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8.6 Solve rational equations

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By the end of this section, you will be able to:
  • Solve rational equations
  • Solve a rational equation for a specific variable

Before you get started, take this readiness quiz.

If you miss a problem, go back to the section listed and review the material.

  1. Solve: 1 6 x + 1 2 = 1 3 .
    If you missed this problem, review [link] .
  2. Solve: n 2 − 5 n − 36 = 0 .
    If you missed this problem, review [link] .
  3. Solve for y in terms of x : 5 x + 2 y = 10 for y .
    If you missed this problem, review [link] .

After defining the terms expression and equation early in Foundations , we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations . We have simplified many rational expressions so far in this chapter. Now we will solve rational equations.

The definition of a rational equation is similar to the definition of equation we used in Foundations .

Rational equation

A rational equation    is two rational expressions connected by an equal sign.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

Rational Expression Rational Equation 1 8 x + 1 2 1 8 x + 1 2 = 1 4 y + 6 y 2 − 36 y + 6 y 2 − 36 = y + 1 1 n − 3 + 1 n + 4 1 n − 3 + 1 n + 4 = 15 n 2 + n − 12

Solve rational equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

Here is an example we did when we worked with linear equations:

. .
We multiplied both sides by the LCD. .
Then we distributed. .
We simplified—and then we had an equation with no fractions. .
Finally, we solved that equation. .
.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then we will have an equation that does not contain rational expressions and thus is much easier for us to solve.

But because the original equation may have a variable in a denominator we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation    we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution .

Extraneous solution to a rational equation

An extraneous solution to a rational equation    is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, c , by writing x ≠ c next to the equation.

How to solve equations with rational expressions

Solve: 1 x + 1 3 = 5 6 .

Solution

The above image has 3 columns. It shows the steps to find an extraneous solution to a rational equation for the example 1 divided by x plus one-third equals five-sixths. Step one is to note any value of the variable that would make any denominator zero. If x equals 0, then I divided by x is undefined. So we’ll write x divided zero next to the equation to get 1 divided by x plus one-third equals five-sixths times x divided by zero. Step two is to find the least common denominator of all denominators in the equation. Find the LCD of 1 divided by x one-third, and five-sixths. The x is 6 x. Step three is to clear the fractions by multiplying both sides of the equation by the LCD. Multiply both sides of the equation by the LCD, 6 x to get 6 times 1 divided by x plus one-third equals 6 x times five-sixths. Use the Distributive Property to get 6 x times 1 divided by x plus 6 x times one-third equals 6 x times five-sixths. Simplify – and notice, no more fractions and we have 6 plus 2 x equals 5 x. Step 4 is to solve the resulting equation. Simplify to get 6 equals 3 x and 2 equals x. Step 5 is to check. If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation. We did not get 0 as an algebraic solution. We substitute x equals 2 into the original equation to get one-half plus one-third equals five-sixths, then three-sixths plus two-sixths equals five-sixths and finally, five-sixths equal five-sixths.
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Solve: 1 y + 2 3 = 1 5 .

− 15 7

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Solve: 2 3 + 1 5 = 1 x .

15 13

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The steps of this method are shown below.

Solve equations with rational expressions.

  1. Note any value of the variable that would make any denominator zero.
  2. Find the least common denominator of all denominators in the equation.
  3. Clear the fractions by multiplying both sides of the equation by the LCD.
  4. Solve the resulting equation.
  5. Check.
    • If any values found in Step 1 are algebraic solutions, discard them.
    • Check any remaining solutions in the original equation.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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