EQUATIONS INVOLVING FRACTIONS (RATIONAL EQUATIONS)

Note:

• A rational equation is an equation where at least one denominator contains a variable.

• When a denominator contains a variable, there is usually a restriction on the domain. The variable cannot take on any number that would cause any denominator to be zero.

• The first step is solving a rational equation is to convert the equation to an equation without denominators. This new equation may be equivalent (same solutions as the original equation) or it may not be equivalent (extraneous solutions).

• The next step is to set the equation equal to zero and solve.

• Remember that you are trying to isolate the variable.

• Depending on the problem, there are several methods available to help you solve the problem.

If you would like an in-depth review of fractions, click on Fractions.

Solve for x in the following equation.

Problem 5.4 a:        

$$\frac{3x-7}{x^{2}-7x+6}-\displaystyle \frac{5x-5}{ x^{2}+4x-5}=\displaystyle \frac{x+20}{x^{2}-x-30}$$

Answer:     $x=3,\quad 5\bigskip\bigskip$

Rewrite the equation such that all the denominators are factored:

$$\begin{eqnarray*}&& \\ \frac{3x-7}{\left( x-1\right) \left( x-6\right) }-\frac{... ...t) } &=&\frac{x+20}{\left( x-6\right) \left( x+5\right) } \\ && \end{eqnarray*}$$

Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq 1,\ 6,$    the second fraction is valid if      $x\neq -5,\ 1,\quad$and the third fraction is valid is     $x\neq 6,\ -5$.    If 1, 6 or -5 turn out to be the solutions, you must discard them as extraneous solutions.

Multiply both sides of the equation by an expression that represents the lowest common denominator. The expression $\left( x-1\right) \left( x-6\right) \left( x+5\right)$ is the smallest expression because it is the smallest expression that is divisible by all three denominators.

$$\begin{eqnarray*}&& \\ \left( x-6\right) \left( x-1\right) \left( x+5\right) \c... ...x+20}{\left( x-6\right) \left( x+5\right) }\right) && \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ && \\ \left( x-6\right) \left( x-1\right) \left( x+5\ri... ...t( x-6\right) \left( x+5\right) }\right) && \\ && \\ && \\ && \end{eqnarray*}$$

This equation can be written as

$$\begin{eqnarray*}&& \\ &&\frac{\left( x-6\right) \left( x-1\right) \left( x+5\r... ...left( x-6\right) \left( x+5\right) }\right) \\ && \\ && \\ && \end{eqnarray*}$$

Multiply the fractions where indicated.

$$\begin{eqnarray*}&& \\ &&\frac{\left( x-6\right) \left( x-1\right) \left( x+5\r... ...ht) }{\left( x-6\right) \left( x+5\right) } \\ && \\ && \\ && \end{eqnarray*}$$

Rearrange the factors in the numerators and rewrite the equations as

$$\begin{eqnarray*}&& \\ &&\frac{\left( x-1\right) \left( x-6\right) \left( x+5\r... ...ht) }{\left( x-6\right) \left( x+5\right) } \\ && \\ && \\ && \end{eqnarray*}$$

Rewrite the equation once again as

$$\begin{eqnarray*}&& \\ \frac{\left( x-1\right) \left( x-6\right) }{\left( x-1\r... ...ot \frac{\left( x-1\right) \left( x+20\right) }{1} && \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ && \\ 1\cdot \frac{\left( x+5\right) \left( 3x-7\right)... ...1}=\frac{\left( x-1\right) \left( x+20\right) }{1} && \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ \left( x+5\right) \left( 3x-7\right) -\left( x-6\right) ... ...t) =\left( x-1\right) \left( x+20\right) && \\ && \\ && \\ && \end{eqnarray*}$$

Simplify the last equation and solve for x.

$$\begin{eqnarray*}&& \\ \left( x+5\right) \left( 3x-7\right) -\left( x-6\right) ... ...=&\left[ \left( x-1\right) \left( x+20\right) \right] \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ \left[ 3x^{2}+8x-35\right] -\left[ 5x^{2}-35x+30\right] ... ...\\ && \\ 3x^{2}+8x-35-5x^{2}+35x-30 &=&x^{2}+19x-20 \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ -2x^{2}+43x-65 &=&x^{2}+19x-20 \\ && \\ && \\ -3x^{2}... ...\ \left( x-3\right) \left( x-5\right) &=&0 \\ && \\ && \\ && \end{eqnarray*}$$

$\left( x-3\right) \left( x-5\right) =0$ when x-3=0 and/or x=5=0.

$$\begin{eqnarray*}&& \\ If\quad x-3 &=&0,x=3 \\ && \\ If\quad x-5 &=&0,x=5 \\ && \\ && \\ && \end{eqnarray*}$$

There are two real solutions:    x=3,x=5

Check the two answers in the original equation.

Check the solution x=3 by substituting 3 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.        

• Left Side: $\qquad \displaystyle \frac{3x-7}{x^{2}-7x+6}-\displaystyle \frac{5x-5}{x^{2}+4x-5}$
  
  $$\begin{eqnarray*}&& \\ &=&\displaystyle \frac{3\left( 3\right) -7}{\left( 3\rig... ...eft( 3\right) ^{2}+4\left( 3\right) -5} =-0.9583333333333 \\ && \end{eqnarray*}$$
  
  

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{x+20}{x^{2}-x-30}=\displaystyle \frac{\left( 3\right) +20}{ \left( 3\right) ^{2}-\left( 3\right) -30}=-0.9583333333$.

  
  

Since the left side of the original equation is not equal to the right side of the original equation after we substitute the value 3 for x, then x=3is a solution.

Check the solution x=5 by substituting 5 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.        

• Left Side: $\qquad \displaystyle \frac{3x-7}{x^{2}-7x+6}-\displaystyle \frac{5x-5}{x^{2}+4x... ...tyle \frac{ 5\left( 5\right) -5}{\left( 5\right) ^{2}+4\left( 5\right) -5}=-2.5$

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{x+20}{x^{2}-x-30}=\displaystyle \frac{\left( 5\right) +20}{ \left( 5\right) ^{2}-\left( 5\right) -30}=-2.5$.

  
  

Since the left side of the original equation is not equal to the right side of the original equation after we substitute the value 5 for x, then x=5is a solution.

You can also check your answer by graphing

$$\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{3x-7}{x^{2}-7x+6}-\displayst... ...{5x-5}{x^{2}+4x-5}-\displaystyle \frac{x+20}{ x^{2}-x-30} \\ && \end{eqnarray*}$$

$.\quad$(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis in two places: 3and 5.

This means that there are two real solutions and the solutions are x=3,5.

If you would like to review the solution to problem 5.4b, click on solution.

If you would like to go back to the equation table of contents, click on contents.

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