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

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.

Solve

Cross multiply Distribute on left & right Set equation equal to 0 Divide all terms by 5 Factor Solve each parenthesis

Check for extraneous

2

5 15

2 2 1x x

2

15

2 2

5

1x x

(5)(x2 - 1) = (2x – 2)(15)5x2 – 5 = 30x – 305x2 – 30x + 25 = 0x2 – 6x + 5 = 0

(x – 5)(x – 1) = 0

x = 1

x = 5 Extraneous solution

Would give a denominator of 0

Solve

2

2 2

2 4x x

2

2

2

2

4x x

(-2)(x - 4) = (x2 – 2)(2)-2x + 8 = 2x2 – 4

0 = 2x2 + 2x – 120 = x2 + x – 6

0 = (x + 3)(x – 2)

x = 2

x = -3

Solve Find the LCD Multiply all terms

by LCD

Solve for x

1 2 1

2 5 2x x

5 – 4 = 5x

1 = 5x

1/5 = x

10x1 2 1

2 5 210 10 10

x xx x x

Solve

4 31

1x x

4(x + 1) – 3(x) = 1(x)(x + 1)4x + 4 – 3x = x2 + xx + 4 = x2 + x

LCD: x(x + 1)

( 1)4 3

1( 1) ( 1)1

x x xx

x x xx

0 = x2 – 4

0 = (x + 2)(x – 2)

x = 2

x = -2

AssignmentPage 514 – 515Problems 1 – 21, odd problemsSHOW YOUR WORK!!