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# Solving Rational Equations

## Equations whith Fractions

When an equation includes fractions, first eliminate all denominators by multiplying both sides of the equation by a common denominator, a number that can be divided (with no remainder) by each denominator in the equation. When an equation involves fractions with variable denominators, it is necessary to check all solutions in the original equation to be sure that no solution will lead to a zero denominator.

EXAMPLE

Solve each equation. Solution

The denominators are 10, 15, 20, and 5. Each of these numbers can be divided into 60, so 60 is a common denominator. Multiply both sides of the equation by 60 and use the distributive property. (If a common denominator cannot be found easily, all the denominators in the problem can be multiplied together to produce one. Add -9r and 8 to both sides. Check by substituting into the original equation. Solution

Begin by multiplying both sides of the equation by x to get 3 - 12x = 0. This equation could be solved by using the quadratic formula with a = -12, b = 0 and c = 3. Another method, which works well for the type of quadratic equation in which b = 0 is shown below.

3 - 12x = 0 Verify that there are two solutions, -1/2 and 1/2. Solution

Factor k + 2k as k(k + 2). The least common denominator for all the fractions is k(k + 2). Multiplying both sides by k(k + 2) gives the following.

2(k + 2) - 3k(k) = k

2k + 4 -3k = k Distributive property

-3k - k + 4 = 0 Add -k; rearrange terms

3k + k - 4 = 0 Multiply by -1

(3k - 4)(k + 1) = 0 Factor

3k - 4 = 0 or k + 1 = 0  Verify that the solutions are 4/3 and -1.

CAUTION

It is possible to get, as a solution of a rational equation, a number that makes one or more of the denominators in the original equation equal to zero. That number is not a solution, so it is necessary to check all potential solutions of rational equations. These introduced solutions are called extraneous solutions.

EXAMPLE

Solve Solution

The common denominator is x(x-3). Multiply both sides by x(x-3) and solve the resulting equation. Checking this potential solution by substitution in the original equation shows that 3 makes two denominators 0. Thus 3 cannot be a solution, so there is no solution for this equation.