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.

Example 6:        

$$\frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4}=\displaystyle \frac{7x+1}{x-2}$$

Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq 3,$    the second fraction is valid if      $x\neq -4,\quad$and the third fraction is valid is    $x\neq 2.$    If x=3,-4,    or    2$\quad$ turn out to be the solutions, you must discard them as extraneous solutions.

The first step is always to rewrite the problem so that every denominator is factored. In this problem, the denominator does not need factoring.

$$\begin{eqnarray*}\displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4} &=&\displaystyle \frac{7x+1}{x-2} \end{eqnarray*}$$

Multiply both sides by the least common multiple         $\left( x-3\right) \left( x+4\right) \left( x-2\right) \qquad$(the smallest number that all the denominators will divide into evenly). This step will eliminate all the denominators in the equation. The resulting equation may be equivalent (same solutions as the original equation) or it may not be equivalent (extraneous solutions),

$$\begin{eqnarray*}\displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4} &=&... ...) \left( x-2\right) \left( \displaystyle \frac{7x+1}{x-2}\right) \end{eqnarray*}$$

which is equivalent to

$$\begin{eqnarray*}\frac{\left( x-3\right) \left( x+4\right) \left( x-2\right) }{1... ...eft( x-2\right) }{1}\left( \displaystyle \frac{7x+1}{x-2}\right) \end{eqnarray*}$$

which can be rewritten as

$$\begin{eqnarray*}\frac{\left( x-3\right) \left( x+4\right) \left( x-2\right) \le... ...ight) \left( x-2\right) \left( 7x+1\right) }{\left( x-2\right) } \end{eqnarray*}$$

which can be rewritten again as

$$\begin{eqnarray*}\frac{\left( x-3\right) }{\left( x-3\right) }\cdot \frac{\left(... ...frac{\left( x-3\right) \left( x+4\right) \left( 7x+1\right) }{1} \end{eqnarray*}$$

which can be rewritten yet again as

$$\begin{eqnarray*}1\cdot \left( x+4\right) \left( x-2\right) \left( 3x-1\right) +... ...right) &=&\left( x-3\right) \left( x+4\right) \left( 7x+1\right) \end{eqnarray*}$$

$$\begin{eqnarray*}\left( x+4\right) \left( 3x^{2}-7x+2\right) +7x\left( x^{2}-5x+... ...x\right) \left( 7x^{2}+29x+4\right) -3\left( 7x^{2}+29x+4\right) \end{eqnarray*}$$

$$\begin{eqnarray*}3x^{3}-7x^{2}+2x+12x^{2}-28x+8+7x^{3}-35x^{2}+42x &=&7x^{3}+29x... ...x^{3}+8x^{2}-83x-12 \\ && \\ && \\ 3x^{3}-38x^{2}+99x+20 &=&0 \end{eqnarray*}$$

At this point, let us try to solve using the Rational Zero Theorem. If you have forgotten the procedure, let's review it briefly.

• If there is a rational solution, it will come from the set of number represented by fractions of the type $$\frac{a}{b}$$ where a is a factor of 20 and b is a factor of 3.

• Using Synthetic Division, we find that one of the answers is 4 and we can now rewrite the equation
  
  $$\begin{eqnarray*}3x^{3}-38x^{2}+99x+20 &=&0 \\ && \\ &&as \\ && \\ \left( x-4\right) \left( 3x^{2}-26x-5\right) &=&0 \end{eqnarray*}$$
  
  

• Recall that the only way $\left( x-4\right) \left( 3x^{2}-26x-5\right) =0$ is if one of the factor equals zero. We know that x-4=0 because x=4 is a solution. For what values of x will 3x²-26x-5=0?
  
  $$\begin{eqnarray*}3x^{2}-26x-5 &=&0 \\ && \\ && \\ x &=&\frac{-\left( -26\righ... ...t( 3\right) } \\ && \\ && \\ x &=&\frac{26\pm \sqrt{736}}{6} \end{eqnarray*}$$
  
  

• The answers are $x=\displaystyle \frac{26+\sqrt{736}}{6},\displaystyle \frac{26-\sqrt{736}}{6} ,4\bigskip\bigskip .$

Let us check the three answers with the original equation.

Check the solution x=4 by substituting x=4 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: $\qquad \displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4}=\displaystyl... ...style \frac{ 11}{1}+\displaystyle \frac{28}{8}=\displaystyle \frac{116}{8}=14.5$

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{7x+1}{x-2}=\displaystyle \frac{7\left( 4\right) +1}{\left( 4\right) -2}=\displaystyle \frac{29}{2}=14.5$

  
  

Since the left side of the original equation equals the right side of the original equation after we substitute the value 4 for x, we have verified that x=4 is a solution.

Check the solution $x=\displaystyle \frac{26+\sqrt{736}}{6}\smallskip\approx 8.85488665542$ by substituting x=8.85488665542 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: $\qquad \displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4}=\displaystyl... ...ft( 8.85488665542\right) }{\left( 8.85488665542\right) +4}\approx 9.18821998875$

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{7x+1}{x-2}=\displaystyle \frac{7\left( 8.85488665542\right) +1}{ \left( 8.85488665542\right) -2}\approx 9.18821998875$

  
  

Since the left side of the original equation equals the right side of the original equation after we substitute the value 4 for x, we have verified that x=8.85488665542 is a solution.

Check the solution $x=\displaystyle \frac{26-\sqrt{736}}{6}\smallskip\approx -0.18821998875$ by substituting x=-0.18821998875 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: $\qquad \displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x}{x+4}=\displaystyl... ... -0.18821998875\right) }{\left( -0.18821998875\right) +4}\approx 0.145113344583$

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{7x+1}{x-2}=\displaystyle \frac{7\left( -0.18821998875\right) +1 }{\left( -0.18821998875\right) -2}\approx 0.145113344583$

  
  

Since the left side of the original equation equals the right side of the original equation after we substitute the value -0.18821998875 for x, we have verified that x=-0.18821998875 is a solution.

You can also check your answer by graphing $\quad \displaystyle \frac{3x-1}{x-3}+\displaystyle \frac{7x }{x+4}-\displaystyle \frac{7x+1}{x-2}\smallskip\smallskip .\quad$(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 three places: 4,-0.18821998875 and 8.85488665542. This also verifies our answers.

If you would like to test yourself by working some problems similar to this example, click on Problem

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

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