Solving Rational Equations

Learning Objective

Introduction

Equations that contain rational expressionsA fraction with a polynomial in the numerator and/or denominator. are called rational equationsAn equation that contains one or more rational expressions.. We can solve these equations using the techniques for performing operations with rational expressions and for solving algebraic equations.

Solving Rational Equations Using Common Denominators

One method for solving rational equations is to rewrite the rational expressions in terms of a common denominator. Then, since we know the numerators are equal, we can solve for the variable. To illustrate this, let’s look at a very simple equation:

`3/4=x/4`

`x=3`

Since the denominator of each expression is the same, the numerators must be equivalent as well. This means that `x=3`.

This is true for rational equations with polynomials too:

`(2x-5)/(x-4)=11/(x-4)`

`2x - 5 = 11`

`x=8`

Again, since the denominators are the same, we know the numerators must also be equal. So we can set them equal to one another and solve for `x`.

We should check our solution in the original rational expression:

`(2x-5)/(x-4)=11/(x-4)`

`(2(8)-5)/(8-4)=11/(8-4)`

`11/4=11/4`

The solution checks, and since `x = 8` does not result in division by `0`, the solution is valid.

When the terms in a rational equation have unlike denominators, solving the equation will involve some extra work. Here’s an example:

Example

Problem

Solve the equation: `(x+2)/8=3/4`

 

 

 

There are no excluded values because the denominators are both constants.

 

`(x+2)/8=3/4*2/2`

 

 

`(x+2)/8=6/8`

 

Find a common denominator and rewrite each expression with that denominator.

The common denominator is `8`.

 

`x + 2 = 6`

 

`x = 4`

 

Since the denominators are the same, the numerators must be equal for the equation to be true. Solve for `x`.

 

`(4+2)/8=3/4`

 

`6/8=3/4`

 

`3/4=3/4`

 

Check the solution by substituting `4` for `x` in the original equation.

 

 

 

 

Answer

`x = 4`

 

 

         

Another way of solving rational equations is to multiply both sides of the equation by the common denominator. This eliminates the denominators and turns the rational equation into a polynomial equation. Here is the same equation we just solved:

Example

Problem

Solve the equation: `(x+2)/8=3/4`

 

 

 

There are no excluded values because the denominators are both constants.

 

`(x+2)/8*8=3/4*8`

 

Multiply both sides by the least common denominator.

 

`x+2=24/4`

`x + 2 = 6`

 

Simplify.

 

 

 

`x + 2 - 2 = 6 - 2`

`x = 4`

 

Solve for `x`.

Answer

`x = 4`

 

 

         

Now that we understand the techniques, let’s look at an example that has variables in the denominator too. Remember that whenever there are variables in the denominator, we need to find any values that are excluded from the domainThe set of all possible inputs of a function which allow the function to work. because they'd make the denominator zero.

To solve this equation, we can multiply both sides by the least common denominator:

Example

Problem

Solve: `7/(x+2)+5/(x-2)=(10x-2)/(x^2-4)`

 

 

 

`x + 2 = 0`

`x = -2`

 

`x - 2 = 0`

`x = 2`

 

`(x + 2)(x - 2) = 0`

`x = -2, 2`

 

First determine the excluded valuesA value for a variable that is not allowed in an expression, such as a variable in a rational expression that would make the denominator equal zero.. These are the values of `x` that result in a `0` denominator.

 

 

 

denominators:

`x + 2`

`x - 2`

`x^2 - 4 = (x - 2)(x + 2)`

 

least common denominator:

`(x - 2)(x + 2)`

 

 

 

 

Find the common denominator of `x-2``x + 2`, and `x^2 - 4`.

 

Since `(x - 2)` and `(x + 2)` are both factors of `x^2 - 4`, the least common denominator is `(x - 2)(x + 2)` or `x^2 - 4`.

`(x+2)(x-2)(7/(x+2)+5/(x-2))=(10x-2)/(x^2-4)*(x+2)(x-2)`

 

`(7(x+2)(x-2))/(x+2)+(5(x+2)(x-2))/(x-2)=(10x-2)/(x^2-4)*(x^2-4)`

 

Multiply both sides of the equation by the common denominator.

 

 

`7(x-2)+5(x+2)=10x-2`

`7x - 14 + 5x + 10=10x - 2`

`12x - 4=10x - 2`

 

Simplify.

 

 

 

`12x - 10x - 4 = 10x - 10x - 2`

`2x - 4 = -2`

`2x - 4 + 4 = -2 + 4`

`2x = 2`

`x = 1`

 

 

 

 

Solve for `x`.

 

 

 

Check to be sure that the solution is not an excluded value. (It is not.)

 

`7/(x+2)+5/(x-2)=(10x-2)/(x^2-4)`

 

`7/(1+2)+5/(1-2)=(10*1-2)/(1^2-4)`

 

`7/3+5/(-1)=(10-2)/(1-4)`

 

`7/3-15/3=8/-3`

 

`-8/3=-8/3`

 

 

Check the solution in the original equation.

 

 

 

 

 

 

 

 

 

Answer

`x = 1`

 

 

 

Solve the equation `4/m=3/(m-2), m!=0" or "2`.  

 

A) `m = 2`

 

B) no solution

 

C) `m = 8`

 

We've seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don't work in the original form of the equation. These types of answers are called extraneous solutionsA solution that results from solving an equation that is not a valid solution in the original equation.. These solutions are artifacts of the solving process and not real answers at all. That's why we should always check solutions in the original equations—we may find that they yield untrue statements or produce undefined expressions.

Solve the equation:

 

`1/(x-6)+x/(x-2)=4/(x^2-8x+12)`

 

A) `x = -1`

 

B) `x = -1`, `6`

 

C) `x = -4`, `3`

 

D) no solution

 

Summary

We solve rational equations by finding a common denominator. We can then follow either of two methods. We can rewrite the equation so that all terms have the common denominator and we can solve for the variable with just the numerators. Or we can multiply both sides of the equation by the common denominator so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to `0`.