hour). How could you solve the exercise using the formula?
2.04
when x = 1, you get 3(1) + 0 = 3. when x = 2, you get 3(2) + 0 = 6. when x = 4, you get 3(4) + 0 = 12. As you saw in the previous exercise, 1, 2, 3, 4, and 5 all make (x − 1) · (x − 2) · (x − 3) · (x − 4) · (x − 5) equal to 0, so the two expressions will have the same value.
Simplifying Expressions
Check Your Understanding 1. Here are the expressions associated with each step:
d r= t d 0.5 hr d (6 mi/hr) · (0.5 hr) = · (0.5 hr) 0.5 hr 3 mi = d 6 mi/hr =
Mathematics I Solutions Manual
• • • • • • •
Choose any number: x. Multiply by 2: 2x. Add 7: 2x + 7. Multiply by 5: 5(2x + 7) = 10x + 35. Add 25: 10x + 35 + 25 = 10x + 60. Divide by 10: (10x+60) = 10(x+6) = x + 6. 10 10 Subtract your starting number: x + 6 − x = 6.
• Chapter 2, page 33
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Choose any number: x. Multiply by 2: 2x. Add 7: 2x + 7. Multiply by 5: 5(2x + 7) = 10x + 35.
This trick is very similar to exercise 1. In that exercise, you added 25, which gave you 10x + 60, and when you divided by 10 to get x + 6. After subtracting x, 6 was the ending number. Here though, you want a ending number of 7, so you want x + 7 left over after you divide by 10, which means you want 10x + 70 before you divide by 10. So, instead of adding 25 you should add 35. Here is the rest of the trick. • • • •
Missing Step: Add 35: 10x + 35 + 35 = 10x + 70. Divide by 10: x + 7. Subtract your starting number: 7. Your ending number is 7.
3. Trace the number trick through the steps. • • • • • •
Choose any number: x. Multiply by 3: 3x. Add 5: 3x + 5. Multiply by 4: 12x + 20. Add 16: 12x + 36. Divide by 12: x + 3.
(a) If you tell the person to subtract 3, then they will be left with x, their starting number. (b) If you tell the person to subtract their starting number, the x will disappear and they will be left with 3.
6. (a) It does equal 4x + 2y, because 6y − 4y = 2y and you are adding that to 4x. (b) It does equal 4x + 2y, because 4(x + y) equals 4x + 4y and subtracting 2y from that leaves you with 4x + 2y. (c) No, it does not equal 4x + 2y. For example, if you let x = 1 and y = 0, then this expression equals 0, but 4x + 2y equals 4(1) + 2(0) = 4. They are not equivalent expressions. (d) No, for the same reason as part (c). (e) Yes, because x + x + x + x + x + x = 6x, −y − y − y − y = −4y, and 2(3y − x) = 6y − 2x. So, when you sum it all up, you have 6x − 4y + 6y − 2x = 4x + 2y. 7. (a) The area is 2(5x + 9) = 10x + 18, and the perimeter is 2(5x + 9) + 4 = 10x + 22. (b) The area is 11(x − 3) = 11x − 33, and the perimeter is 2(x − 3) + 22 = 2x + 16. (c) The area is 2(2x + 9) = 4x + 18, and the perimeter is 2(2x + 9) + 4 = 4x + 22. (d) The area is 5(8 − 3x) = 40 − 15x, and the perimeter is 2(8 − 3x) + 10 = 26 − 6x. 8. Follow the number tricks all the way through: • Choose a Number: x. • Add 6: x + 6. • Multiply by 3: 3x + 18. • Subtract 4: 3x + 14. • Multiply by 2: 6x + 28. • Add 2: 6x + 30. • Divide by 6: x + 5. • What’s your ending number? x + 5.
• Choose a Number: x. • Multiply by 2: 2x. • Subtract 6: 2x − 6. • Multiply by 5: 10x − 30. • Add 50: 10x + 20. • Divide by 10: x + 2. • Subtract your starting number: 2. • What’s your ending number? 2. (a) He can figure it out using the Trick 1, because the ending number will always be 5 more than the starting number. (b) The ending number is 2, no matter what the starting number is. So, when someone tells him that the ending number is 2, it doesn’t help him to determine the starting number.
x + 2x + 3x + 4x + 5x x in each case: 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) 15 (This can be checked on a calculator.) x + 2x + 3x + 4x + 5x x x 2x 3x 4x 5x = + + + + x x x x x 1 2 3 4 5 = + + + + 1 1 1 1 1 = 1 + 2 + 3 + 4 + 5 = 15
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On Your Own
4. Evaluate the expression:
(a) (b) (c) (d) (e) (f) (g) (h)
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(b) The area is 7(x − 4) = 7x − 28, and the perimeter is 7 + (x − 4) + 7 + (x − 4) = 14 + 2x − 8 = 6 + 2x. (c) The area is (6x − 8) · 12 = 3x − 4, and the perimeter is (6x − 8) + 12 + (6x − 8) + 12 = 12x − 16 + 1 = 12x − 15. (d) The area is 9(10 − 2x) = 90 − 18x, and the perimeter is (10 − 2x) + 9 + (10 − 2x) + 9 = 20 − 4x + 18 = 38 − 4x.
The trick is that you get six every time. It will always work, because when you choose x, which could be any number, and perform the steps, the end result is 6. Since the ending number doesn’t have x in it, the answer will be six no matter what you choose as a starting number. 2. Choose x as a starting number and see what happens. • • • •
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9. If you start by simplifying the expression, that makes all of the other parts easier.
5. (a) The area is 3(4x + 2) = 12x + 6, and the perimeter is (4x + 2) + 3 + (4x + 2) + 3 = 8x + 10. Mathematics I Solutions Manual
2(3m + 5) − 5(m + 1) − 4 = 6m + 10 − 5m − 5 − 4 = 6m − 5m + 10 − 5 − 4 =m+1 • Chapter 2, page 34
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So, the expression simplifies to m + 1. You can use this to answer the rest of the exercises. (a) (b) (c) (d) (e) (f) (g)
If m = 3 then m + 1 = 4. If m = 17 then m + 1 = 18. If m = −2 then m + 1 = −1. If m = 4 then m + 1 = 5. If m = 13 then m + 1 = 43 . 4 7 If m = − 11 then m + 1 = 11 . As above, the expression 2(3m + 5) − 5(m + 1) − 4 simplifies to m + 1.
(a) (b) (c) (d) (e) (f) 16. (a) (b)
13. (a) (b) (c) (d) (e) (f) (g) (h)
14. (a) (b) (c) (d) (e) (f) (g) (h)
(i)
(c) (d)
1 · (1 + 1) = (1) · (2) = 2 2 · (2 + 1) = (2) · (3) = 6 3 · (3 + 1) = (3) · (4) = 12 4 · (4 + 1) = (4) · (5) = 20 11 · (11 + 1) = (11) · (12) = 132 −3 · (−3 + 1) = (−3) · (−2) = 6 −7 · (−7 + 1) = (−7) · (−6) = 42 It’s because the product of an even number and any other number is an even number. x can be either even or odd, and there is no third alternative. If x is even, then x · (x + 1) is even. If x is odd, then x + 1 must be even, and so x · (x + 1) must also be even. 2(1 + 1) = 2(2) = 4 2(2 + 1) = 2(3) = 6 2(3 + 1) = 2(4) = 8 2(4 + 1) = 2(5) = 10 2(11 + 1) = 2(12) = 24 2(−3 + 1) = 2(−2) = −4 2(−7 + 1) = 2(−6) = −12 The product of an even number and any integer will always be even. Since 2 is even, the product of it and any integer is even. Since x is an integer, x + 1 is an integer, and so 2 · (x + 1) is an even integer. Adding 1 to a number changes whether it’s even or odd. For example, 2 is even, but 2 + 1 = 3 is odd. Since 2x is two times a number, it’s even. Adding 1 to 2x makes it odd. So, 2x + 1 is always odd.
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x + 2x + 3x + 4x + 5x − 12x = 15x − 12x = 3x x + 2x + 3x + 4x + 5x − 13x = 15x − 13x = 2x x + 2x + 3x + 4x + 5x − 14x = 15x − 14x = x x + 2x + 3x + 4x + 5x − 15x = 15x − 15x = 0 x + 2x + 3x + 4x + 5x − 16x = 15x − 16x = −x x + 2x + 3x + 4x + 5x − 17x = 15x − 17x = −2x 1−22 = 1−4 = −3 =3 1−2 −1 −1 1−32 1−9 −8 = −2 = −2 = 4 1−3 1−(−3)2 = 1−9 = −8 = −2 1−(−3) 1+3 4 1−(−11)2 1−121 −120 = 12 = 12 = 1−(−11)
2 1− 12 = (e) 1 1− 2
(f) (g) 17. (a) (b) (c) (d)
1− 14
1− 12
=
3 4 1 2
=
6 4
=
−10 3 2
198 1−x2 1−x
=1+x (2 + 1)2 − 22 = (3)2 − (2)2 = 9 − 4 = 5 (3 + 1)2 − 32 = (4)2 − (3)2 = 16 − 9 = 7 (−3 + 1)2 − (−3)2 = (−2)2 − (−3)2 = 4 − 9 = −5 (−11 + 1)2 − (−11)2 = (−10)2 − 121 = 100 = − 121 2 2 −21 2 (e) 12 + 1 − 12 = 32 − 14 = 94 − 14 = 2 (f) 395 (g) (x + 1)2 − x2 = 2x + 1
2.05
Maintain Your Skills
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15. For each exercise, notice that x + 2x + 3x + 4x + 5x = 15x.
10. 7(6t + 2) + 3 − 5(t + 1) = 42t + 14 + 3 − 5t − 5 = 37t − 12 Therefore, the answer is B. 11. If you buy the clothes at separate stores, the taxes are 0.06s for the shirt and 0.06p for the pants, so the total is 0.06s + 0.06p. If you buy them at the same store, you pay taxes on the total, or 0.06(s + p). By the Distributive Property, the two expressions are equal. 12. (a) If 28 students went on the trip, the balance would be 300 − 90 − 6(28) = 42. If 29 students went on the trip, the balance would be 300 − 90 − 6(29) = 36. If 30 students went on the trip, the balance would be 300 − 90 − 6(30) = 30. (b) 35, because a greater number of students would cause the balance to be negative.
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Rephrasing the Basic Rules
Check Your Understanding a ÷ b = a · ( b1 ) c − d = c + (−d) ab = (−a)(−b) If ab = 1, then b = a1 and a = b1 . Because if a = 0 is true, then a = 0 or b = 0 is true, so the proof is complete. (b) You know this because every real number except 0 has a reciprocal. (c) The left side becomes b and the right side becomes 0. (d) This proves the Zero Product Property because it shows that either a or b has to equal zero. That is, if a doesn’t equal zero, then b has to equal zero.
1. (a) (b) (c) (d) 2. (a)
On Your Own 3. (a) The definition uses x and y. Replace the x with 4 and y with 6 to get 4 ♥ 6 = 3(4) + 6 = 12 + 6 = 18 (b) 6 ♥ 4 = 3 · 6 + 4 = 18 + 4 = 22 (c) Look at the solutions to parts a and b to see that 4 ♥ 6 = 6 ♥ 4. So ♥ is not commutative. 4. (a) Yes, ♠ is commutative. You can check by seeing if −3(x + y) = −3(y + x) is true. And, since the two numbers are inside the parentheses, and they are added (and addition is commutative), then ♠ is also commutative.
Mathematics I Solutions Manual
• Chapter 2, page 35
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(b) No, ♠ isn’t associative. It’s a little tougher to show this using variables, so pick some numbers. You need three, so try 1, 2, and 3. First, test (1♠2)♠3. (1♠2) = −3(1 + 2) = −3(3) = −9 So substitute −9 for 1♠2, and continue. (−9)♠3 = −3((−9) + 3)) = −3(−6) = 18 So (1♠2)♠3 = 18. Now, test 1♠(2♠3). (2♠3) = −3(2 + 3) = −3(5) = −15 So substitute −15 for 2♠3, and continue. 1♠(−15) = −3(1 + (−15)) = −3(−14) = 42
5. (a) (b) (c) (d) (e)
(f)
6. (a) (b) (c) (d) (e) (f) 7. (a) (b) (c) (d)
In order for the operation to be associative, it has to be true for any numbers you choose. Since 18 = 42, you have shown that ♠ is not associative. 2(x + 2) − (x + 2) = x + 2 2(x + 2) − (x + 2) + 4(x + 2) − 3(x + 2) = (x + 2) + (x + 2) = 2(x + 2) = 2x + 4 2(x + 2) − (x + 2) + 4(x + 2) − 3(x + 2) + 6(x + 2) − 5(x + 2) = 3(x − 2) = 3x + 6 2(x + 2) − (x + 2) + 4(x + 2) − 3(x + 2) + 6(x + 2) − 5(x + 2) + 8(x + 2) − 7(x + 2) = 4(x + 2) = 4x + 8 2(x + 2) − (x + 2) + 4(x + 2) − 3(x + 2) + 6(x + 2) − 5(x+2)+8(x+2)−7(x+2)+10(x+2)−9(x+2) = 5(x + 2) = 5x + 10 You always get 0. Every time you see x, it is in parentheses as part of the expression x + 2. If you substitute −2 for x, you have (−2) + 2, which equals 0. You then have a whole bunch of numbers, multiplied by 0, then added. Each multiplication result will be 0, and if you add any number of zeroes, you’ll get 0. 4(x + 2) + 11 = 4x + 8 + 11 = 4x + 19 x + 2(5 + 2x) = x + 10 + 4x = 5x + 10 9(2x − 5) − 3 = 18x − 45 − 3 = 18x − 48 5(x − 1) + 8(x + 1) = 5x − 5 + 8x + 8 = 13x + 3 7(x + 1) + (7x + 7) = 7x + 7 + 7x + 7 = 14x + 14 7(x + 1) + (−1)(7x + 7) = 7x + 7 − 7x − 7 = 0 2(x + 4) + 7 = 2x + 8 + 7 = 2x + 15 13 + 3(1 + 2x) = 13 + 3 + 6x = 16 + 6x 3(2x − 5) − 8 = 6x − 15 − 8 = 6x − 23 4(x + 3) + 7(x + 3) = 4x + 12 + 7x + 21 = 4x + 7x + 12 + 21 = 11x + 33 Or, you could say,
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4(x + 3) + 7(x + 3) = (4 + 7)(x + 3) = 11(x + 3) = 11x + 33 (e) 6(3 − 2x) − 3(x + 1) = 18 − 12x − 3x − 3 = 15 − 15x (f) 4(x − 7) − 2(2 − 3x) = 4x − 28 − 4 + 6x = 10x − 32 8. (a) • Choose a number: x. • Add 6: x + 6. • Multiply by 3: 3(x + 6) = 3x + 18. • Subtract 10: (3x + 18) − 10 = 3x + (18 − 10) = 3x + 8. • Multiply by 2: 2(3x + 8) = 6x + 16. • Add 50: (6x + 16) + 50 = 6x + (16 + 50) = 6x + 66. • Divide by 6: 6x+66 = 6(x+11) = x + 11. 6 6 (b) In the third and fifth steps you use the Distributive Property, in the last step you use the Multiplicative Inverse Rule. In the fourth and sixth steps you use the Associative Property of Addition. (c) You’re left with x + 11. To get rid of the “+11” part, you just have to add the additive inverse of 11, which is −11. That’s equivalent to subtracting 11. 9. a ⊗ b = ab + a, but b ⊗ a = ba + b = ab + b. In general ab + a = ab + b, and so ⊗ is not commutative. In checking associativity, we have a ⊗ (b ⊗ c) = a ⊗ (bc + b) = a(bc + b) + a = abc + ab + a and (a ⊗ b) ⊗ c = (ab + a) ⊗ c = (ab + a)c + (ab + a) = abc + ac + ab + a = abc + ab + a + ac Since in general abc + ab + a = abc + ab + a + ac, the operation ⊗ is not associative. Therefore, the correct answer is D. 10. If c is the cost of a CD, the first expression works out to (c · (0.8)) + (c · (0.8))0.05. (Take the twenty % discount first, then add the 5% sales tax.) If you add tax first and then take the discount you’ll get: (c + 0.05c) · 0.8. Using the Distributive Property on the second expression gives the following equation.
Mathematics I Solutions Manual
(c + 0.05c) · 0.8 = c · 0.8 + 0.05c · 0.8 Applying the commutative and associative properties, rewrite the right side of the equation. c · 0.8 + 0.05c · 0.8 = (c · (0.8)) + (c · (0.8))0.05 Now, look at the right side of this equation. That’s the same expression you found when you took the discount first, and then added tax. In other words, it doesn’t matter whether they calculate the discount first or the tax first. The answer is the same. Try it with numbers, to confirm the identity. • Chapter 2, page 36
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Maintain Your Skills 11. (a) (b) (c) (d)
12. (a) (b) (c) (d)
13. (a) (b) (c) (d)
2A
40 2130 918,270 Remember your any-order, any-grouping principles of arithmetic. If you shuffle things around, you see that in each exercise, you can multiply 5 · 2 = 10 first. It’s easy, then, to multiply 4 · 10 to get 40, 213 · 10 to get 2130, and 91,827 · 10 to get 918,270. 1700 2200 19,700 Just as in Exercise 10, shuffling the multiplication lets you multiply 25 · 4 first to get 100. You can then finish the problem easily by multiplying the remaining number by 100: 17 · 100 = 1700, 22 · 100 = 2200, and 197 · 100 = 19,700 540 810 11,130 Once again, using the any-order, any-grouping principles makes this exercise much easier, especially given that 10 × 13 = 10. So by 13 rearranging the multiplication, you can end up with 3 simple problems: 54 · 10 = 540, 81 · 10 = 810, and 1113 · 10 = 11,130.
