2.7 Solving Equations in One Variable

2.7 Solving Equations in One Variable Video Tutorial

http://bit.ly/tfJSRM

Equations involving fractions are rational equations and may be written in the form: f ( x) 0 g ( x) Solutions -- When f ( x) and g ( x) are polynomial functions with no common factors, then solutions of the equation are the zeros of f ( x) . Extraneous Solutions -- When we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions. For this reason we must check each solution of the resulting equation in the original equation. I. Clearing Fractions Solve the equation x 

3 4 x

Confirm numerically by substituting the solutions for the variable in the original equation.

Now practice with this equation…. x  2 

15 x

Confirm numerically by substituting the solutions for the variable in the original equation.

II. Solving a Rational Equation 1 Solve the equation x 0 x4

Confirm Graphically

III. !!!Eliminating Extraneous Solutions!!! Solve the following equation…

2x 1 2   2 x 1 x  3 x  4x  3

Check numerically by replacing x with each of the solutions.

Now try this one on your own… x 3 3 6   2 0 x x  2 x  2x

Next take a run at this one…

2

3 12  2 x  4 x  4x

IV. Application How much pure acid must be added to 50mL of a 35% acid solution to produce a mixture that is 75% acid.

x = mL of 100% acid added. x + 50 = mL of final solution

Now practice with this one…

To produce 500mL of a 25% acid solution, how much (x) of a 40% acid solution is required when mixed with a 20% solution?

x = mL of 40% acid solution. 500 – x = mL of 20% acid solution.

Now use your math tools for this one… You need a 200 yd2 pasture for your horse and want to limit fencing costs by minimizing the perimeter. What are the dimensions of the pasture?