1. Essential Question: What must you do to find the solutions of a rational equation?

2.  Back when we solved radical (square root) equations, we had to check for extraneous solutions. We’ll have to do that again for rational equations, but only to make sure any denominators never equal 0.  There are two ways to solve rational equations. 1) With two rational equations, place them on opposite sides of an equal sign, cross multiply and solve. 2) When there are more than two rational equations, multiply all terms by the LCD.

3.  Solve   Cross multiply  Distribute on left & right  Set equation equal to 0  Divide all terms by 5  Factor  Solve each parenthesis Check for extraneous (5)(x 2 - 1) = (2x – 2)(15) 5x 2 – 5 = 30x – 30 5x 2 – 30x + 25 = 0 x 2 – 6x + 5 = 0 (x – 5)(x – 1) = 0 x = 1x = 5 Extraneous solution Would give a denominator of 0

4.  Solve  (-2)(x - 4) = (x 2 – 2)(2) -2x + 8 = 2x 2 – 4 0 = 2x 2 + 2x – 12 0 = x 2 + x – 6 0 = (x + 3)(x – 2) x = 2x = -3

5.  Solve  Find the LCD  Multiply all terms by LCD  Solve for x  5 – 4 = 5x 1 = 5x 1 / 5 = x 10x

6.  Solve  4(x + 1) – 3(x) = 1(x)(x + 1) 4x + 4 – 3x = x 2 + x x + 4 = x 2 + x LCD: x(x + 1) 0 = x 2 – 4 0 = (x + 2)(x – 2) x = 2x = -2

7.  Assignment  Page 514 – 515  Problems 1 – 21, odd problems  SHOW YOUR WORK!!