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(a) (b) (c) (d)
2 · w + l feet 2 · 8 + 12 = 16 + 12 = 28 feet 2 · 5 + 20 = 10 + 20 = 30 feet 2 · w + (w + 9) or 3 · w + 9 feet
2. (a) 4(2x+3)+2(x+1)−7 = 4(2·1+3)+2(1+1)−7 = 4(5) + 2(2) − 7 = 20 + 4 − 7 = 17 (b) 4(2x + 3) + 2(x + 1) − 7 = 4(2 · 6 + 3) + 2(6 + 1) − 7 = 4(12 + 3) + 2(7) − 7 = 4(15) + 14 − 7 = 60 + 14 − 7 = 67 (c) 4(2x+3)+2(x+1)−7 = 4(2(−2)+3)+2(−2+1)− 7 = 4(−4 + 3) + 2(−1) − 7 = 4(−1) + (−2) − 7 = −4 + (−2) − 7 = −4 + (−2) + (−7) = −13 (d) 4(2x+3)+2(x+1)−7 = 4(2· 12 +3)+2( 12 +1)−7 = 4(1 + 3) + 2( 32 ) − 7 = 4(4) + 3 − 7 = 16 + 3 − 7 = 12 (e) 4(2x + 3) + 2(x + 1) − 7 = 8x + 12 + 2x + 2 − 7 = 10x + 7. Substitute x = 1 and you get 10(1) + 7 = 17. Substitute x = 6 and you get 10(6) + 7 = 60+ 7 = 67. The answers are the same because the expressions represent the same number. This will always be true. 3. Answers may vary. For example: Variables are useful because they let you describe a general situation more easily and conveniently. 4. Answers may vary. One way is to create steps so that the original number has no effect on the final number. For
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example, if you multiply the number by 2, and, later subtract 2 times the original number. Then you can always know the final number. Here is an example: • Pick a number. • Multiply the number by 2. • Add 7. • Subtract 2 times your original number. • Add 5. • Multiply by 3. The result will always be 36. 5. (a) 4 → 20 → 22 → 66 → 58 (b) x → 5x → 5x + 2 → 3(5x + 2) → 3(5x + 2) − 8
INVESTIGATION 2B
2.06
EQUATIONS
Getting Started
For You to Explore 1. Here’s one way to give the directions: “Walk to your right on Causeway St. Causeway St. becomes Staniford St. Walk up the hill. When you reach Cambridge St., turn right. Stay on the left side of Cambridge St. to avoid the construction. Keep walking straight, then walk across the Longfellow Bridge from Boston into Cambridge.”
MATHEMATICAL REFLECTIONS
1. Cheng does not need to fence the side of the rectangle against the house, so he needs fence for two short sides and one long side.
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2. Answers may vary. The route stayed the same, but all the direction of the turns changed. Essentially, the individual steps were reversed. 3. Think of your starting number as x. Then your result is 5x − 10. You want to know if his choices will give you x back again. (a) (b) (c) (d) (e) (f) (g)
No. No. No. Yes, because (5x−10+10) is equal to x. 5 2(5x−10+10) Yes, because is equal to x. 10 No. Try it with x = 20. He’ll get 19 back. Yes, because 5x−10 + 2 is equal to x. 5
4. (a) This operation can be undone by subtracting 5 from the result. (b) This operation can be undone by multiplying the result by 10. (c) This operation can’t be undone. The result is always zero, so there is no way to know what the starting number was. (d) This can be undone by adding 28, then dividing by 3. The order is important here! (e) This operation can’t be undone. Different numbers can give the same result. For example, the numbers 17, 26, and 440 all give the result 8, and there’s no way to know what the starting number was if you only knew that the sum of the digits was 8.
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(f) This can be undone by adding 11 six times in a row, or by adding 11 × 6. 5. Your ending number will be 3 + 5k, where k is the number of times you added 5. So, all Spiro has to do is subtract 3 from your ending number (which gives him 5k) and then divide by 5 to get k, which is the number of times you added 5.
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performed 5 times. (f) Doing this operation over and over again gives −5 the first time, then 5, then −5, and so on. There is no way to find out how many times the operation was performed, but you do know it was performed an odd number of times. (g) Spiro needed to add at least 990 (1000 − 10), and the number of times the operation is performed must be at least 990 ≈ 141.43 times. Since the number of 7 operations has to be an integer, the smallest integer that works is 142.
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10. No, Spiro will not always be able to find your starting number. For example: • Start with 60. Adding 10 gives 70. Seventy divided by 7 is 10, and the remainder is 0. • Start with 39. Adding 10 gives 49. Forty-nine divided by 7 is 7, and the remainder is 0. • Start with 4. Adding 10 gives 14. Fourteen divided by 7 is 2, and the remainder is 0.
On Your Own 6. When you leave the house turn right and go until you see the police station on the corner. Take a left there, onto Belmont. Take a left onto Western, then a right onto Addison. Four blocks later, take a left onto Clark, and you’ll see the baseball field. 7. Answers will vary depending on what directions you look for, but should look like the directions from exercise 6. 8. (a) The starting number was 9. (b) The starting number was 4. (c) The starting number was −6. (d) Except for zero, dividing a number by itself always results in 1, so there is no way to find the starting number. (e) There is no way to find the starting number without knowing what Jamal’s favorite number is. (f) There is no way to find the starting number without knowing what Jamal’s favorite number is. (g) Try undoing each step one-by-one as with the other parts of this exercise. The first step to undo is adding 13. Subtracting 13 gives 24, dividing by 4 gives 6, adding 3 gives 9, multiplying by 10 gives 90, and subtracting 7 gives 83. The reverse of each operation is performed in reverse order from the original. 9. (a) One way to do this exercise is to notice that 93 − 8 = 85, and the number of times 5 must be added to get 85 is 85 = 17, so the answer is 17 times. 5 (b) Each operation adds a digit to the number: 1, 10, 100, and so on. Since 1,000,000 has six zeros, the operation was performed 6 times. (c) 130 has to be subtracted to go from 30 to -100, so the operation is performed 130 = 65 times. 2 (d) The result is always 0 after each operation, so it could have been performed 5 or 7 or 672 times. There is no way to find out how many times the operation was performed. (e) One way to do this exercise is to find the ratio between the original number and the result: 100 5 1 = 32. Since 2 = 32, the operation must be
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If you started with 60, 39, or 4, the final answer is 0, and Spiro won’t automatically be able to guess the original number. Maintain Your Skills 11. • I and III are identical, because they both amount to multiplying by 4. • II and VI are identical, because they both amount to multiplying by 2, then adding 3. • IV and V are identical, because they both amount to dividing by 6.
2.07
Reversing Operations
Check Your Understanding 1. These steps each reverse one part of the process, and they are performed in the opposite order. 2. Suppose your partner gave you the value 310. To reverse the entire process, reverse each step of the process in the opposite order. Take the result and . . . • • • • •
First, divide by 10 to get 31. Second, subtract 7 to get 24. Third, divide by 8 to get 3. Fourth, multiply by 4 to get 12. Finally, subtract 6 to get the original number, 6.
You could also call your partner’s original number n, then follow the process with n: n → n + 6 → n+6 → 4 8( n+6 ) = 2n + 12 → 2n + 19 → 10(2n + 19) = 4 20n + 190. Then, if the result is 310, the value of n can be found by solving the equation 20n + 190 = 310. If the result was r, the original number can be found by solving the equation 20n + 190 = r for the variable n. 3. (a) Take the starting number and add 13. This operation can be reversed by subtracting 13 from the result. (b) Take the starting number and divide by −2. This operation can be reversed by multiplying the result by −2. (c) Take the starting number, multiply by 5, subtract 12, then multiply by 3. This operation can be reversed by dividing by 3, adding 12, then dividing by 5. These steps undo each of the steps in the operation and are performed in reverse order. (d) Take the starting number, multiply by 15, then subtract 36. This operation can be reversed by adding 36, then dividing by 15. 4. If the ending number was 22, the number had to be 27 before Dana subtracted 5, since 27 − 5 = 22. The
Mathematics I Solutions Manual
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number you have to start with to get 27 after adding 7 would be 20, since 20 + 7 = 27. Finally, the number that you multiply 2 by to get 20 would be 10. So, the starting number must have been 10. To check, perform the steps:
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(−3)4 = −3 · −3 · −3 · −3 = (−1 · 3) · (−1 · 3) · (−1 · 3) · (−1 · 3) = (−1 · −1 · −1 · −1) · (3 · 3 · 3 · 3) = (−1)4 · 34 = 1 · 34 = 34
5. Using the same logic as in the solution to Exercise 4, you get 13 . You can check by following the steps: × 12 = 4 4 − 9 = −5
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positive number:
10 × 2 = 20 20 + 7 = 27 27 − 5 = 22
1 3
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So for any even exponent p, xp = (−x)p no matter what number x is. If p is an odd exponent, then xp and (−x)p will be different (as long as x isn’t zero).
On Your Own
6. (a) Here is the completed table: Input: x −4 −3 −2 −1 0 1 2 3 4
Output: x2 16 9 4 1 0 1 4 9 16
(b) This cannot be reversible, since there is more than one input that gives the same output. One example is −3 and 3, which each give the output 9. If you only knew the output value was 9, there would be no way to decide whether the input was −3 or 3. (c) Here is the completed table: Input: x −4 −3 −2 −1 0 1 2 3 4
Output: x3 −64 −27 −8 −1 0 1 8 27 64
It looks as though the operation is reversible, since all outputs are different. In fact, cubing can be reversed by calculating the cube root. (d) You should find that if the power is even, the results for negative numbers are equal to the results for positive numbers. If the power is odd, the results are all different. All odd powers of x: x3 , x5 , x7 , . . . are reversible operations, while all even powers of x: x2 , x4 , x6 , x8 , . . . are not reversible. (e) The product of two negative numbers is a positive number, so the product of four negative numbers is a
7. (a) Take the starting number, square it, then add 6. This operation cannot be reversed, since squaring is not reversible. (b) Take the starting number, multiply it by 3, then subtract 28. This operation can be reversed by adding 28, then dividing by 3. Note the order in these steps is important. (c) Take the starting number, subtract 8, then divide by 4. This operation can be reversed by multiplying by 4, then adding 8. (d) Take the starting number, multiply it by 3, then subtract 33, then add 5. This operation can be reversed by subtracting 5, then adding 33, then dividing by 3. 8. To get from his starting number to 24, Marty performed the following operations: • Add 5. • Multiply by 8. To backtrack, start with the 24 and then: • Divide by 8. • Subtract 5. So 24 = 3 and 3 − 5 = −2. So Marty’s starting number 8 was −2. 9. To get from her starting number to 39, Bianca performed the following operations: • Multiply by 4. • Add 11. To backtrack, start with the 39 and then: • Subtract 11. • Divide by 4. 39 − 11 = 28 and 28 =7 4 10. You get from the starting number to 1429 by performing the following operations.
Mathematics I Solutions Manual
• Multiply by 179. • Add 318. So, to get back, reverse each step, and apply them in reverse order. • Subtract 318 from 1429. • Divide whatever’s left by 179. • Chapter 2, page 39
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11. The operation is reversible when each input gives a different output. In Tables B and D, all the outputs are different from one another. In Table A, the inputs −2 through 1 each give the output −3, so this is not a reversible operation. In Table C, the inputs 2, 3, and 4 each give the output 0, so this is not a reversible operation. 12. In the equation, x is multiplied by 53 . To undo multiplication you can divide by the same number. Dividing by a number is the same as multiplying by its reciprocal. Therefore to undo multiplication by 53 , you can do either IV, divide by 53 , or I, multiply by 35 . The answer is D. 13. (a) Since the answer is always 0, the only way to restrict this would be to allow only a single number as input! There is no other helpful restriction. (b) Yes, this can work if you restrict the domain to numbers whose sums are all different. For example, you could restrict to all the numbers in the 30’s (30, 31, 32, . . . ). Since all these numbers have different sums to their digits, the operation would become reversible. Another possible restriction would be all numbers containing only the digit 1 (1, 11, 111, . . . ). 14. (a) If n is larger than 5, the result will be less than n. If n is less than 5, the result will be larger than n. Only with the input n = 5 will the result be 5. (b) There is no number you can start with that will return itself, since the output will always be 6 more than the input. The equation for this problem is n + 6 = n, which is never true. (c) Start with a positive number. Take its square root. Take the square root of the result. If you continue, no matter what positive number you start with, you should find the value comes closer and closer to 1 each time. √ It is also clear that n = 0 will return 0 (since 0 = 0), but it may not have been as √ obvious to find it. The equation for this problem is n = n. The solutions are 0 and 1. (d) If n is positive, 3n + 12 will always be larger than n. This means that if the output equals the input, n must be negative. You can create a table to find that −6 as the input does indeed return −6. (e) The equation for this problem is n2 − 6 = n, and it can be solved by building a table. n −4 −3 −2 −1 n 0 1 2 3 4
n2 − 6 10 3 −2 −5
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Maintain Your Skills 15. Choose any number. Using your calculator, add 3 to the number and divide the total by 2. Suppose you start with 14. You’ll want to enter something like (14+3)/2 You can then change your formula to refer to the previous answer, like this: (ANS+3)/2 If you keep hitting the Enter key, you’ll get closer and closer to 3. Why does this happen? Well, try it with algebra. If your starting number is x, then the result after the first time is 12 (x + 3), which if you multiply out gives you 0.5x + 1.5. Take that result, and plug it in for x in the original equation, and then simplify. 1 ((0.5x 2
+ 1.5) + 3) = 12 (0.5x + (1.5 + 3)) = 12 (0.5x + 4.5) = 12 (0.5x) + 12 (4.5) = 0.25x + 2.25
Continue this a couple of times by taking each result, and plugging it in for x in the original equation 12 (x + 3) , and see what happens. 0.5(0.5x + 1.5) + 1.5 = 0.25x + 2.25 0.5(0.25x + 2.25) + 1.5 = 0.13x + 2.63 0.5(0.13x + 2.63) + 1.5 = 0.07x + 2.82 0.5(0.07x + 2.82) + 1.5 = 0.04x + 2.91 0.5(0.04x + 2.91) + 1.5 = 0.02x + 2.96 0.5(0.02x + 2.96) + 1.5 = 0.01x + 2.98 The number you multiply x by (remember, x is the original number) gets smaller and smaller . . . eventually it will go away. The other number is getting bigger, but seems to be slowing down at just about 3. In fact, it turns out that 3 is the fixed point of the expression. 16. Following the same process as you did in Exercise 14, you can use your calculator to see that the answers get closer to 5. As before, you can try it with algebra. If your starting number is x, then the result after the first time is 12 (x + 5), which if you multiply out gives you 0.5x + 2.5. Take that result, and plug it in for x in the original equation, and then simplify. Continue this a couple of times and see what happens.
n2 − 6 −6 −5 −2 3 10
0.5(0.5x + 2.5) + 2.5 = 0.25x + 3.75 0.5(0.25x + 3.75) + 2.5 = 0.13x + 4.38 0.5(0.13x + 4.38) + 2.5 = 0.07x + 4.69 0.5(0.07x + 4.69) + 2.5 = 0.04x + 4.85 0.5(0.04x + 4.85) + 2.5 = 0.02x + 4.93 0.5(0.02x + 4.93) + 2.5 = 0.01x + 4.97 Just like last time, the number you multiply x by (remember, x is the original number) gets smaller and
Mathematics I Solutions Manual
• Chapter 2, page 40
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smaller . . . eventually it will go away. The other number is getting bigger, but seems to be slowing down at just about 5. In fact, it turns out that 5 is the fixed point of the expression. 17. Following the same process as you did in Exercise 14, you can use your calculator to see that the answers get closer to 1. As before, you can try it with algebra. If your starting number is x, then the result after the first time is 12 (x + 1), which if you multiply out gives you 0.5x + 0.5. Take that result, and plug it in for x in the original equation, and then simplify. Continue this a couple of times and see what happens.
To backtrack, add 1 to 38 to get 39, then divide by 5 to = 7.8. get the answer, n = 39 5 7. (a) The equation is 33 + 6b = 75. The backtracking steps are: • Start with 75. • Subtract 33. • Divide by 6. = 7, so The result is b. So 75 − 33 = 42, and 42 6 when a = 11, b = 7. (b) The equation is 3a + 72 = 75. The backtracking steps are: • Start with 75. • Subtract 72. • Divide by 3.
