Algebra 1 Section 13.6

Definition A rational equation is an equation containing rational expressions.

Rational Equations As we learned before, we can multiply both sides of the equation by the LCD of all the terms and then solve the resulting equation.

Example 1 x – 7 7 3 x + 2 – = 1 21 ( )21 x – 7 7 21 3 x + 2 – 21 = 1 21 ( )21 x – 7 7 21 3 x + 2 – 21 = 21(1) 3(x – 7) – 7(x + 2) = 21

Example 1 3(x – 7) – 7(x + 2) = 21 3x – 21 – 7x – 14 = 21

Example 2 1 2x 5 12 + = - 3x LCD = 12x 1 2x 5 12 + = 3x - 12x 1 2x 12x + = - 3x LCD = 12x 1 2x 5 12 + = 3x - 12x 1 2x 12x 5 12 + 12x 1 3x - 12x =

Example 2 1 2x 12x 5 12 + 12x 1 3x - 12x = 6 + 5x = -4 5x = -10 x = -2 = -4 5x = -10 x = -2

Rational Equations When you multiply by a variable expression to clear the denominators, the new equation may introduce solutions for which the original equation is undefined.

Rational Equations You must check that your solution is not an excluded value for any of the expressions in the original rational equation. Any solution that does not check is extraneous.

Example 3 5 x + 1 x – 6 -2 = 5 x + 1 x – 6 -2 = 5(x – 6) = -2(x + 1)

This solution checks in the original equation. Example 3 5x – 30 = -2x – 2 7x – 30 = -2 7x = 28 x = 4 This solution checks in the original equation.

Since this is a proportion, cross-multiplying could be used. Example 3 5 x + 1 x – 6 -2 = Since this is a proportion, cross-multiplying could be used. 5(x – 6) = -2(x + 1)

Example 4 x – 12 x – 10 x2 – 10x 20 = 2 x – x – 12 x – 10 x(x – 10) 20 The LCD is x(x – 10).

Example 4 x – 12 x – 10 x(x – 10) 20 = 2 x – x(x – 12) = 2(x – 10) –

Example 4 x2 – 14x + 40 = 0 (x – 10)(x – 4) = 0 x = 10, 4 10 is an extraneous solution because it makes a denominator equal to zero. x = 4 is the only solution.

Example 5 3 + n 5 + n 3 5 = 6 7 The LCD is 7(5 + n). 3 + n 5 + n = 6 7

Example 5 7(3 + n) = 6(5 + n) 21 + 7n = 30 + 6n n = 9 The solution checks.

Homework: pp. 558-560