Mathematics » Rational Expressions and Equations » Solve Rational Equations

Solving Rational Equations

Solving 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 hadan 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 \(x\ne c\) next to the equation.

Example: How to Solve Equations with Rational Expressions

Solve: \(\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.\)

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.

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.

We always start by noting the values that would cause any denominators to be zero.

Example

Solve: \(1-\frac{5}{y}=-\frac{6}{{y}^{2}}.\)

Solution

 .
Note any value of the variable that would makeany denominator zero..
Find the least common denominator of all denominators inthe equation. The LCD is\({y}^{2}\). 
Clear the fractions by multiplying both sides ofthe equation by the LCD..
Distribute..
Multiply..
Solve the resulting equation. First writethe quadratic equation in standard form..
Factor..
Use the Zero Product Property..
Solve..
Check. 
We did not get 0 as an algebraic solution. 
. 

Example

Solve: \(\frac{5}{3u-2}=\frac{3}{2u}.\)

Solution

 .
Note any value of the variable that would make anydenominator zero..
Find the least common denominator of all denominatorsin the equation. The LCD is\(2u\left(3u-2\right)\). 
Clear the fractions by multiplying both sides of theequation by the LCD..
Remove common factors..
Simplify..
Multiply..
Solve the resulting equation..
We did not get 0 or \(\frac{2}{3}\) as algebraic solutions. 
. 

When one of the denominators is a quadratic, remember to factor it first to find the LCD.

Example

Solve: \(\frac{2}{p+2}+\frac{4}{p-2}=\frac{p-1}{{p}^{2}-4}.\)

Solution

 .
Note any value of the variable that would make anydenominator zero..
Find the least common denominator of all denominatorsin the equation. The LCD is\(\left(p+2\right)\left(p-2\right)\). 
Clear the fractions by multiplying both sides of theequation by the LCD..
Distribute..
Remove common factors..
Simplify..
Distribute..
Solve..
 .
 .
We did not get \(2\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}-2\) as algebraic solutions. 
. 

Example

Solve: \(\frac{4}{q-4}-\frac{3}{q-3}=1.\)

Solution

 .
Note any value of the variable that would make anydenominator zero..
Find the least common denominator of all denominatorsin the equation. The LCD is\(\left(q-4\right)\left(q-3\right)\). 
Clear the fractions by multiplying both sides of theequation by the LCD..
Distribute..
Remove common factors..
Simplify..
Simplify..
Combine like terms..
Solve. First write in standard form..
Factor..
Use the Zero Product Property..
We did not get 4 or 3 asalgebraic solutions. 
. 

Example

Solve: \(\frac{m+11}{{m}^{2}-5m+4}=\frac{5}{m-4}-\frac{3}{m-1}.\)

Solution

 .
Factor all the denominators, so we can note any value of the variable the would make any denominator zero..
Find the least common denominator of all denominatorsin the equation. The LCD is\(\left(m-4\right)\left(m-1\right)\). 
Clear the fractions..
Distribute..
Remove common factors..
Simplify..
Solve the resulting equation..
 .
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero.The algebraic solution is an extraneous solution. There is no solution to this equation. 

The equation we solved in the example above had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.

Example

Solve: \(\frac{n}{12}+\frac{n+3}{3n}=\frac{1}{n}.\)

Solution

 .
Note any value of the variable thatwould make any denominator zero..
Find the least common denominator of all denominatorsin the equation. The LCD is\(12n\). 
Clear the fractions by multiplying both sides of theequation by the LCD..
Distribute..
Remove common factors..
Simplify..
Solve the resulting equation..
 .
 .
 .
Check. 
\(n=0\) is an extraneous solution. 
. 

Example

Solve: \(\frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4.\)

Solution

 .
Factor all the denominators,so we can note any value ofthe variable that would makeany denominator zero..
Find the least common denominator.The LCD is\(\left(y-6\right)\left(y+6\right)\). 
Clear the fractions..
Simplify..
Simplify..
Solve the resulting equation..
 .
 .
 .
 .
Check. 
\(y=-6\) is an extraneous solution. 
. 

Example

Solve: \(\frac{x}{2x-2}-\frac{2}{3x+3}=\frac{5{x}^{2}-2x+9}{12{x}^{2}-12}.\)

Solution

 .
We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD..
Note any value of the variable that would make anydenominator zero..
Find the least common denominator.The LCD is \(12\left(x-1\right)\left(x+1\right)\) 
Clear the fractions..
Simplify..
Simplify..
Solve the resulting equation..
 .
 .
 .
Check. 
\(x=1\) and \(x=-1\) are extraneous solutions.
The equation has no solution.
 

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