0.5(0.04x + 0.97) + 2.5 = 0.02x + 0.99 0.5(0.02x + 0.99) + 2.5 = 0.01x + 1.00
The result is a. So 75 − 72 = 3, and b = 12, a = 1. (c) Here is the completed table:
0.5(0.01x + 1.00) + 2.5 = 0.00x + 1.00
Solving Equations by Backtracking
Check Your Understanding 1. The steps (starting from y) are: • Add 7.
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• Multiply by 5. • Subtract 1.
0.5(0.13x + 0.88) + 2.5 = 0.07x + 0.94 0.5(0.07x + 0.94) + 2.5 = 0.04x + 0.97
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6. The steps (starting from n) are:
0.5(0.5x + 0.5) + 0.5 = 0.25x + 0.75 0.5(0.25x + 0.75) + 2.5 = 0.13x + 0.88
Just like before, the number you multiply x by (remember, x is the original number) gets smaller and smaller . . . eventually it will go away. The other number got bigger until it stuck at 1. In fact, it turns out that 1 is the fixed point of the expression.
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3 3
= 1, so when
a b 0 12.5 1 12 2 11.5 3 11 4 10.5 5 10 (d) You can describe the relationship in a number of ways. You can say that as a goes up by 1, b goes down by 12 . Or, you can say as b grows by 1, a drops by 2. 8. (a) To show that 1 and 8 are solutions to the equation, replace the x with each value.
To backtrack, subtract 7 from 3 to get the answer, y = −4. 2. The steps (starting from x) are: • Multiply by 2. • Subtract 7.
x3 + 26x = 11x2 + 16 ?
13 + 26 · 1 = 11 · 12 + 16 27 = 27
To backtrack, add 7 to 110 to get 117, then divide by 2 to get the answer, x = 117 = 58.5. 2 3. The steps (starting from z) are: • Divide by 12. To backtrack, multiply 10 by 12 to get the answer, z = 120. 4. The steps (starting from w) are:
?
83 + 26 · 8 = 11 · 82 + 16 720 = 720
(b) Answers will vary, but most other values for x will not result in an equality. For example, try x = −1 and x = 10: ?
(−1)3 + 26 · (−1) = 11 · (−1)2 + 16 −27 = −5
• Multiply by −2. • Add 15. To backtrack, subtract 15 from 31 to get 16, then divide by −2 to get the answer, w = −8. 5. The steps (starting from n) are: • Subtract 13. • Divide by 12. To backtrack, multiply 3.5 by 12 to get 42, then add 13 to get the answer, n = 55. Mathematics I Solutions Manual
X
?
103 + 26 · 10 = 11 · 102 + 16 1260 = 1116 X (c) The third solution is x = 2. You can find this solution by testing values. As you pick numbers greater than 8, it becomes clear that the left side gets steadily larger (more positive) than the right. For numbers less than 1, the left side gets steadily more • Chapter 2, page 41
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• • • •
negative than the right. So the third number must be between 1 and 8. To verify that x = 2 works, replace x with 2: ?
23 + 26 · 2 = 11 · 22 + 16 60 = 60
On Your Own 9. Both methods work. Since all the simplification that Sasha did followed the basic rules of expressions, she knows that the equation she started with and the equation she ended with were equivalent. 10. (a) Here’s a series of statements that represent the expression.
(a) Take 17. Add 1 to get 18. Divide by 3 to get 6. Subtract 2 to get 4. Check the answer. 3(a + 2) − 1 = 17 ?
3((4) + 2) − 1 = 17 ?
3(6) − 1 = 17 ?
18 − 1 = 17 17 = 17
(b) Take 8. Add 1 to get 9. Divide by 3 to get 3. Subtract 2 to get 1. (c) Take −19. Add 1 to get −18. Divide by 3 to get −6. Subtract 2 to get −8. (d) Take 0. Add 1 to get 1. Divide by 3 to get 13 . Subtract 2 to get − 53 or 1 23 . 12. The method for solving this exercise is similar to the other solutions in this lesson. Here are the basics of the solution. First of all, Andrew’s equation can be written as −4(a + 1) + 2 = 22 −4(a + 1) + 2 = 22 −4(a + 1) = 20 20 (a + 1) = = −5 −4 a = −6
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Divide by 2: 6 ÷ 2 = 3 Add 15: 3 + 15 = 18 Divide by 3: 18 ÷ 3 = 6 Subtract 8: 6 − 8 = −2
n=
100 +4 f
(d) Yes. Since the backtracking steps include “divide by f ,” then f cannot be zero. You could also reason that Jamal’s favorite number could not be zero because in the original problem, one of the steps was “multiply by my favorite number.” Multiplication by zero is not a reversible operation.
(b) Here’s a series of statements that represent the reverse of the operations.
11. You can follow the steps from part (b) of Exercise 10 to solve the equations.
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Joya’s starting number was –2, so the answer is A. 14. (a) The starting number would be 8. (b) The starting number would be 14. (c) Starting with 100, you would divide by f (Jamal’s favorite number), then add 4. If Jamal’s favorite number is written by f , the starting number is given by
• Take a number and add 2. • Multiply the results of the previous step by 3. • Subtract 1 from the results of the previous step.
• Take a number and add 1. • Divide the results of the previous step by 3. • Subtract 2 from the results of the previous step.
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Maintain Your Skills 15. (a) The backtracking steps are “Start with 93. Subtract 13, then divide by −5. The result is x.” So 80 93 − 13 = 80, and −5 = −16. So x = −16. (b) In order to backtrack this equation, you can describe the steps like this: “Pick a number. Multiply that number by 5. Subtract that total from 13. The answer is 93.” You may not have thought of how to reverse that last step until now. Here’s a way to think about it: if you’re subtracting a number from 13, your result is what you have left over. In other words, you’ve “split” 13 into two numbers that add together to make 13. If you subtract a from 13, for instance, and get b, then to reverse it, you’d subtract b from 13 to get a. So to reverse the steps, you can say, “Start with 93. Subtract that number (93) from 13, then divide by 5.” So 13 − 93 = −80, and −80 = −16, so x = −16. 5 (c) The backtracking steps are “Start with −45. Subtract 12, then divide by −7.” So −45 − 12 = −57, and −57 = 57 , so x = 57 . −7 7 7 (d) The backtracking steps are “Start with −45. Subtract that number from 12, then divide by 7.” So 12 − (−45) = 57, and 57 cannot be reduced, so 7 . x = 57 7 (e) The backtracking steps are “Start with 4. Add 17, 21 then divide by −3.” So 4 + 17 = 21, and −3 = −7, so x = −7. (f) The backtracking steps are “Start with 4. Subtract that number from −17, then divide by 3.” So −17 − 4 = −21, and −21 = −7, so x = −7. 3
Andrew’s starting number was −6. 13. Work backward to find Joya’s starting number. Her ending number is 6. Mathematics I Solutions Manual
You may notice that each pair of equations gave the same answer. That’s because (a) and (b) are equivalent equations, as are (c) and (d), and so are (e) and (f). Take a closer look at (a) and (b). You can use your basic rules to transform the first equation into the second. And since the • Chapter 2, page 42
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To backtrack, subtract 5 from 11 to get 6, then divide 6 by 2 to get the answer x = 62 = 3. (b) The steps (starting from m) are
−5x + 13 = 93 13 + (−5x) = 93 by any-order, any-grouping 13 − 5x = 93 since subtraction is adding the opposite
• Multiply by 3 • Subtract 4
You can follow the same steps to show that the other two pairs are also equivalent. . 16. (a) Divide by 10, then subtract 3; 43 10 (b) Subtract 3, then divide by 10; 7. (c) Divide by 10, then subtract 3; 43 . 10 (d) Subtract 3, then divide by 10; 7. (e) Multiply by 10, then subtract 3; 727. (f) Subtract 3, then multiply by 10; 700. (g) Multiply by 10, then add 3; 733. 10 (h) Divide by 73, then divide by 3; 219 .
To backtrack, add 4 to −7 to get −3, then divide −3 by 3 to get the answer m = −3 = −1. 3 (c) The steps (starting from y) are • Divide by 3 • Add 1 To backtrack, subtract 1 from 5 to get 4, then multiply 4 by 3 to get the answer y = 4 · 3 = 12. (d) The steps (starting from y) are • Add 1 • Divide by 3
MATHEMATICAL REFLECTIONS
1. (a) Use the formula and solve for C. K = C + 273.15 K − 273.15 = C + 273.15 − 273.15 K − 273.15 = C (b) Substitute 0 for K in the formula you found in part (a).
4.
5. 6. 7.
C = K − 273.15 C = 0 − 273.15 C = −273.15
8.
So absolute zero is −273.15◦ C. (c) Substitute your answer from part (b) into the formula from page 131.
To backtrack, start with the 7 and then perform the following operations.
F = 95 (−273.15) + 32 F = −491.67 + 32
• Add 8 • Divide by 3
F = −459.67
So, 7 + 8 = 15 and was 5.
◦
F = 95 C + 32 F = 95 (K − 273.15) + 32
2. (a) (b) (c) (d) 3. (a)
To backtrack, multiply 5 by 3 to get 15, then subtract 1 from 15 to get the answer y = 15 − 1 = 14. (a) 3(8 − 2) = 3(6) = 18 (b) 32 = 3 + 6 → 9 = 9 and (−2)2 = (−2) + 6 → 4=4 2(3(−4) + 1) = 2(−12 + 1) = 2(−11) = −22 = 22 An equation is a mathematical sentence that states that two quantities are equal. An inefficient way is to use guess and check. A better way is to use guess-check-generalize to find an equation, then solve the equation using backtracking. Of course, some problems you can solve directly using backtracking without ever writing out an equation. To get from her original number to 7, Breanna performed the following operations. • Multiply by 3 • Subtract 8
F = 95 C + 32
So absolute zero is −459.67 F . (d) To find the formula from kelvins to degrees Fahrenheit, you need to combine the two formulas you already have. Take your basic formula to convert ◦ C to ◦ F, then replace the C with K − 273.15.
page 43
• Multiply by 2 • Add 5
basic rules do not change the value of an expression, the two equations must be equivalent.
2B
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INVESTIGATION 2C
2.09
15 3
= 5. Breanna’s original number
SOLVING LINEAR EQUATIONS
Getting Started
F = 95 K − 491.67 + 32
For You to Explore
F=
1. If Sasha starts with the number x she gets the following equation.
9 K 5
− 459.67
Subtract 3 from the number. Add 6 to the number. Subtract 5 from the number and divide by 2. Multiply the number by 3 and subtract 2. The steps (starting from x) are
Mathematics I Solutions Manual
2x + 3 − 8 = 13 You can solve this by backtracking. Start with 13, then: • Add 8. (And you get 21.) • Chapter 2, page 43
“000200010271723958_CH02_p031-072”
• Subtract 3. (And you get 18.) • Divide by 2. (And you get 9.) Sasha’s original number was 9. Check that it works. 2x + 3 − 8 = 13 2(9) + 3 − 8 = 13 18 + 3 − 8 = 13 21 − 8 = 13 13 = 13 2. Again, you can solve this problem by backtracking. Start with 8 and reverse each operation, starting at the bottom. Start with 8. • Subtract 3. (And you get 5.) • Divide by 21 . (And you get 10, because dividing by 12 is the same as multiplying by 2.) • Add 11. (And you get 10 + 11 = 21.) So, 21 was Tony’s original number. 3. Again, you can use backtracking. Start with −9. • Add 19. (And you get 10.) • Divide by −2 (And you get −5.) • Subtract 4. (And you get −9.) 4. Backtracking won’t work for this solution. If you follow Casey’s instructions, you’ll get the equation 7x + 18 = 10x But there are plenty of ways to reason about it which will work. For instance, you could notice that the difference between the two sides of the equation is that, in addition to the shared 7x, there’s an 18 on the left and a 3x on the right. So, 18 must equal 3x, which means x = 6. If you check that answer 7(6) + 18 = 60 = 10(6) you’ll see that it’s correct. 5. Again, this exercise can’t be solved by backtracking. If you call Anna’s number x you get the equation 3x − 4 − 3 = 2x You could reason that subtracting 7 is the same as subtracting an x. So, x must equal 7. If you plug x = 7 into the equation 3x − 4 − 3 = 2x, you’ll see that it works. 6. Again, you can’t use backtracking. You get the following equation. 5x + 6 = 3x + 12 What’s the difference between the two sides? Well, the left side has an additional 2x (since 5x − 3x = 2x) and the right side has an additional 6 (since 12 − 6 = 6), so that means that 2x = 6. Which means that x = 3. 7. Here the equation generated by James’s number is 4(x + 1) − 11 = 2x. To begin this exercise, it would be useful to simplify the left side. 4(x + 1) − 11 = 4x + 4 − 11 = 4x − 7. So you have
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8. The equation generated by this story is 6(x + 1) + 1 = x + 18. Simplifying the left side, you get 6x + 7 = x + 18 How do you get from the left side to the right side? You subtracted 5x and added 11 and you ended up equal. So −5x + 11 = 0 which means that 11 = 5x or x = 11 . 5 9. Focus on the larger rectangle first. Its length is 4 and its width is (x + 3). You can multiply these together to get the area A. A = 4(x + 3) Now, look at the two smaller rectangles. The length of both rectangles is 4. The rectangle on the left has a width of x, so its area is 4x. The rectangle on the right has a width of 3, so its area is 4 · 3, or 12. Adding the areas of the two smaller rectangles will also give you the area of the larger rectangle. A = 4x + 4 · 3 = 4x + 12 On Your Own 10. (a) Start with 29, subtract 11, then divide by 3. • 29 − 11 = 18 • 18 =6 3 a = 6. (b) Start with −15, subtract 5, divide by −2, then add 15. • −15 − 5 = −20 • −20 = 10 −2 • 10 + 15 = 25 p = 25. 11. Basically, there’s nowhere to start the backtracking process. Usually, you would take the result and apply the reverse operations to get back to the original number. But here the result is an expression, so applying reverse operations to it will just leave you with another expression (and not the original number). 12. There are lots of different ways to solve the equation. One way to think about it is that to go from left to right you are “trading” 6 for 2t. That means 6 = 2t, which means t = 3. 13. You could simplify the equation and try backtracking to see if you get the solution −2. The fastest way, though, is to remember that in order to determine if a number is a solution to an equation, substitute it for r and see if you get a true statement. (a) 6r + 2 = 12 + r
4x − 7 = 2x
?
6(−2) + 2 = 12 + (−2) ?
−12 + 2 = 12 + (−2) −10 = 10 So r = −2 is not a solution to this equation. (b) 3r + 2 + 10r = 7 + 7r + (−17) ?
3(−2) + 2 + 10(−2) = 7 + 7(−2) + (−17) ?
Again, you could reason about this exercise by seeing that subtracting 7 is the same as subtracting 2x, so 7 = 2x or x = 72 . Mathematics I Solutions Manual
−6 + 2 + (−20) = 7 + (−14) + (−17) −24 = −24 So r = −2 is a solution to this equation. • Chapter 2, page 44
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(c) r + 11 − 3r = 15 + 2r ?
(−2) + 11 − 3(−2) = 15 + 2(−2) ?
(−2) + 11 + 6 = 15 + (−4) 15 = 11 So r = −2 is not a solution to this equation. (d) 7(r + 2) + 8 = 4r + 16
?
7(0) + 8 = (−8) + 16 ?
So r = −2 is a solution to this equation. 14. As with exercise 13, test whether s = 43 is a solution by substituting 43 for s for each equation and simplifying. If the result is a true statement, then s = 43 is a solution. If the result is a false statement, then it is not. (a) 4s = s + 4 4
4 3
?
=
4 3
16. This is like Exercise 9. The areas of the four rectangles, starting from the upper-left and moving clockwise, are a · a = a2 , ab, b · b = b2 , and another ab. You got each area by multiplying the length times the width. The area of the entire square is the sum of the areas of the rectangles that form it; so its area is a2 + ab + b2 + ab = a2 + 2ab + b2 . The area of the square is also its length (a + b) times its width (a + b). So another expression for the area of the square is
+4
16 ? 4 = 3 + 12 3 3 16 16 = 3 3
(a + b)(a + b).
So s = 43 is a solution to this equation. (b) 9s − 2 = 5s + 10 3 ? 9 43 − 2 = 5 43 + 10 3 ? 20 3 ? = 30 3
12 − 2 = 10
+
10 3
Maintain Your Skills 17. For each exercise, you are looking for the value that gives you back 1. In chapter 1, you read about multiplicative inverses, or the reciprocal of a fraction. In these exercises, the value on the left side is always 1, so you want to find the inverse, or reciprocal, of whatever number is multiplied to x. The reciprocal of 8 is 18 . 1 The reciprocal of −19 is − 19 . 11 13 The reciprocal of 13 is 11 · 11 . You For this one, you first need to multiply 10 11 13 10 can cancel the two 11’s and end up with 13 , and the reciprocal of 10 is 13 . 13 10 (e) As in the previous one, you first need to multiply 5 11 · . You have nothing to cancel, so you multiply 7 12 numerators to get the new numerator and denominators to get the new denominator.
