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.

Example 1: $$\frac{2x-1}{x+1}=\displaystyle \frac{2x}{x-1}+\displaystyle \frac{5}{x}$$

Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq -1,$ the second fraction is valid if $x\neq 1,\quad$and the third fraction is valid if$x\neq 0$.If $\pm 1$ $\quad$ $0\quad$ 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.

$$\begin{eqnarray*}\displaystyle \frac{2x-1}{x+1} &=&\displaystyle \frac{2x}{x-1}+\displaystyle \frac{5}{x} \end{eqnarray*}$$

Multiply both sides of the equation by the least common multiple $\left( x-1\right) \left( x+1\right) \left( x\right) \qquad$(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),

$$\begin{eqnarray*}\displaystyle \frac{2x-1}{x+1} &=&\displaystyle \frac{2x}{x-1}+\displaystyle \frac{5}{x} \end{eqnarray*}$$

$$\begin{eqnarray*}\left( x-1\right) \left( x+1\right) \left( x\right) \left( \dis... ...frac{2x}{x-1}+\displaystyle \frac{5}{x}\right) \\ && \\ && \\ \end{eqnarray*}$$

$$\begin{eqnarray*}\left( x-1\right) \left( x+1\right) \left( x\right) \left( \dis... ...1\right) \left( x\right) \left( \displaystyle \frac{5}{x}\right) \end{eqnarray*}$$

which is equivalent to

$$\begin{eqnarray*}\frac{\left( x-1\right) \left( x+1\right) \left( x\right) }{1}\... ...ght) \left( x\right) }{1}\left( \displaystyle \frac{5}{x}\right) \end{eqnarray*}$$

which can be rewritten as

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

which can be rewritten again as

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

which can be rewritten yet again as

$$\begin{eqnarray*}1\cdot \frac{\left( x-1\right) \left( x\right) \left( 2x-1\righ... ...ht) }{1} +1\cdot \frac{\left( x-1\right) \left( x+1\right) 5}{1} \end{eqnarray*}$$

$$\begin{eqnarray*}\left( x-1\right) \left( x\right) \left( 2x-1\right) &=&\left( ... ... x+1\right) +5\left[ \left( x-1\right) \left( x+1\right) \right] \end{eqnarray*}$$

$$\begin{eqnarray*}x\left( 2x^{2}-3x+1\right) &=&2x^{3}+2x^{2}+5\left( x^{2}-1\right) \\ && \\ && \\ 2x^{3}-3x^{2}+x &=&2x^{3}+2x^{2}+5x^{2}-5 \end{eqnarray*}$$

$$\begin{eqnarray*}-3x^{2}+x &=&7x^{2}-5 \\ && \\ && \\ 0 &=&10x^{2}-x-5 \end{eqnarray*}$$

Solve for x using the quadratic formula:

$$\begin{eqnarray*}0 &=&10x^{2}-x-5 \\ && \\ && \\ x &=&\frac{-\left( -1\right)... ...t( 10\right) } \\ && \\ && \\ x &=&\frac{1\pm \sqrt{201}}{20} \end{eqnarray*}$$

The exact answers are $x=\displaystyle \frac{1\pm \sqrt{201}}{20}$ and the approximate answers are x=0.758872343938 and x=-0.658872343938.

Check the answer in the original equation.

Check the solution $x=\displaystyle \frac{1+\sqrt{201}}{20}\smallskip$ by substituting x=0.758872343938 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: $\qquad \displaystyle \frac{2x-1}{x+1}$

  
  

  $$\begin{eqnarray*}&\approx &\displaystyle \frac{2\left( 0.758872343938\right) -1}... ...t( 0.758872343938\right) +1} \\ && \\ &\approx &0.294361719689 \end{eqnarray*}$$
  
  

  
  

• Right Side: $\qquad \displaystyle \frac{2x}{x-1}+\displaystyle \frac{5}{x}$

  
  

  $$\begin{eqnarray*}&\approx &\displaystyle \frac{2\left( 0.758872343938\right) }{\... ...eft( 0.758872343938\right) } \\ && \\ &\approx &0.294361719685 \end{eqnarray*}$$
  
  

  
  

It does not check exactly because we rounded the answer. However, it checks close enough to tell us that the answer $x=\displaystyle \frac{1+\sqrt{201}}{20}$ is a solution to the original problem.

Check the solution $x=\displaystyle \frac{1-\sqrt{201}}{20}\smallskip$ by substituting x=-0.658872343938 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: $\qquad \displaystyle \frac{2x-1}{x+1}$

$$\begin{eqnarray*}&\approx &\displaystyle \frac{2\left( -0.658872343938\right) -1... ...( -0.658872343938\right) +1} \\ && \\ &\approx &-6.79436171969 \end{eqnarray*}$$

Right Side: $\qquad \displaystyle \frac{2x}{x-1}+\displaystyle \frac{5}{x}$

$$\begin{eqnarray*}&\approx &\displaystyle \frac{2\left( -0.658872343938\right) }{... ...ft( -0.658872343938\right) } \\ && \\ &\approx &-6.7943617969. \end{eqnarray*}$$

Since the left side of the original equation equals the right side of the original equation after we substitute the value -0.658872343938 for x, we have verified that $x=\displaystyle \frac{1-\sqrt{201}}{20}$ is a solution.

You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{2x-1}{x+1}- \displaystyle \frac{2x}{x-1}-\displaystyle \frac{5}{x}.\smallskip\quad$(formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; the intercept(s) will be the real solution(s). Note that the graph crosses the x-axis in two places: -0.658872343938 and 0.758872343938.

If you would like to work another example, click on Example

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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