10 = 10
(a) (b) (c) (d)
So s = 43 is a solution to this equation. (c) 5(s − 1) − 1 = 2s − 23 ? 5 34 − 1 − 1 = 2 43 − 23 ? 5 43 − 33 − 1 = 83 − 23 ? 5 13 − 1 = 63 2 3
= 2
So s = 43 is not a solution to this equation. (d) 2(s + 1) + 5 = 7s + 13 ? 2 34 + 1 + 5 = 7 43 + 13 ? 2 43 + 33 + 5 = 28 + 13 3 7 ? 2 3 + 5 = 29 3 14 3 14 3
So s =
4 3
? 29 3 ? 29 = 3 = 29 3
+5=
+
15 3 29 3
page 45
(a) 34 + 8 = 42 (b) 34 · 2 = 68 (c) 34 − 7 = 27 (You would get the same as Colleen got in her original game because adding −7 is the same as subtracting 7.)
?
8=8
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15. The key is to find out what Colleen has at the end of her second-to-last step, before she takes that last step. In the original game, the last step was to add −7, and she ended up with 27. So the number she had before her last step is 34. To find the new answers, perform the steps listed to the number 34.
7((−2) + 2) + 8 = 4(−2) + 16
0 + 8 = (−8) + 16
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5 11 5 · 11 55 · = = 7 12 7 · 12 84 The reciprocal of
55 84
is
84 . 55
2.10 When Backtracking Does Not Work Check Your Understanding
is a solution to this equation.
1. If you remove the 5, you’ll see that 2l = 30, which means that l = 15. 2. Ignoring the 5j which the top and bottom have in common, you are left with 2j = 10 which gives j = 5.
Mathematics I Solutions Manual
• Chapter 2, page 45
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3. If you remove 2n from both sides and 7 from both sides, you are left with 2n = 3 which means n = 32 . 4. Remove an x from both sides and 2 from both sides and you get 5x = 8 which means that x = 85 . 5. If 3x = −15 then x = −5. 6. Remove an 11 from both sides and you are left with 2y = 16 which means y = 8. 7. Remove an a from both sides and a 5 from both sides and you are left with a = 3. 8. Remove a u from both sides and you are left with 3u − 5 = 16. From here you can use backtracking. Or you could “remove” a −5 from both sides to get 3u = 21 which means u = 7. 9. Remove an 8m from both sides and remove a 1 from both sides, and you are left with 3m = 27. This means m = 9. 10. Answers may vary. Here are some possible answers.
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37a + 12 + 14a − 9 = 6a + 11 − 14 + 5a 51a + 3 = 11a − 3 Now, solve the equation. 51a + 3 = 11a − 3 51a + 3 − 11a = −3 40a + 3 = −3 40a = −3 − 3 40a = −6 a=
(d) One possible answer is 6 = 2x. You can get that by recognizing that six is two times three, so if x = 3, then six is two times x, which can be written as 6 = 2x. (e) The simplest answer is the equation x = 3.
−3 −6 = 40 20
(d) First simplify the left side by combining like terms. s − 2s + 3s − 4s + 5s = s + 100 3s = s + 100 Now, solve the equation. 3s = s + 100
2·3+2=6+2=8
7 · 3 + 5 + 2(3 − 1) = 21 + 5 + 2(2) = 26 + 4 = 30
page 46
(c) First simplify both sides by combining like terms.
(a) The equations 2(x + 1) = 8 and 2(x + 1) = x + 5 are possible correct answers. You can make your task easier by distributing the 2 to get 2x + 2 on the left side. Then ask, “What can go on the right side to get x = 3?” Just substitute 3 for x and see what you get.
So, 8 can go on the right side of the equation if x = 3 is a solution. The equation with x + 5 on the right side was found using trial and error. (b) One possible answer is 3x − 1 = 8, which you can get by substituting 3 for x. (c) One possible answer is 7x + 5 + 2(x − 1) = 30, because
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3s − s = 100 2s = 100 s = 50 On Your Own 12. Ignore a 2 from both sides and an r from both sides and you are left with 5r = 10 or r = 2. 13. Ignore a 2d from both sides and a 4 from both sides and you are left with 6 = 3d or 2 = d. 14. If 3p = 36 then p = 12. 15. 7x − 8 = x + 16 6x − 8 = 16 6x = 24
11. (a) 5r + 11 = 95 − r 5r + 11 + r = 95 6r + 11 = 95 6r = 95 − 11
x=4 16. (a) 3x + 5 = 26 3x + 5 − 5 = 26 − 5 3x = 21 21 3x = 3 3 x=7
6r = 84 r = 14 (b) 14 − 23x = 60x + 180 14 = 60x + 180 + 23x 14 = 83x + 180 14 − 180 = 83x −166 = 83x −2 = x
Mathematics I Solutions Manual
(b) 3(x − 1) + 5 = 26 3(x − 1) + 5 − 5 = 26 − 5 3(x − 1) = 21 3x − 3 = 21 3x − 3 + 3 = 21 + 3 • Chapter 2, page 46
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3x = 24 3x 24 = 3 3 x=8 (c) 3(x + 3) + 5 = 26 3(x + 3) + 5 − 5 = 26 − 5 3(x + 3) = 21 3x + 9 = 21 3x + 9 − 9 = 21 − 9 3x = 12 3x 12 = 3 3 x=4
(e) 3(2x + 1) + 5 = 26 3(2x + 1) + 5 − 5 = 26 − 5 3(2x + 1) = 21 3(2x + 1) 21 = 3 3 2x + 1 = 7 2x + 1 − 1 = 7 − 1 2x = 6 2x 6 = 2 2 x=3
page 47
x + 8 = 2(x + 2) x + 8 = 2x + 4 8=x+4 4=x The fact that 4 = x implies that Emma is four years old right now. James is six years older than that, so he is ten years old. You can check that it works by checking to see that in two years, Emma will be 6 and James will be 12, so yes, he will be twice as old as she is. 18. Let x be the cost of one pair of shoes. Shantell bought 7 pairs of shoes, so she spent 7x dollars. Since she has $17 left over, the total amount of money that Shantell had to start with was 7x + 17 dollars. Yahaira bought 4 pairs of shoes, so she spent 4x dollars. Since she has $71 left over, the total amount of money that Yahaira had to start with was 4x + 71 dollars. Since the each started with the same amount of money, you get the following equation. 7x + 17 = 4x + 71 Now you can solve this equation for x to find out the cost of one pair of shoes. 7x + 17 − 17 = 4x + 71 − 17 7x = 4x + 54 7x − 4x = 4x + 54 − 4x 3x = 54 3x 54 = 3 3 x = 18
(f) 3(5 − 2y) + 5 = 26 3(5 − 2y) + 5 − 5 = 26 − 5 3(5 − 2y) = 21 3(5 − 2y) 21 = 3 3 5 − 2y = 7 5 − 2y − 5 = 7 − 5 −2y = 2 −2y 2 = −2 −2 y = −1
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17. (a) Emma’s age is x, so James’s age is six more than x. That is, you add 6 to x to get James’s age, so his age is x + 6. (b) For example, if she is 3 years old now (which she is not), she’ll be 5 in two years. You can get that by adding 2. So if she’s x years old now, she’ll be x + 2 years old in two years. (c) There are two correct answers. He will be x + 8 years old, but he will also be 2(x + 2) years old. (d) Since he’s x + 6 years old now, he will be x + 6 + 2 = x + 8 years old in two years. Also in two years, he will be double Emma’s age at that point. That’s double x + 2, which can be written as 2(x + 2). (e)
(d) 3(x − 7) + 5 = 26 3(x − 7) + 5 − 5 = 26 − 5 3(x − 7) = 21 3(x − 7) 21 = 3 3 x−7=7 x−7+7=7+7 x = 14
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The cost of one pair of shoes is $18. The answer is C. 19. (a) Remove a 2x from both sides and you get 3x − 7 = 2. By backtracking you get x = 3. (b) Remove a 3g from both sides and you get −8 = 2g − 20. By backtracking you get g = 6. (c) Remove a (−3b) from both sides and you get 1 = 5b − 8. By backtracking you get b = 95 . (d) Remove a (−7k) from both sides and you get 15 + 3k = 12. By backtracking you get k = −1.
Mathematics I Solutions Manual
• Chapter 2, page 47
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20.
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(d) 6x + 1 = 19 + x 6x + 1 − 1 = 19 − 1 + x 6x = 18 + x
5x − 27 = 2x + 48 2
2
5x − 27 + 27 = 2x2 + 48 + 27 2
6x − x = 18 + x − x 5x = 18 5x 18 = 5 5 18 x= 5
5x2 = 2x2 + 75 5x2 − 2x2 = 2x2 + 75 − 2x2 3x2 = 75 3x2 75 = 3 3 x2 = 25 x = ±5
(e) 7x + 1 = 19 + x 7x + 1 − 1 = 19 − 1 + x 7x = 18 + x 7x − x = 18 + x − x 6x = 18 6x 18 = 6 6 x=3
Maintain Your Skills 21. (a) 3x + 1 = 19 + x 3x + 1 − 1 = 19 − 1 + x 3x + 0 = 18 + x 3x = 18 + x 3x − x = 18 + x − x 2x = 18 2x 18 = 2 2 x=9
(f) You have to subtract 1 and x from both sides of each equation. You divide by a different number each time in the last step of solving the equation. 22. (a) 3(x − 1) = 2(x + 1) 3x − 3 = 2x + 2 3x − 3 + 3 = 2x + 2 + 3 3x = 2x + 5 3x − 2x = 2x − 2x + 5
(b)
x=5 (b) 4x + 1 = 19 + x 4x + 1 − 1 = 19 − 1 + x
4(x − 1) = 2(x + 1) 4x − 4 = 2x + 2 4x − 4 + 4 = 2x + 2 + 4
4x = 18 + x 4x − x = 18 + x − x 3x = 18 3x 18 = 3 3 x=6 (c)
4x = 2x + 6 4x − 2x = 2x − 2x + 6 2x = 6 2x 6 = 2 2 x=3 (c)
5x + 1 = 19 + x 5x + 1 − 1 = 19 − 1 + x 5x = 18 + x 5x − x = 18 + x − x 4x = 18 4x 18 = 4 4 x = 29 Mathematics I Solutions Manual
5(x − 1) = 2(x + 1) 5x − 5 = 2x + 2 5x − 5 + 5 = 2x + 2 + 5 5x = 2x + 7 5x − 2x = 2x − 2x + 7 3x = 7 3x 7 = 3 3 x = 73 • Chapter 2, page 48
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(d)
2.11 The Basic Moves for Solving Equations 6(x − 1) = 2(x + 1) 6x − 6 = 2x + 2 6x − 6 + 6 = 2x + 2 + 6
Check Your Understanding 1. (a)
6x = 2x + 8 6x − 2x = 2x − 2x + 8 4x = 8 4x 8 = 4 4 x=2
6x + 12 = 11x − 33 6x + 12 − 6x = 11x − 33 − 6x 12 = 5x − 33 12 + 33 = 5x − 33 + 33 45 = 5x 45 5x = 5 5 9=x
(e) 7(x − 1) = 2(x + 1) 7x − 7 = 2x + 2 7x − 7 + 7 = 2x + 2 + 7 7x = 2x + 9 7x − 2x = 2x − 2x + 9
(b) 8x + 13 = −4x + 11 8x + 13 + 4x = −4x + 11 + 4x 12x + 13 = 11 12x + 13 − 13 = 11 − 13 12x = −2 12x −2 = 12 12 x = − 16
5x = 9 5x 9 = 5 5 x = 95 (f) 8(x − 1) = 2(x + 1) 8x − 8 = 2x + 2 8x − 8 + 8 = 2x + 2 + 8 8x = 2x + 10 8x − 2x = 2x − 2x + 10 6x = 10 6x 10 = 6 6 5 x= 3 (g) The answer to part (a) is 5, which can be written as 15 . The answer to part (b) is 3, which can be written as 62 , and so on. The answers to parts (a) through (f) can be written as
(c) This problem is the same as the previous one. You can shuffle the left or the right side as much as you want without changing the solutions of the equation, as long as you don’t shuffle the equal sign. (d) If you distribute the 6 on the left side, and the 11 on the right side, you end up with the same equation as in part (a). 6(a + 2) = 11(a − 3) 6a + 12 = 11a − 33 .. . (e) 2z + 9z = 4z + 45 + 2z 11z = 6z + 45 11z − 6z = 6z + 45 − 6z 5z = 45 5z 45 = 5 5 z=9
5 6 7 8 9 10 , , , , , and . 1 2 3 4 5 6 The pattern is that each time the number on the left side is increased the way it is in the exercise, the numerator and the denominator of each number both increase by 1. In general, the solution of the equation a(x − 1) = 2(x + 1) is x=
a+2 a−2
as long as a doesn’t equal 2 (to avoid dividing by zero). Mathematics I Solutions Manual
(f) The exact same expression appears on both sides of the equation, so it has to be true for any value of n. But what happens if you do the basic moves? 3n + 13 = 3n + 13 3n + 13 − 3n = 3n + 13 − 3n 13 = 13 13 − 13 = 13 − 13 0=0 • Chapter 2, page 49
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You could stop once you have 13 = 13, since there are no variables anymore, so the equation is true for all values of n. In other words, 3n + 13 = 3n + 13 is a true equation. 2. The solution should look something like this.
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page 50
7. 4z + 2 = z − 4 has one solution. 4z + 2 = z − 4 4z + 2 − z = z − 4 − z 3z + 2 = −4 3z + 2 − 2 = −4 − 2 3z = −6 3z −6 = 3 3 z = −2
• Subtract 48x from both sides. • Add 15 to both sides. • Divide by the coefficient of x (which is 25). This will give you x = some number. 3. The exercise asks you to change the last step in your game. You can figure out this exercise in two ways. Reverse the last step to find out that you know what the result is after the second-to-last step. You ended with 13 after adding 10, so to undo the “add 10,” you subtract 10 from 13 and get 3. So, if the second-to-last number is 3, adding 8 to it produces 3 + 8, which equals 11. 4. (a) Subtract 6 from both sides. (b) Subtract t from both sides. (c) Add 100 to both sides. (d) Subtract 3t from both sides. (e) Subtract 5t from both sides. (f) Multiply both sides by 5. 5. Since all the equations came about by applying the basic moves, they will all give the same answer for x, which is 3.5 or 72 . 6. (a) The instruction list from the expression on the left side would be: • • • •
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2x + 2 = 2x + 7 has no solution. 2x + 2 = 2x + 7 2x + 2 − 2 = 2x + 7 − 2 2x = 2x + 5 2x − 2x = 2x + 5 − 2x 0=5 But that last line is a false statement, since 0 = 5. So there’s no value of x that will give you a true statement. 3s + 12 = 3(s + 3) + 3 has infinitely many solutions. 3x + 12 = 3(x + 3) + 3 3x + 12 = 3x + 9 + 3 3x + 12 = 3x + 12 3x + 12 − 12 = 3x + 12 − 12
Take a number x. Subtract 7. Multiply by 4. Add 13.
3x = 3x 3x − 3x = 3x − 3x 0=0
So to backtrack, start with the result, 27. Undo the last step by subtracting 13 and get 14. Undo the previous step by dividing by 4, and get 72 . Undo the previous step by adding 7, and get 21 . So the answer 2 is 21 . 2 (b) Using the basic moves, 4(x − 7) + 13 = 27 4(x − 7) + 13 − 13 = 27 − 13 4(x − 7) = 14 4(x − 7) 14 = 4 4 x − 7 = 72 x−7+7= x=
7 + 2 21 2
And 0 = 0 is indeed a true statement. So no matter what x equals, you can always use basic moves to end up with 0 = 0, which means any value of x is a solution to the equation. On Your Own 8. 7f − 19 = 4f + 41 3f − 19 = 41 3f = 60 f = 20 9.
7
(c) Answers may vary. A sample is given. The two processes are similar in that the backtracking steps in part (a) are exactly the operations performed in part (b). They are different in that when you use backtracking, you are only performing the operations on the number on the right side of the equation. When you use basic moves to solve, you must always perform the operation on both sides of the equation.
4r + 6 = 2r − 17 2r + 6 = −17 2r = −23 r = − 23 2 10.
Mathematics I Solutions Manual
4a + 1 = 11a + 8 1 = 7a + 8 −7 = 7a −1 = a • Chapter 2, page 50
“000200010271723958_CH02_p031-072”
11. (a) Subtract 4 from both sides. (b) Subtract 29 from both sides. (c) Subtract 17 from both sides. (d) Multiply both sides by 3. (e) Divide both sides by 2. (f) Add 3x to both sides. (g) Subtract 36 from both sides. (h) Subtract x from both sides. 12. Since all the equations came about by applying the basic moves, they will all give the same answer for x, which is 9.5 or 19 . 2 13. (a) • Choose any number: x. • Multiply by 3: 3(x) = 3x. • Add 5: (3x) + 5 = 3x + 5. • Multiply by 4: 4(3x + 5) = 12x + 20. • Add 16: (12x + 20) + 16 = 12x + (20 + 16) = 12x + 36. • Divide by 12: 12x+36 = 12(x+3) = x + 3. 12 12 • Subtract your original number: (x + 3) − x = x + (3 − x) = x + (−x + 3) = (x + (−x)) + 3 = 0 + 3 = 3. (b) Yes. The ending number is always 3. 14. (a) No numbers have that property. (b) You end up with the equation 12 = 13, or 0 = 1, or something similar. 15.
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Solutions of Linear Equations
Check Your Understanding 1. 2r + 1 = 5r − 11 12 = 3r 4=r 2. 4 − w = 3w + 1 3 = 4w 3 =w 4 3. 16t + 9 = 2(8t + 1) 16t + 9 = 16t + 2 9 = 2 The equation has no solutions. 4. 5u + 8 = 40 − 2u 7u = 32 32 u= 7
17 − (5 − p) = 2(5p − 16) 12 + p = 10p − 32 44 = 9p 44 p= 9
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5. 3 r + 4 = 10 2 3 r=6 2 r=4
The correct answer is C. 16. (a) Every number has that property. (b) You end up with the equation 0 = 0 or something similar. 6. Maintain Your Skills
4 + 11e = 2 + 5e + (2 + 6e) 4 + 11e = 4 + 11e 0=0
17. (a) x = 5. 3x + 2 = 22 − x 4x + 2 = 22 4x = 20 x=5 (b) x = 5. Subtracting 1 from each side gives you the original equation. (c) x = 5. Subtracting 2 from each side gives you the original equation. (d) x = 5. Subtracting 5 from each side gives you the original equation. (e) x = 5. Adding 11 to each side gives you the original equation. (f) x = 5. Subtracting 99 from each side gives you the 7 original equation. (g) They all have the same solution. They are all the same basic equation 3x + 2 = 22 − x, but the same number is added to or subtracted from both sides.
The equation has infinitely many solutions. 7. Your answers may vary. Here are some possible answers. (a) 25x + 13 = 88, 2x + 1 = 7, or x = 3 are possible answers. (b) 3x + 4 = 2x + 9 is a possible answer. (c) 14x + 21 = 8x − 45 is a possible answer. (d) 7x − 10 = 4x − 5 is a possible answer. (e) 13x + 11 = 13x + 65 is a possible answer. This equation has no solutions. 8. (a) (b) (c) (d)
Subtract 15 from both sides. Subtract 6x and 2y from both sides. Subtract 6x from both sides. Subtract 6x from both sides, then divide both sides by 2. 9. Divide both sides by 10 to get the equation 7a + 5 = 9. Then add 10 to obtain 7a + 15 = 19.
Mathematics I Solutions Manual
• Chapter 2, page 51
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10. The key is to find out what you have just before your second-to-last step. The ending number is 36, and the last two steps are to add 11 and multiply by 3. To find out the number just before the second-to-last step, you divide 36 by 3, and then subtract 11 from the result. • Divide by 3: 36 ÷ 3 = 12 • Subtract 11: 12 − 11 = 1 The number just before the second-to-last step is 1. To find the answers to parts (a), (b), and (c), perform the steps listed to the number 1. (a) • Add 8: 1 + 8 = 9. • Multiply by 8: 9 · 8 = 72. The ending number is 72. (b) • Multiply by 5: 1 · 5 = 5. • Subtract 20: 5 − 20 = −15. The ending number is −15. (c) • Add 19: 1 + 19 = 20. • Multiply by 5: 20 · 5 = 100. The ending number is 100. 11. (a) x + y + 6 must equal 7 + 6 = 13, not 12, so it’s false. (b) x + y − 4 must equal 7 − 4 = 3, which is true. (c) 3(x + y) must equal 3 · 7 = 21, not 28, so it’s false. (d) The equation is true if x = 3 and y = 4, or when x = 4 and y = 3. It is false any other time. (e) x+y must equal 77 = 1, which is true. 7 (f) The position around equality does not matter, so it’s true. (g) The equations x + y = 7 and x − y = 3 are both true only when x = 5 and y = 2. One or the other is false with any other combination. (h) Suppose you state the original equation “in terms of y,” meaning, “Start with y. Add x. The result is 7.” You can use backtracking to say, “Start with 7. Subtract x. The result is y.” This last statement is what the equation represents, so it’s true. (i) One possible combination would be x = 3 and y = 4, which makes this statement true. But another combination could be x = 10 and y = −3, which makes this statement false. So it may be true or false.
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Both sides of the equation are the same, so every number is a solution. 14. 4(n − 1) = 6n − 12 4n − 4 = 6n − 12 4n − 4 − 4n = 6n − 12 − 4n −4 = 2n − 12 −4 + 12 = 2n − 12 + 12 8 = 2n 8 2n = 2 2 4=n 15. (a) 3x + 5 = 25 3x + 5 − 5 = 25 − 5 3x = 20 3x 20 = 3 3 x = 20 3 (b) 3(x − 1) + 5 = 25 3(x − 1) + 5 − 5 = 25 − 5 3(x − 1) = 20 3(x − 1) 20 = 3 3 x − 1 = 20 3 x−1+1= x=
3(x + 3) + 5 = 25 3(x + 3) + 5 − 5 = 25 − 5 3(x + 3) = 20 3(x + 3) 20 = 3 3 x + 3 = 20 3
12.
−4a = 10 −4a 10 = −4 −4 a = − 52 13. 11 − 2(d − 1) = −2d + 13 11 − 2d + 2 = −2d + 13 −2d + 13 = −2d + 13 Mathematics I Solutions Manual
+1
(c)
On Your Own 2a + 1 = 6a + 11 2a + 1 − 6a = 6a + 11 − 6a −4a + 1 = 11 −4a + 1 − 1 = 11 − 1
20 3 23 3
x+3−3= x=
20 3 11 3
−3
(d) 3(x − 7) + 5 = 25 3(x − 7) + 5 − 5 = 25 − 5 3(x − 7) = 20 3(x − 7) 20 = 3 3 x − 7 = 20 3 x−7+7= x= • Chapter 2, page 52
20 3 41 3
+7
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(e)
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Equation II is not true for any values of t, therefore it has no solutions. 3(2x + 1) + 5 = 25 3(2x + 1) + 5 − 5 = 25 − 5
III: 5(t − 1) = −5(t + 4) + 15 5t − 5 = −5t − 20 + 15
3(2x + 1) = 20 3(2x + 1) 20 = 3 3 2x + 1 = 20 3 2x + 1 − 1 = 2x =
20 3 17 3 17 3
5t − 5 = −5t − 5 5t = −5t 10t = 0 t=0
−1
Equation III is true only when t = 0. It has exactly one solution. The answer is D.
2x = 2 2 x = 17 6 (f)
Maintain Your Skills 3(5 − 2y) + 5 = 25 3(5 − 2y) + 5 − 5 = 25 − 5
18. All solutions follow the same pattern, except for part (d). Parts (a) and (d) are worked out in detail. (a) d = 4
3(5 − 2y) = 20 3(5 − 2y) 20 = 3 3 20 5 − 2y = 3 5 − 2y − 5 = −2y =
20 3 5 3 5 3
5d − 2 = 2d + 10 3d = 12 d=4
−5
(b) d = 6. Similar solution to above. (c) d = 12. Similar solution to above. (d) There is no solution.
−2y = −2 −2 y = − 56
5d − 2 = 5d + 10 0 = 12
16. (a) true (Multiply both sides of the equation by 2). (b) false (Need to add 20 to the left side and 11 to the right.) (c) true (Add 20 to both sides of the equation.) (d) true (Multiply both sides by −1). (e) false (Should be 3r − 7 = 0, if you subtract 9 from each side.) (f) true (Multiply both sides of the equation by 2, and distribute the left side. (g) true (Subtract 2 from each side.) (h) false (r = 73 .) 17. First, solve each equation. I: 4s − 4 = 3(s − 3) + 5 + s 4s − 4 = 3s − 9 + 5 + s 4s − 4 = 4s − 4 0=0 Equation I is true for all values of s. II: − 3(t + 2) = 8(t + 1) − 11(t − 2) −3t − 6 = 8t + 8 − 11t + 22 −3t − 6 = −3t + 30 −6 = 30
d = −12 d = −6 d = −4 Part (d) has a different number of solutions than the rest, because there’s a 5d on both sides that cancels out. All that’s leftover are the constants, −2 and 10. Since −2 = 10, this equation has no solutions. (i) The pattern is that the solution increases as the coefficient of d on the right side of the equation increases. Once the coefficient of d reaches 5, there are no solutions. Then suddenly, the solutions are the same as before, except now they’re negative.
(e) (f) (g) (h)
19. Notice that for all parts of this exercise, there is a 4c on both sides of the equation. If you subtract 4c from both sides, you are left with just the numbers, and no variables. So the only equation that would have any solution would be the one where those numbers are the same, and that equation would always be true. If the numbers are different, the equation would always be false, and so it has no solution. The equation in part (d) is always true, and so its solution set is all real numbers. All the other equations have no solution.
Mathematics I Solutions Manual
• Chapter 2, page 53
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7.
2.13
Focus on the Distributive Property
1 −(−3 − q) = − (6q − 26) 2 3 + q = −3q + 13 4q = 10 4q 10 = 4 4 5 q= 2
Check Your Understanding 1. 2(x + 3) = 15 2x + 6 = 15 2x = 9 9 x= 2
8. 4(e + 4) = 2(17 − 2e) 4e + 16 = 34 − 4e 4e + 16 + 4e = 34 − 4e + 4e
2. 5(j − 3) = 10(j − 2) 5j − 15 = 10j − 20 5 = 5j
8e + 16 = 34 8e + 16 − 16 = 34 − 16 8e = 18 8e 18 = 8 8 9 e= 4
1=j 3. 3(k − 4) + 6 = 2 3k − 12 + 6 = 2 3k − 6 = 2 3k = 8 8 k= 3 4.
9. 2(2x + 1) − 3(x − 5) = 18 4x + 2 − 3x + 15 = 18 x + 17 = 18 x + 17 − 17 = 18 − 17 x=1 10.
−(x + 3) = 7 −x − 3 = 7 −x = 10 x = −10 5. −(s + 2) = 4(s + 1) −s − 2 = 4s + 4 −6 = 5s −6 5s = 5 5 6 − =s 5 6. −2(2w + 3) = 10 −4w − 6 = 10 −4w = 16 −4w 16 = −4 −4 w = −4
2 =z − z+ 3 2 −z − = z 3 2 −z − − z = z − z 3 2 −2z − = 0 3 2 2 2 −2z − + = 0 + 3 3 3 2 −2z = 3 2 −2z = 3 −2 −2 1 z=− 3 11. A different method besides expansion boxes is to try it with numbers. For example, let a = 2 and b = 2. Then (a + b)2 = (2 + 2)2 = 42 = 16, but a2 + b2 = 22 + 22 = 4 + 4 = 8. They are not equal in that case, so they’re definitely not always equal. 12. There are four columns and two rows in the array. If you add the columns first, each column would total a + b. Since there are 4 columns, the total sum of the array would be 4(a + b).
Mathematics I Solutions Manual
• Chapter 2, page 54
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What if you add the rows first? The first row contains 4 a’s, so its sum would be 4a. The sum of the second row would be 4b, making the sum of the entire array 4a + 4b. From the any-order, any-grouping principles, you know that it does not matter what order you add the numbers in. The sum should be the same. So the first method must get the same result as the second, or in other words,
19.
4(a + b) = 4a + 4b.
20.
13. The reasoning is similar to Exercise 12. You have four columns, each with value (a + b + c), so the value of the table is 4(a + b + c). But you also have some rows. One has value 4a, the other is 4b, and the last one is 4c. So, the value of the entire table is 4a + 4b + 4c, the sum of the value of the rows. So,
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28 = −14x + 21 7 = −14x 7 − =x 14 1 − =x 2 −4(p + 2) + 8 = 2(p − 1) − 7p + 15 −4p − 8 + 8 = 2p − 2 − 7p + 15 −4p = −5p + 13 p = 13 21.
4(a + b + c) = 4a + 4b + 4c
2(5 − n) = 6 + 6n 10 − 2n = 6 + 6n
14. The array x y z
x y z
x y z
x y z
x y z
x y z
page 55
4 = 8n 1 n= 2
x y z
can be used to illustrate the fact that 7(x + y + z) = 7x + 7y + 7z.
22. Use expansion boxes or other methods to write each expression without parentheses: (a) The answer is 24x − 84.
On Your Own 15.
−7 −84
2x 24x
12 (b) The answer is 12 − 3x.
5(2a − 1) = 60 10a − 5 = 60 10a = 65 65 a= = 6.5 10
−x −3x
4 12
3
(c) The answer is 2x2 + 5x. +5
2x
16.
2
+5x
3x
−18
2x
−12
14x
−3
8x
− 12 7
2x
x (d) The answer is 2x − 12.
−(d − 4) = 13 −d + 4 = 13 d = −9
2 3
(e) The answer is 8x −
17. 15 = −5(x − 4) 15 = −5x + 20 −5 = −5x 1=x
12 . 7
4 7
(f) The answer is −12x + 24. Be careful! (−4) · (−6) = +24. 3x −12x
18. 3(x − 2) + 2(x − 2) = 40 3x − 6 + 2x − 4 = 40 5x − 10 = 40 5x = 50 x = 10 Mathematics I Solutions Manual
−4
−6 +24
(g) The answer is −3x2 − 21x + 42. Be careful! (−3) · (−14) = +42. x2 −3
−3x
• Chapter 2, page 55
2
+7x
−14
−21x
+42
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(h) The answer is −10x − 15. − 56
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(c)
12x
+18
−10x
−15
2(x + 1) + 5 = x + 2 ?
2(−3 + 1) + 5 = −3 + 2 ?
(i) The answer is − 72 x3 + 11x. Be careful! (− 12 ) · (−22) = +11. − 12 x
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7x2
−22
− 72 x3
+11x
2(−2) + 5 = −1 ?
−4 + 5 = −1 1 = −1, No (d)
23.
2x + 3 − 4x = −3x ?
2(−3) + 3 − 4(−3) = −3(−3)
−z(2a − 4) = −2az + 4z = 4z − 2az
?
−6 + 3 − (−12) = 9 ?
The correct answer is C. 24. (a) Incorrect. x = −10. Jade misdistributed in her first step. −(x − 3) = −x + 3, not −x − 3. (b) Correct. (c) Incorrect. v = 19 . Jade misdistributed in her first step. 8 −3(v − 8) = −3v + 24, not −3v − 24.
−6 + 3 + 12 = 9 9 = 9, Yes 3. (a) 6x + 3 = 2x + 7 6x − 2x + 3 = 2x − 2x + 7 4x + 3 = 7
Maintain Your Skills 25. (a) (b) (c) (d) (e) (f) (g)
2C
4x + 3 − 3 = 7 − 3 4x = 4 4x ÷ 4 = 4 ÷ 4 x=1
x=7 x = 107 x = 47 x = 97 x = 15 x = −93 Whatever number you subtract from x (within the parentheses), the solution increases by that much. So, when you subtract 100, the solution goes from 7 to 107.
(b) 4m − 3 = m + 12 4m − m − 3 = m − m + 12 3m − 3 = 12 3m − 3 + 3 = 12 + 3 3m = 15 3m ÷ 3 = 15 ÷ 3 m=5
MATHEMATICAL REFLECTIONS
1. Answers may vary. For example, 2x + 3 = 2(x + 1) is an example of an equation with no solution. 2x + 3 = x + (x + 3) is an example of an equation with infinitely many solutions. 2. (a)
4. (a) 5x − 6 = 3x + 4 5x − 3x − 6 = 3x − 3x + 4 2x − 6 = 4 2x − 6 + 6 = 4 + 6
4x + 1 = 2x − 5 ?
4(−3) + 1 = 2(−3) − 5
2x = 10 2x ÷ 2 = 10 ÷ 2 x=5
?
−12 + 1 = −6 − 5 −11 = −11, Yes (b)
(b) x − 4 = −2x ?
(−3) − 4 = −2(−3)
x + 2 + 3x = 4(x − 1) 4x + 2 = 4x − 4 4x − 4x + 2 = 4x − 4x − 4
−7 = 6, No Mathematics I Solutions Manual
2 = −4 → No solution • Chapter 2, page 56
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(c)
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INVESTIGATION 2D WORD PROBLEMS 2(3x + 4) + 2 = 6x + 10 6x + 8 + 2 = 6x + 10 6x + 10 = 6x + 10 6x − 6x + 10 = 6x − 6x + 10 10 = 10 → All real numbers
2.14
Getting Started
For You to Explore 1. (a) No, 12 isn’t Chiko’s number.
(12 + 4) + (12 − 4) + (12 · 4) + (d)
12 4
= 16 + 8 + 48 + 3 = 75 3 − x = 2x + 9 3 − x − 2x = 2x − 2x + 9 3 − 3x = 9 3 − 3 − 3x = 9 − 3
And 75 = 60 (b) No, 8 isn’t Chiko’s number.
8 (8 + 4) + (8 − 4) + (8 · 4) + 4 = 12 + 4 + 32 + 2 = 50
−3x = 6 −3x ÷ (−3) = 6 ÷ (−3) x = −2
And 50 = 60 (c) Start with a number b. Applying the same steps to b gives b (b + 4) + (b − 4) + (4b) + 4
5. 2(m + 3) − 4(2m + 1) = m − 3(2 + m) 2m + 6 − 8m − 4 = m − 6 − 3m −6m + 2 = −2m − 6 −6m + 2m + 2 = −2m + 2m − 6 −4m + 2 = −6
as the total of the four calculations. Since the total has to equal 60, an equation to solve is b (b + 4) + (b − 4) + (4b) + = 60 4
−4m + 2 − 2 = −6 − 2 −4m = −8
Solve this equation by combining like terms on the left side, then dividing. b (b + 4) + (b − 4) + (4b) + = 60 4 b + b + 4b + 0.25b = 60
−4m ÷ (−4) = −8 ÷ (−4) m=2 6. Answers will vary. You can illustrate the equation on a number line or illustrate the numbers and variables with squares and triangles. You can use the basic moves that never change the solutions of an equation, such as adding the same value to both sides of the equation. 7. Answers will vary. First, select a fixed size for the variable. On top of the number line, line up lengths that represent your variable (for example, if you have 5x, line up 5 copies of the length x). Then, line up a length that represents the constant (for example, if you have 5x + 3, you want to add a length of 3). On the bottom, repeat the process for the left side of the equation. 8.
6.25b = 60 6.25b 60 = 6.25 6.25 b = 9.6 2. The equation is similar to Chiko’s in Exercise 1, using 5’s instead of 4’s. Solving the equation is also similar. n (n + 5) + (n − 5) + (5n) + = 144 5 1 n + n + 5n + n = 144 5 36 n = 144 5
5x = x + 24 4x = 24
5 36 5 · n= · 144 36 5 36 n = 20
x=6 Mathematics I Solutions Manual
• Chapter 2, page 57
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−0.1c −7 = −0.1 −0.1 c = 70
3. (a) Answers will vary. One guess is 100 years old. To check this, you’d add one sixth of 100, one twelfth of 100, one seventh of 100, five, one half of 100, and four:
So Cameron’s books cost $70.
100 100 100 100 2 + + +5+ + 4 = 98 6 12 7 2 7 Since the total isn’t 100, this is an incorrect guess. (b) Guesses will vary, but the steps should be the same. To check a guess, follow these steps. Step 1: Guess Diophantus age at his death. Step 2: Divide your guess by 6. Step 3: Divide your guess by 12. Step 4: Divide your guess by 7. Step 5: Divide your guess by 2. Step 6: Add up the total from Steps 2, 3, 4, and 5. Then add an additional 5 and an additional 4 to that total. Step 7: Check to see if the total in Step 6 equals your guess from Step 1. (c) If the guess is n, follow the same steps, replacing the guessed number with n. The goal is for the result of all the calculations to equal n.
On Your Own 5. (a) Sometimes true, since it is only true when the input number is 3, and false every other time. (b) Sometimes true, since it is only true when n = 3, and false every other time. The written equation is equivalent to the statement in part (a). (c) Never true, since the result from subtracting will always be six less than the result from adding, they can never give the same answer. (d) Sometimes true, since it is only true when n = 20, and false every other time. To solve this equation, simplify it by adding and subtracting the same quantities from each side. First, subtract 3n from each side, then add 6.
n n n n + + +5+ +4=n 6 12 7 2 (d) One of the best ways to find n is to multiply through the fractions on the left side by a common denominator. The least common denominator is 84: n n n n + + +5+ +4=n 2 n6 n12 n7 n 84 + + + 5 + + 4 = 84n 6 12 7 2 14n + 7n + 12n + 420 + 42n + 336 = 84n 75n + 756 = 84n 75n + 756 − 75n = 84n − 75n 756 = 9n 84 = n So Diophantus lived to age 84, according to the story. It would’ve been a great guess to pick 84 from the beginning, since it’s the smallest number that makes all the fractions in the problem work out evenly. 4. Many solution methods are possible here, including guess-and-check and equation solving. If c is the cost of Cameron’s books, one equation is c − 0.1c + 15 = c + 8 And, solving the equation gives: c − 0.1c + 15 = c + 8 c − 0.1c + 15 − c = c + 8 − c −0.1c + 15 = 8 −0.1c + 15 − 15 = 8 − 15 −0.1c = −7 Mathematics I Solutions Manual
5n − 6 = 3n + 34 5n − 6 − 3n = 3n + 34 − 3n 2n − 6 = 34 2n − 6 + 6 = 34 + 6 2n = 40 n = 20 (e) Sometimes true, since it is only true when n = 20, and false every other time. This is a simplified version of the equation in part (d). (f) Always true, since this is an example of the distributive property of multiplication. In both cases, the result is 6 more than twice the original number. (g) Never, because the written equation is equivalent to the statement in part (c). Simplifying this equation by subtracting the same quantity from each side gives an equation that is never true. n−3=n+3 n−3−n=n+3−n −3 = 3 Since −3 and 3 are not equal, the equation cannot be true no matter what value is chosen for n. (h) Always true, because the written equation is equivalent to the statement in part (f). Simplifying this equation by using the distributive property gives an equation that is always true. 2n + 6 = 2(n + 3) 2n + 6 = 2 · n + 2 · 3 2n + 6 = 2n + 6 Since the quantities on either side are exactly the same expression, they must always be equal, no matter what value is chosen for n. • Chapter 2, page 58
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(i) Sometimes true, because it is true when n = 1 and when n = 9, and false every other time. Since the only numbers whose square is 16 are 4 and −4, the value of n − 5 must be either 4 or −4. If n − 5 = 4, then n is 9. If n − 5 = −4, then n is 1. 6. Let Vanessa’s old wage be w. She used to make 32w per week. Now she makes 40(w + 2) per week. She makes an additional 200 dollars per week. So you get the following equation.
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(a) x = 5 3x − 7 = 13 − x 4x − 7 = 13 4x = 20 x=5 (b) Now, you know that whatever is in the parentheses must equal 5. So a + 2 = 5 which means a = 3. (c) Whatever is in the parentheses must equal 5. So 5b = 5 which means b = 1. (d) Whatever is in the parentheses must equal 5. So 8 − c = 5 which means c = 3. (e) Whatever is in the parentheses must equal 5. So 12 + 2d = 5 which means d = − 72 . (f) You don’t need to distribute the parentheses, since the expression in parentheses is just subsituting for x in the first equation. Since x = 5 was the solution to the first equation, whatever is in the parentheses must equal 5 in all of the other equations.
32w + 200 = 40(w + 2) 32w + 200 = 40(w + 2) 32w + 200 = 40w + 80 120 = 8w 15 = w So Vanessa’s old wage (w) was $15 per hour. 7. The solution to this exercise is included in the solution to Exercise 6 above. The equations in parts (a), (b), (d), (e), and (i) were true sometimes. They were true, respectively, when
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10. (a) 5x − 3 = x + 14 5x − 3 − x = x + 14 − x
(a): The input number was 3. (b): The input number was 3. (d): n = 20. (e): The input n = 20. (i): The input number is 1 or 9.
4x − 3 = 14 4x − 3 + 3 = 14 + 3 4x = 17 4x 17 = 4 4 17 x= 4
8. (a) Since there is a basic move that takes one equation and turns it into the other, the two equations have the same solution. (b) 7n − 13 = 4n + 35 7n − 13 − 4n = 4n + 35 − 4n 3n − 13 = 35 3n − 13 + 13 = 35 + 13 3n = 48 3n 48 = 3 3 n = 16
(b) 7a − 3 = 3a + 14 7a − 3 − 3a = 3a + 14 − 3a 4a − 3 = 14
(c)
5m − 19 = m − 2 5m − 19 − m = m − 2 − m
(c) This time, start by expanding each side using the distributive property.
4m − 19 = −2 4m − 19 + 19 = −2 + 19 4m = 17
13(n − 1) = 5(2n + 7) 13n − 13 = 10n + 35 13n − 13 − 10n = 10n + 35 − 10n 3n − 13 = 35 At this point, this is the same equation as a step in the previous part, so the remaining steps are identical. The solution is n = 16. Maintain Your Skills 9. Once you solve the first equation, you can use that to solve the other equations. Mathematics I Solutions Manual
At this point, you have the same equation as you did in line 3 above, except with the variable a instead of x. Even so, the remainder of the solution is the same, . so the end result will be a = 17 4
Again, this equation is the same as you had in line 5 of part (a) above. (d) −2q − 6 = −6q + 11 −2q − 6 + 6q = −6q + 11 + 6q 4q − 6 = 11 4q − 6 + 6 = 11 + 6 4q = 17 • Chapter 2, page 59
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This equation is the same as line 5 of part (a) above, and the solution follows in the same way as part (a). (e) Each equation in parts (a) through (d) are equivalent to the equation 4x = 17. Like Exercise 9, each equation has the same form. But unlike Exercise 9, each equation has the same solution since the expression substituted is just a lone variable, and not a more complex expression. (f) Answers will vary, but one possible equation is 9t + 20 = 5t + 37. All correct answers must be equivalent to the equation 4x = 17.
2.15
Building Equations
Check Your Understanding 1. (a) It is an easy guess since the result works out nicely. This happens because the problem asks you to calculate one-fourth of the number, and one-fourth of 4 is 1. (b) The result is 5. (c) The result is 15, which is three times larger than the result when using 4. Note that 5 15 = 4 12 so the results are proportional to the inputs. (d) To check a guess you multiply your guess by a fourth, then add that to the guess. So if your guess is n then calculate n + 14 n. Compare this with 210. This gives the equation x + 14 x = 210 Solve this equation. x + 14 x = 210 5 x 4
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4 5
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(e) n = 10. 8(35 + n) = 360 35 + n = 45 n = 10 3. This solution is very similar to the previous solution. The equation to solve is 9(35 + c) − 210 = 150 The solution is c = 5, so Katie will need 5 new customers. 4. (a) The cost of the CDs would be 6 · 15.99 = 95.94, before the discount. The discount would be 95.94 · 0.15 = 14.39 (rounded to the nearest penny). So the total cost of the order would be 95.94 − 14.39 + 12.00 = 93.55, which is under $100, so yes, Tony can buy 6 CDs that each cost $15.99. (b) The cost of the CDs would be 7 · 15.99 = 111.93, before the discount. The discount would be 111.93 · 0.15 = 16.79 (rounded to the nearest penny). So the total cost of the order would be 111.93 − 16.79 + 12.00 = 107.14, which is over $100, so no, Tony can not buy 7 CDs that each cost $15.99. (c) The cost of the CDs would be 2 · 15.99 + 5 · 11.99 = 31.98 + 59.95 = 91.93, before the discount. The discount would be 91.93 · 0.15 = 13.79 (rounded to the nearest penny). So the total cost of the order would be 91.93 − 13.79 + 12.00 = 90.14, which is under $100, so yes, Tony can buy 2 CDs that cost $15.99 and 5 CDs that cost $11.99. (d) Call the total retail price p. In part (a), p = 95.94, in part (b), p = 111.93, and in part (c), p = 91.93. In each case, you multiplied the retail price by 0.15, subtracted that number from the original price, and added 12 to get the actual cost. Since Tony wants to spend exactly $100 after the discount and shipping, the equation would be: p − 0.15p + 12 = 100
= 210
x=
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0.85p + 12 = 100 0.85p = 88 88 p= 0.85 p ≈ 103.53
· 210 = 168
2. (a) Last summer Katie made 35 · $6 = $210 mowing lawns. With 4 new customers, Katie will make 39 · $8 = $312 this summer, which is only $102 more, so no, Katie will not meet her goal. (b) With 13 new customers, Katie will have 48 customers. She will make 48 · $8 = $384 this summer, which is $174 more. She will exceed her goal. (c) With 8 new customers, Katie will have 43 customers. She will make 43 · $8 = $344 this summer, which is $134 more. She will not reach her goal. (d) If Katie gets n new customers, the total number of new customers she will have is 35 + n. Each customer pays $8, so she will make 8(35 + n) dollars. This needs to be $150 more than $210, so Katie needs to make $360. One possible equation is 8(35 + n) = 360 and another is 8(35 + n) − 210 = 150
So Tony needs to buy $103.53 worth of CDs before discount and shipping in order to use all of his gift certificate. 5. Let A equal Anna’s age now. In 36 years her age will be A + 36, so A + 36 = 5A 36 = 4A 9=A Her current age is 9 years old. On Your Own 6. Use the guess-check-generalize method to find the equation. To check if p is the correct answer, you would first calculate 12 p (because’s he’s halfway through.) Then
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2
The check If Arnold is 30, then 17 years ago, he was 30 − 17 = 13.
The steps Subtract 17 from Arnold’s age.
2 3
Multiply Arnold’s age by 23 .
?
13 = 20 No.
4
Try another guess.
So if Cameron buys $150 worth of books, the final cost is the same whether or not he joins the club. 10. (a) You can show that Travis’s answer is wrong by substituting it into the original equation. After you simplify, if you get a true statement, then his answer is correct. If not, then his answer is wrong. x+5−
of Arnold’s current age is 2 · 30 = 20. 3
3
35 3
6·
1 10
7 ? = 6 35 ? 6 · 3 + 5 − 76 = ? 35 + 6 · 5 − 6 · 76 = 3 ?
+5−
12 + 35 6 6 · 12 +
35 6
6 · 12 + 6 ·
35 6
70 + 30 − 7 = 72 + 35 93 = 107
If you apply the same steps to the variable a, you will get the equation: a − 17 =
x = 12 + 12 x 1 35 ? + 5 − 10 = 12 + 12 35 3 3 35 3
Compare the results of Steps 1 and 2.
So Travis’s answer is incorrect. (b) Travis made his mistake on the first line of his solution. He wrote:
2 a 3
This equation has solution a = 51. 8. x − 5 = 56 x + 2 Solve the equation for x. x−5=
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r − 0.1r + 15 = r −0.1r + 15 = 0 −0.1r = −15 −15 r= −0.1 r = 150
2 1 p + 84 = p 2 3 2 1 84 = p − p 3 2 1 84 = p 6 504 = p
1
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the equation to check would be r − 0.1r + 15 = r.
you’d calculate 23 p. Now, 12 p + 84 should equal 23 p. That gives you the equation.
7. Guess that Arnold is 30.
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x+5− 11 10
1 10
x = 12 + 12 x
x + 5 = 12 + 12 x
1 11 But x − 10 x = 10 x. He added when he should have subtracted. The correct second line would be
5 x+2 6
5 x− x=2+5 6 1 x=7 6 x = 42
9 10
x + 5 = 12 + 12 x
(c) Here’s the complete corrected solution. x+5− 9 10
Allen is 42 years old today. The correct answer choice is C. 9. To check which way is cheaper, you need to compare the actual retail cost of books with the cost after discount but adding the $15 membership fee. (a) With the membership, the discount would be 70 · 0.1 = 7. The final cost would be 70 − 7 + 15 = 78, so it is cheaper not to join the club. (b) The final cost would be 250 − 250(0.1) + 15 = 250 − 25 + 15 = 240, so it is cheaper to join the club. (c) The final cost would be 100 − 100(0.1) + 15 = 100 − 10 + 15 = 105, so it is cheaper not to join the club. (d) If the retail cost is r, the cost after joining the club would be r − r(0.1) + 15. To check for what the actual retail cost of books would be if he ended up spending the same whether or not he joined the club,
9 10
1 10
x = 12 + 12 x
x + 5 = 12 + 12 x
x + 5 − 12 x = 12 + 12 x − 12 x 2 5
x + 5 − 5 = 12 − 5 2 5
x=7 x=
35 2
11. One way to find this is by guessing and checking. For example, if the best score was 90, the average would be
Mathematics I Solutions Manual
15 · 82 + 90 = 82.5 16 which is too low. In general the equation to solve is 15 · 82 + s = 83 16 and the solution to this equation is s = 98. A second way to think about this is that if the best score was 82, the average would remain 82. The best • Chapter 2, page 61
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score needs to be 82, plus enough to bring the entire average to 83. Since there are 16 students, the best score needs to be 16 points above 82 to bring the average up one entire point. 12. Taylor spent 100 − 16 = 84 dollars on the CDs, and if seven CDs are worth $84, then each is worth 84 = 12 7 dollars.
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by zero! In practical terms, it also wouldn’t make sense for there to be a $12 discount if the item were already free. (c) The resulting equation is d − 12 = (d) The resulting equation is 10(d − 12) = 8d
Maintain Your Skills
10d − 120 = 8d
13. (a)
(e) Solve the equation by subtracting 10d from each side, then dividing by −2.
5x − 3 = x + 15 4x = 18 18 9 x= = 4 2
10d − 120 = 8d 10d − 120 − 10d = 8d − 10d −120 = −2d −120 −2d = −2 −2 60 = d
(b) 6x − 3 = x + 15 5x = 18 18 x= 5
You could also solve this equation by subtracting 8d from each side and adding 120 to each side.
(c) 10x − 3 = x + 15 9x = 18 18 x= =2 9
10d − 120 = 8d 10d − 120 − 8d = 8d − 8d 2d − 120 = 0 2d − 120 + 120 = 0 + 120
(d)
2d = 120 2d 120 = 2 2 d = 60
99x − 3 = x + 15 98x = 18 18 9 x= = 98 49 (e) 181x − 3 = x + 15 180x = 18 18 1 x= = 180 10 (f) The solution gets smaller and smaller as A gets larger. 18 (g) x = A−1
2.16
8 d 10
Solving Word Problems
Check Your Understanding 1. (a) If d is the original cost, then (d − 12) is the cost after 8 the discount. The fraction 10 represents 80%, which 80 or as the could also have been written as 45 or as 100 decimal 0.8. The equation says that the ratio between the price with discount and the original price is the same as 80%. (b) The value of d cannot be zero, since it would lead to the fraction −12 . This is an undefined fraction, since 0 there is no number that gives −12 when you multiply
The supplies cost $60 before the discount, and $48 after. Testing these values shows that $48 is 80% of $60. 3 2. (a) The probability that it will rain on Monday is 10 30 (since 30%, written in fraction form, is 100 , which 3 reduces to 10 ). (b) It is certain that it either will rain or will not rain on Monday. So the probability that it will rain or will not rain is equal to 1 (since it is 100 percent sure that it will rain or will not rain). Since the probability that it 3 will rain is 10 , the probability that it will not rain is 3 7 1 − 10 = 10 . (c) As in the last part, you need to subtract the probability that it will rain from 1 to get the probability that it will not rain. So the probability it 1 9 will not rain tomorrow is 1 − 10 = 10 . (d) If the probability of rain is p, the probability of no rain is (1 − p). 3. If you let p be the probability that it will rain, then (from exercise 2), you know the probability that it will not rain is 1 − p. The meteorologist said that there is a 50% greater chance of it raining than not raining, so the chance of it raining is 0.5 more than the chance of it not
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• Chapter 2, page 62
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raining. So the equation you want to solve is p = (1 − p) + 0.5 p = 1.5 − p p + p = 1.5 − p + p 2p = 1.5 2p 1.5 = 2 2 p = 0.75
6.4 + 8.0 + 8.1 + 6.5 + 8.2 + 7.6 ≈ 7.467 6 This is almost, but not quite, enough for Heidi to make the team. (c) Answers will vary. Steps should be the same as above. (d) Start by adding all of Heidi’s score = 37.2. Now, to test if x is the minimum score Heidi needs to make the team, add 37.2 + x and take the average (i.e. divide by 6.) That total should equal 7.5. This gives you the following equation: 37.2 + x = 7.5 6 (e) Use the equation from part (d). If x is Heidi’s sixth score, the output needs to be 7.5. An equation is
x = 7.8 So 7.8 is the cutoff score. Another way to do this is to recognize that if Heidi’s average is to be 7.5, she must score a total of 7.5 · 6 = 45 points in the six events.
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8.0 + 8.1 + 6.5 + x = 7.5 4 Solve this equation by multiplying each side by 4 and combining numeric terms on the left side: 8.0 + 8.1 + 6.5 + x = 4 · 7.5 22.6 + x = 30 x = 7.4 So, with the new rules, Heidi only needs to score at least 7.4. (b) To find the highest average, imagine that Heidi scores a perfect 10 in the last event. That score is sadly thrown out, but the four scores to average would be 6.5, 8.0, 8.1, and 8.2, giving an average of 7.7. To find the lowest average, a 0 score in the last event would be thrown out and the four scores 6.4, 6.5, 8.0, and 8.1 would be averaged to 7.25.
On Your Own 7. Let s be the number of students Ms. Meyer was going to bring on the field trip. Using the guess-check-generalize method you’ll arrive at the following equation. 9s = 6(s + 47) Solving this equation gives s = 94. So Ms. Meyer can now bring 47 more than this, so she can bring 94 + 47 = 141 students. 8. This solution follows the same steps as the other solutions in the lesson. Set up the equation using the guess-check-generalize method. If m is the number of nickels that Chi has, you arrive at the equation m − 100 = 0.95m This equation has the solution m = 2000, so Chi has 2000 nickels. 9. Guess that Bianca is 20. • Eight years from now, she’ll be 20 + 8 = 28. • 50% more than her current age is 1.5 · 20 = 30. • Are those two the same? No. You can apply the same steps to a variable, say b. If Bianca is b years old right now.
37.2 + x = 7.5 6 To solve this equation, multiply each side by 6 and combine numeric terms on the left side: 37.2 + x = 6 · 7.5 37.2 + x = 45
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6. (a) The lowest score, 6.4, and the highest score, 8.2, are rejected. The other 4 scores, including the unknown x, are used to make an average:
There is a 75% chance of rain on Wednesday. The chance of no rain would be 25%, which confirms the answer is correct (there is a 50% higher chance of rain). 4. No, the conclusion is not reasonable. Since it’s possible for it to not rain on both days, you can’t just add the percentages together. Put another way, suppose you flipped two coins. Each coin comes up “heads” 50 percent of the time, but it doesn’t mean that if you flip two coins that one or the other will always be heads. 5. (a) To find the mean, add up the scores and divide by the number of scores (5): 6.4 + 8.0 + 8.1 + 6.5 + 8.2 = 7.44 5 The median score of 8.0 is found by writing the scores in increasing order and taking the third score (which is the middle number in the ordered list). (b) To find Heidi’s new mean, add up the scores and divide by the number of scores (6):
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• Eight years from now, she’ll be b + 8 • 50% more than her current age is 1.5 · b = 1.5b. • Are those two the same? b + 8 = 1.5b? That’s our equation. b + 8 = 1.5b has the solution b = 16, so Bianca is 16. 10. Due to the numbers in this problem, it is difficult to make a guess that works out evenly (the least common denominator is 523 · 415 = 217,045). Guessing a nice, round number is usually a good choice then. If the trip is 3000 miles, the flight east takes 3000 ≈ 523 5.736 hours, or 5 hours and 44 minutes. The flight west
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takes 3000 ≈ 7.229 hours, or 7 hours and 14 minutes. The 415 total time is 12 hours, 58 minutes, which is too long. If the trip is 2000 miles, the flight east takes 2000 ≈ 3.824 hours, or 3 hours and 49 minutes. The flight 523 west takes 2000 ≈ 4.819 hours, or 4 hours and 49 415 minutes. The total time is 8 hours, 38 minutes, which is too short. An eagle eye to the numbers here might notice that the 2000-mile flights are exactly 23 as long as the 3000-mile flights. If the trip is t miles, an expression for the total time of the two trips is t t + 523 415 and an equation to solve is t t + = 11 523 415 This equation is messy, so a calculator is definitely a good idea. t t + = 11 523 415 t t 415 · 523 + = 415 · 523 · 11 523 415 415t + 523t = 2387495 938t = 2387495 t ≈ 2545.30 To the nearest mile, each flight is 2545 miles long. Another way to solve this is to use proportional reasoning, once you recognize that the flight time is proportional to the distance traveled. As a benchmark, the flight of 3000 miles lasts about 12.965 hours. Finding the distance for an 11-hour trip can be done by solving the proportion 3000 miles t miles = 12.965 hours 11 hours The value of t here is found by multiplying 3000 by 11 and dividing by 12.965, giving (to the nearest mile) the result t = 2545 miles. 11. He must pay for the food, plus the tax, plus the tip. The food costs 6.75 + 3.95; the tax is (6.75 + 3.95)(0.08); the tip is (6.75 + 3.95)(0.15). You can combine the last two. The correct answer choice is C. 12. (a) You can test Derman’s answer of x = 32 by substituting it for x and seeing if, after you simplify the equation, you get a true or false statement. 20 − 3(x − 2) = 9x − 4 ?
20 − 3( 32 − 2) = 9 · ? 27 2 ? 27 = 2 = 19 2
20 − 3(− 12 ) = 40 2
+
3 2 43 2
3 2
−4
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(c) 20 − 3(x − 2) = 9x − 4 20 − 3x+6 = 9x − 4 26 − 3x = 9x − 4 30 = 12x 2.5 = x Check that this is the correct solution. 20 − 3(x − 2) = 9x − 4 20 − 3(2.5 − 2) = 9 · (2.5) − 4 20 − 3(0.5) = 22.5 − 4 18.5 = 18.5 It works! x = 2.5 Maintain Your Skills 13. The third one; if you try to solve it you’ll get 5 = 11, which doesn’t make any sense. 14. The second one; if you try to solve it you’ll get 14 = −25, which doesn’t make any sense.
2.17
More than One Variable—Solving in Terms of Each Other
Check Your Understanding 1. 6x + 3y = 5x − 13 + 2y 6x + 3y − 2y = 5x − 13 6x + y = 5x − 13 y = −x − 13 2. (a) The second equation is easier to use here, since when you substitute 0 for x, you’re left right away with the equation y = 13 . 4 (b) Again, the second equation is easier to use. , which Substituting 10 for x gives you y = 20+13 4 simplifies to y = 33 . 4 (c) This time, the first equation is easier to use. When you substitute 0 for y, you’re left with the relatively simple equation −13 = 2x, which takes one step to solve.
−4 −
−13 = 2x −13 2x = 2 2 13 − =x 2
8 2
Since this is a false statement, x = 32 can’t be the correct solution. (b) Derman misdistributed −3(x − 2). (See below for the correct solution.) Mathematics I Solutions Manual
(d) As you have before, to find out if something is a solution, make the substitution. Replace each x with • Chapter 2, page 64
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20 and each y with 13, and simplify.
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4. (a) Substitute 10 for x, and 6 for y.
?
5x + 6y = 90
?
5(10) + 6(6) = 90
2 · 20 + 3 · 13 = 7 · 13 − 13
?
40 + 39 = 91 − 13
?
79 = 78
50 + 36 = 90 86 = 90
It turns out that the statement is not true, so (20, 13) is not a solution to the equation. (e) You could substitute the values for x and y, but you don’t have to. Since the two equations are equivalent, if (20, 13) is not a solution to one, it cannot be a solution to the other. 3. (a) w 4 = h 3 4 w ·h= ·h 3 h w = 34 h (b) w 4 = h 3 w 4 ·h= ·h h 3 w · 34 = 43 h · 34 3 4
w=h
(c) Substitute 15 for w. h= h= h=
3 w 4 3 · 15 4 45 4
Since the ending statement is false, those values do not make the equation true. (b) Substitute 9 for x, then solve the equation for y. 5x + 6y = 90 5(9) + 6y = 90 45 + 6y = 90 6y = 45 45 15 y= = 6 2 makes it true, and any other number you So y = 15 2 choose makes it false. (c) Answers will vary. (d) Yes, the points are related. They will all fall on one line (namely, the line whose equation is 5x + 6y = 90.) 5. (a) Suppose p is the correct price. Then you know that your guess is somewhere in the range between p − 75 and p + 75. You can look at it as two different equations. Your guess is greater than or equal to the low end of the range, and your guess is less than or equal to the high end. In symbols, that looks like: 2599 ≥ p − 75
w= w=
2599 ≤ p + 75
These aren’t equations, but inequalities. Still, you can solve each of these using the basic moves.
The TV will be 45 inches tall. 4 (d) Substitute 21 for h. w=
and
4 h 3 4 · 21 3 84 = 28 3
So the TV will be 28 inches wide. (e) From part (d), you know that a TV that is 21 inches tall will be 28 inches wide. If d is the length of the diagonal, you can use the Pythagorean theorem to figure out the length of the diagonal. (Remember the equation a2 + b2 = c2 . In this case, a = 21, b = 28, and c = d.) a2 + b 2 = c 2 212 + 282 = d 2 441 + 784 = d 2 1225 = d 2 ±35 = d Since the length of the diagonal cannot be negative, the correct answer must be 35 inches. Mathematics I Solutions Manual
2599 ≥ p − 75 2599 + 75 ≥ p − 75 + 75 2674 ≥ p 2599 ≤ p + 75 2599 − 75 ≤ p + 75 − 75 2524 ≤ p You can write that together as $2524 ≤ p ≤ $2674. (b) If the actual price were $4599, your guess would have to be greater than or equal to $75 less than the actual price, or $4524, and less than or equal to $75 more than the actual price, or $4674. Note that these answers are very similar to those from the previous problem. (c) If g is your guess, then it would have to be greater than or equal to $75 less than the actual price, p, which can be written g ≥ p − 75. It would also have to be less than or equal to $75 more than the actual price p, which can be written as g ≤ p + 75. You can combine both statements like this: p − 75 ≤ g ≤ p + 75. • Chapter 2, page 65
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Another way to think about it is to say that the difference between the guess and the price has to be less than or equal to 75, either above or below (since you don’t care whether your guess is higher or lower than the actual price). In this case, if g is bigger than p, then g − p ≤ 75, and if g is less than p, then p − g ≤ 75. 6. (a) Once you remove the border (which is 4 inches vertically, and 4 inches horizontally) the remaining space should have the proportions of a photograph. Namely, it should be in a 4 to 3 ratio. So, if you take away 4 from the width (w − 4) and you take away 4 from the height (h − 4), then these should be in a 4 to 3 ratio. (w − 4) 4 = (h − 4) 3
4h − 16 +4 3 4h − 16 12 w= + 3 3 4h − 4 w= 3 w=
(c) Substitute 12 for h in the equation. 4h − 4 3 4(12) − 4 w= 3 48 − 4 w= 3 44 w= 3 w=
inches wide. So the piece is 44 3 (d) Substitute 15 for h in the equation. 4h − 4 3 4(15) − 4 w= 3 60 − 4 w= 3 56 w= 3 w=
So the piece is
56 3
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(e) Substitute 20 for h in the equation. 4h − 4 3 4(20) − 4 w= 3 80 − 4 w= 3 76 w= 3 w=
inches wide. So the piece is 76 3 7. (a) No. 4 · 5 + 8 · 10 = 100¢. (b) No. 6 · 5 + 4 · 10 = 70¢. (c) 5n + 10d = 90 (d) 5n + 10d = 90 10d = 90 − 5n 90 5n d= − 10 10 n d =9− 2 (e) No. Because then n2 wouldn’t be an integer, and Chi can’t have a fractional number of dimes.
(b) 4 (w − 4) = (h − 4) 3 (w − 4) 4 · (h − 4) = · (h − 4) (h − 4) 3 4(h − 4) (w − 4) = 3 4h − 16 w−4= 3
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On Your Own 8. (a) This statement is not always true. Try x = 0 and y = 4. It makes the first equation true (4 = 4), but the second equation false (20 = 4). If the 5 were distributed to all terms in the parentheses, so that you had 5y = 5(2x + 4) = 10x + 20, then that statement is always true. (b) This statement is always true, since 25% is equal to 0.25. (c) This statement is not always true. Try a = 3 and b = 4. The first expression gives you 9 + 16 = 25, but the second gives you 7(3 + 4) = 7 · 7 = 49. (d) This statement is always true, since 3a · 4b = 3 · 4 · a · b = 12 · ab = 12ab. (e) This statement is not always true. You can simplify the first expression 0.9m − 2m = (0.9 − 2)m = −1.1m. and −1.1m will only be the same as −2.9m if m = 0. (f) This statement is not always true. If you distribute the −4 in the first expression correctly, you get −4d − 24, which is never the same as −4d + 24. (g) This statement is always true when y = 0, since, in the first expression, the 10 in the numerator cancels with the 10 in the denominator, which gives you the second expression. (h) This statement is not always true. The 3 cannot be canceled, because it’s being added to the variable in the numerator. You could break the fraction apart:
inches wide. Mathematics I Solutions Manual
x+3 x 3 x = + = +1 3 3 3 3 So it is true to say “The fraction x + 1.” 3 • Chapter 2, page 66
x+3 3
is the same as
“000200010271723958_CH02_p031-072”
9. (a) The perimeter is 2l + 2w. (b) Answers will vary, but the expression on the left is the perimeter, and the expression on the right is “10 more than four times the difference between its length and width”. (c) Begin by using the Distributive Property on the right side, then combining like terms:
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(b) To complete this table, substitute the values 0 through 7 for the variable l in the equation − 41l p = 21026 . Then, round up the resulting value of p to the next highest whole number (0 or higher). Here is the completed table: # letter 0 1 2 3 4 5 6 7
2l + 2w = 10 + 4(l − w) 2l + 2w = 10 + 4l − 4w 2l + 2w − 4l = 10 + 4l − 4w − 4l −2l + 2w = 10 − 4w −2l + 2w − 2w = 10 − 4w − 2w −2l = 10 − 6w l = −5 + 3w (d) The relationship is l = 3w − 5. When w is 10, l is 3 · 10 − 5 = 25 centimeters. 10. Nathan will spend $6 on two snow globes, which will leave $9 for key chains. If he buys k key chains, then 2k < 9, so k < 92 . The biggest k can be is 4. The correct answer is B. 11. (a) Trying different numbers of postcard stamps, there is no way to make exactly $2.10 in postage. Three postcard stamps is not enough, and four is too many. (b) Corey will need four postcard stamps, since three will not be enough to make $2.10 in postage. Four postcard stamps gives enough postage. More than four is not needed, and would just be a waste of money. (c) Substitute 3 for l in the equation, and solve for p.
87 26
# postcard 9 7 5 4 2 1 0 0
. When l is 7, the value of p from the formula is −2 25 26 Since Corey cannot use a negative number of stamps, the answer is 0 even though the value from the formula rounds up to −2. (c) Corey’s smartest choice is to use 2 letter stamps and 5 postcard stamps. This costs 2 · $.41 + 5 · $.26 = $.82 + $1.30 = $2.12. This only wastes 2¢ worth of extra postage. All the other choices waste at least 6¢ in extra postage. Maintain Your Skills 13. (a) 5x + 6y = 90 5x + 6y − 5x = 90 − 5x 6y = 90 − 5x (90 − 5x) y= 6
41l + 26p = 210 41(3) + 26p = 210 123 + 26p − 123 = 210 − 123 26p = 87 p=
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5x + 6y = 90
9 = 3 26
The fact that p is a fraction between 3 and 4 explains why 3 postcard stamps is not enough, and 4 are 9 needed. By rounding 3 26 to the next highest integer, the number of postcard stamps can be found by using the equation. 12. (a) To solve for p, first subtract 41l from both sides, then divide by 26. 41l + 26p = 210 41l + 26p − 41l = 210 − 41l 26p = 210 − 41l 26p 210 − 41l = 26 26 210 − 41l p= 26 Note that these are the same steps you just followed in the previous problem, except that the variable l is used instead of a number. Mathematics I Solutions Manual
5x + 6y − 6y = 90 − 6y 5x = 90 − 6y (90 − 6y) x= 5 (b) 11x + 13y = 150 11x + 13y − 11x = 150 − 11x 13y = 150 − 11x (150 − 11x) y= 13 11x + 13y = 150 11x + 13y − 13y = 150 − 13y 11x = 150 − 13y (150 − 13y) x= 11 • Chapter 2, page 67
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14. (a) 5x + 6y + 7z = 90 5x + 6y + 7z − 6y − 7z = 90 − 6y − 7z 5x = 90 − 6y − 7z (90 − 6y − 7z) x= 5
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3. (a) Let n be the number of new customers. So his total number of customers is 10 + n. If he charges $8 per week for each customer, then the total amount he makes from all his customers is 8 · (10 + n). The check is whether that expression is equal to the amount he wants to make, $96. So the equation is 8(10 + n) = 96
(b) 5x + 6y + 7z = 90
Now, solve for n.
5x + 6y + 7z − 5x − 7z = 90 − 5x − 7z 6y = 90 − 5x − 7z (90 − 5x − 7z) y= 6
8(10 + n) = 96 80 + 8n = 96 8n = 16 n=2
(c) 5x + 6y + 7z = 90 5x + 6y + 7z − 5x − 6y = 90 − 5x − 6y 7z = 90 − 5x − 6y (90 − 5x − 6y) z= 7
(b) Let x be how much more he needs to charge. So the total amount per customer would be 8 + x. Since he has 10 customers, the total he’ll make per week is (8 + x) · 10, or 10(8 + x). He wants to make $96 per week, so the equation would be
(d) All the answers follow a similar pattern. The left side of each equation is a fraction. The numerator of the fraction is the number from the right side of the equation minus the terms which you aren’t solving for. (So, if you’re solving for x, you get the number by subtracting the y term and the z term from the number on the right side of the equation.) The denominator is the coefficient of the variable you’re trying to solve for. 2D
MATHEMATICAL REFLECTIONS
10(8 + x) = 96 Now solve for x. 10(8 + x) + 96 80 + 10x = 96 10x = 16 x = 1.60 4. 2x + 3y = x − 4y + 10 2x − 2x + 3y = x − 2x − 4y + 10
1. Let x be the number. Then 2(x + 5) = 5x + 1 2x + 10 = 5x + 1 −3x + 10 = 1 −3x = −9
3y = −x − 4y + 10 3y + 4y = −x − 4y + 4y + 10 7y = −x + 10 −x + 10 y= 7
x=3 2. Guess 10. Six years ago, Bill would have been 10 − 6 = 4. 14 · 10 = 2.5 Compare the results 4 = 2.5. Try another guess. Apply the same steps to the variable a. Guess a. Six years ago, Bill would have been a − 6. 14 as old as he is now, a, would be 14 a. Are the two expressions equal? That gives the equation a − 6 = 14 a Now, solve the equation. a − 6 = 14 a a − 6 − 14 a = 0 a − 14 a = 6 4 3
·
3 a 4 3 a 4
=6
= 43 · 6 a=8
5. If you take a guess (it can be anything), and keep track of your steps, that will give you a recipe for checking any guess. Then, once you know the steps to take to check any guess, you can apply those same steps to a variable. That will result in the appropriate equation for the problem at hand. Finally, once you’ve found the appropriate equation, you solve it. 6. If you have an equation with x’s and y’s and you want to solve for y, you do it basically the same way you would solve for y in an equation with only one variable. Move all of the y terms to one side of the equation, and everything to the other side. Then, divide by the coefficient of y. This will give y = on one side of the equation and an expression involving numbers and x’s on the other side.
Mathematics I Solutions Manual
• Chapter 2, page 68
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7. Using the guess-check-generalize method, you can find that the appropriate equation is x + x4 = 560. x 4
= 560
5 x 4
= 560
x+
x=
· 560 x = 448 4 5
So, x = 448. Checking the answer, you see that 448 + 448 = 448 + 112 = 560. 4
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5. (a) Subtract 3. (b) Divide by 5. (c) This operation cannot be undone since there is more than one number that has the same sum of its digits (43 and 52, for example). (d) Multiply by 10. 6. (a) multiply x by itself, then add 15 (b) add 9 to y, then divide by 7 (c) multiply b by 8, then add 2, then multiply by 15 (d) multiply a by 7, then subtract 12, then add 4 7. (a) The steps (starting from x) are • Multiply by 2. • Add 4.
CHAPTER REVIEW
To backtrack, subtract 4 from 10 to get 6, then divide 6 by 2 to get 3. The answer is x = 3. (b) The steps (starting from x) are
1. Follow the steps to create an expression. • • • • •
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Pick a number: x Multiply by 3: 3x Add 4: 3x + 4 Multiply by 5: 5(3x + 4) Subtract 2: 5(3x + 4) − 2
• Multiply by 3. • Subtract 2.
The final answer is 5(3x + 4) − 2, or 15x + 18 2. (a) If x = 2, then 2(3x + 5) + x = 2(3 · 2 + 5) + 2 = 2(6 + 5) + 2 = 2(11) + 2 = 22 + 2 = 24 (b) If x = 5, then 2(3x + 5) + x = 2(3 · 5 + 5) + 5 = 2(15 + 5) + 5 = 2(20) + 5 = 40 + 5 = 45 (c) If x = −3, then 2(3x + 5) + x = 2(3 · (−3) + 5) + (−3) = 2((−9) + 5) + (−3) = 2(−4) + (−3) = −8 + (−3) = −11 (d) 2(3x + 5) + x = 6x + 10 + x
To backtrack, add 2 to 8 to get 10, then divide 10 by . The answer is x = 10 . 3 to get 10 3 3 (c) The steps (starting from x) are • Divide by 4. • Add 1. To backtrack, subtract 1 from 7 to get 6, then multiply 6 by 4 to get 24. The answer is x = 24. (d) The steps (starting from x) are • Add 2. • Divide by 5. To backtrack, multiply −1 by 5 to get −5 and then subtract 2 from −5 to get −7. The answer is x = −7. 8. To determine whether x = −3 is a solution to each equation, substitute −3 for x and evaluate. (a) 2(−3) + 7 = (−3) + 2 −6 + 7 = −3 + 2 1 = −1
= 7x + 10 If x = 2, then 7x + 10 = 7(2) + 10 = 24 3. (a) commutative property of addition (b) associative property of multiplication (c) additive identity (d) multiplicative identity (e) distributive property 4. (a) 3(x + 5) + 8 = 3x + 15 + 8 = 3x + 23 (b) 2(4x − 3) + 7(x + 2) = (8x − 6) + (7x + 14) = 15x + 8 (c) 3(2x − 8) − 4(x − 5) = (6x − 24) − (4x − 20) = 6x − 24 − 4x + 20 = 2x − 4 (d) 2(3x − 4) − (2x − 1) = (6x − 8) − (2x − 1) = 6x − 8 − 2x + 1 = 4x − 7 Mathematics I Solutions Manual
x = −3 is not a solution to the equation. (b) (−3) − 4 + 3(−3) = 8 − (−3) −3 − 4 − 9 = 8 + 3 −16 = 11 x = −3 is not a solution to the equation (c) 2((−3) − 1) = (−3) − 5 2(−4) = −3 − 5 −8 = −8 x = −3 is a solution to the equation. (d) ((−3) − 4) − (3 − (−3)) = 5(−3) + 2 (−7) − (6) = −15 + 2 −13 = −13 x = −3 is a solution to the equation. • Chapter 2, page 69
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9. (a) 2k + 7 = 3k + 4 (b) 3p = 36 10. (a)
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(b) 90 + 85 + 83 + 90 4 = 87
A=
3x + 2 = x − 12 3x + 2 − x = x − 12 − x 2x + 2 = −12 2x + 2 − 2 = −12 − 2 2x = −14 2x −14 = 2 2 x = −7
(c) For Sue to have an average of 90, her fourth test score must be the solution to the equation 90 = 90+85+83+x . Solve for x. 4 90 + 85 + 83 + x 4 90 + 85 + 83 + x 4 · 90 = 4 · 4 360 = 90 + 85 + 83 + x 360 = 258 + x 90 =
(b) x + 2(3x − 5) = 5x − 6 x + 6x − 10 = 5x − 6 7x − 10 = 5x − 6 7x − 10 − 7x = 5x − 6 − 7x −10 = −2x − 6 −10 + 6 = −2x − 6 + 6
360 − 258 = 258 + x − 258 102 = x So Sue needs to score a 102 on her next test to have an average of 90. Unless Sue’s teacher allows extra credit, this is not possible. 14. (a) 5x + 12y = 80 (b)
−4 = −2x −2x −4 = −2 −2 2=x x=2
5x + 12y = 80 5x + 12y − 5x = 80 − 5x 12y = 80 − 5x 12y 80 − 5x = 12 12 80 − 5x y= 12
(c) 2(x + 1) + 3(2x − 5) = 4(2x + 3) 2x + 2 + 6x − 15 = 8x + 12 8x − 13 = 8x + 12 8x − 13 − 8x = 8x − 12 − 8x −13 = 12 This equation has no solution. 11. (a) The left side simplifies to 7a + 9, so every real number is a solution. (b) The equation simplifies to 11 = 8, so it has no solutions. (c) The equation simplifies to 42 = 7, so it has no solutions. (d) 15q = q + 12, 14q = 12, q = 12 = 67 . It has one 14 solution. 12. (a) x = y8 (b) y + 3 = 8x, x = y +8 3 (c) y8 = x − 3, x = y8 + 3 (d) y = 8x − 24 + 7, y = 8x − 17, y + 17 = 8x, x = y +8 17 13. (a) Sue’s average is the sum of her test grades divided by the number of tests.
CHAPTER TEST Multiple Choice 1. The correct answer is D, 7. Test each number by replacing it for x in the equation. For each of the first three, one of the terms becomes zero, and the Zero Product Property says that anything multiplied by zero is zero. When x = 7, the expression is equal to 4 · 2 · 14 = 112. 2. The correct answer is A, −5(t + 4). Using the Distributive Property on each expression, the first expression gives −5t − 20 while the other three all give −5t + 20. Trying a number for t can also be used. The last three expressions give the same value for any choice of t, while the first is always different. 3. The correct answer is C. By the Distributive Property, the expression is equivalent to
90 + 85 + 83 A= 3 = 86 Mathematics I Solutions Manual
6x − 12 − 10x + 2 which simplifies to −4x − 10 Selecting a value for x would also work. A value for x in the original expression must give the same answer as the simplified version. Picking a simple number like x = 5 gives only one match among the answer choices. • Chapter 2, page 70
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4. The correct answer is C. You can use backtracking only when the variable appears exactly once. 5. The correct answer is B. The red arrows add up to 4t + 8, while the blue arrows add up to 3t + 12. 6. The correct answer is D, 8. Solve the equation by using the Distributive Property, then the basic moves. 3(b − 2) = 2(b + 1) 3b − 6 = 2b + 2 3b − 6 − 2b = 2b + 2 − 2b b−6=2 b−6+6=2+6 b=8
(a) (b) (c) (d)
4x − 15 − 2x = 2x − 5 − 2x 2x − 15 = −5 2x − 15 + 15 = −5 + 15 2x = 10 10 x= or 5 2 14. (a) For Coupon A, find the discount (15% of the original price), then subtract that from the original. The discount is 0.15x, so the price after discount is x − 0.15x, or 0.85x. For Coupon B, the pizza is $2 less than its original price, which is x − 2. (b) The equation uses the expressions from the first part of the problem. Since both coupons give the same price, they are equal: x − 0.15x = x − 2 0.85x = x − 2 0.85x − x = x − 2 − x −0.15x = −2 −0.15x −2 = −0.15 −0.15 −2 x= ≈ 13.33 −0.15
Open Response 8. 2[3(y + 7) − 8] = 2[3y + 21 − 8] = 2[3y + 13] = 6y + 26 9. The expanded expression is −12x − 8y + 24. Below is an expansion box that shows how to simplify the expression.
−4
3x −12x
2y −8y
−6 24
10. Answers will vary, but should revolve around the basic and derived rules. Specifically, the derived move of combining like terms allows you to say that (r + 2r + 3r + 4r) = 10r, making the expression 10r . Then, r multiplicative inverse allows you to write 10r as (10r)( 1r ) r (as long as r isn’t zero). Associativity and the property of the multiplicative inverse (that r · 1r = 1) completes the explanation. 11. (a) 5♣15 = 5(5 − 15) = 5(−10) = −50. (b) x♣3 = 5(x − 3), so the exercise asks for the solution to 5(x − 3) = 20. The solution is x = 7. (c) 3♣y = 5(3 − y), so the exercise asks for the solution to 5(3 − y) = 20. The solution is y = −1, since 3 − y must equal 4. (d) Try it with numbers. 5♣15 = −50, but 15♣5 = 50. They’re not the same, so it cannot be commutative.
You obtain 15 = 15. Yes, it is a solution. 12 = 12. Yes. 16 = 36. No. 6 = 11. No.
5(2x − 3) − 6x = .5(4x − 10) 10x − 15 − 6x = 2x − 5 4x − 15 = 2x − 5
2x + 3y − 2x = 5x + 6 − 2x
Answer choice A is incorrect since there is still an x term on the right side. It is not solved completely for x in terms of other variables.
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13. Solve using the distributive law and combining like terms, then using basic moves to gather terms:
2x + 3y = 5x + 6
3y − 6 = 3x 3y − 6 3x = 3 3 y−2=x
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12. Substitute the value 4 for x.
7. The correct answer is C, x = y − 2. Solve for x by gathering all terms with x on one side, and all other terms on the other.
3y = 3x + 6 3y − 6 = 3x + 6 − 6
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15. Answers will vary. Possible answers include 8z − 48, 2(4z − 24), z + z + z + z + z + z + z + z − 48, 4(z − 12) + 4z, and many others. 16. (a) Choose a number. Multiply by 3. Subtract 5. (b) Choose a number. Subtract 5. Multiply by 3. (c) Choose a number. Add 1. Multiply by 2. Add 6. (d) Choose a number. Multiply by 3. Add 6. Multiply by 2. Subtract 7. Multiply by 4. 17. (a) 10x − 5 − 9 = 10x − 14 (b) 12x + 4 + 2x = 14x + 4 (c) 6x − 15 + 10x + 16 = 16x + 1 (d) 3 + 8x − 10 = 8x − 7 (e) 2x − 6 − x − 4 = x − 10 (f) 5x − 6x + 18 = −x + 18 18. Let x be the number. Then you obtain, stepwise: x + 10, 4x + 40, 4x + 32, 2x + 16, 2x + 14, x + 7.
Mathematics I Solutions Manual
(a) To end up with x, you subtract 7. (b) Subtract the original number to end up with a constant (7). • Chapter 2, page 71
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19. (a) F = 95 (100) + 32 = 9(20) + 32 = 180 + 32 = 212◦ (b) Subtract 32 to get 66.6, then divide by 59 (multiply by 5 ) to obtain 37◦ C. 9 (c) F − 32 = 95 C, C = 59 (F − 32) (d) C = 59 (68 − 32) = 59 (36) = 5(4) = 20◦ 20. (a) She earned $6a for the bracelets and $12b for the necklaces, so 6a + 12b = 168. (b) 12b = 168 − 6a, b = 168 − 6a , b = 14 − 12 a 12 12
Challenge Problem 21. Using the zero property, (x − 3)(x − 7) = 0 has 3 and 7 as its only solutions.
Mathematics I Solutions Manual
• Chapter 2, page 72
