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 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 5.3d: $$\frac{3x+7}{8x-1}=\displaystyle \frac{3x-1}{7x+6}+\displaystyle \frac{ 2x+2}{15x+5}$$

Answer:x=7

Solution:

Recall that you cannot divide by zero. Therefore, the first fraction is valid if , $\quad x\neq \displaystyle \frac{1}{8},$ the second fraction is valid if $x\neq -\displaystyle \frac{6}{7},\quad$ and the third fraction is valid is $x\neq -\displaystyle \frac{1}{3}$ If $\quad \displaystyle \frac{1}{8},$ $-\displaystyle \frac{6}{7},$, or $-\displaystyle \frac{1}{3}\quad$ turn out to be the solutions, you must discard them as extraneous solutions.

Multiply both sides by the least common multiple $\left( 8x-1\right) \left( 7x+6\right) \left( 15x+5\right) \qquad$ (the smallest number that all the denominators will divide into evenly). This step will eliminate all the denominators.

$$\begin{eqnarray*}\displaystyle \frac{3x+7}{8x-1} &=&\displaystyle \frac{3x-1}{7x... ...ystyle \frac{3x-1}{7x+6}+\displaystyle \frac{2x+2}{15x+5}\right] \end{eqnarray*}$$

$$\begin{eqnarray*}\left( 8x-1\right) \left( 7x+6\right) \left( 15x+5\right) \left... ...ft( 15x+5\right) \left[ \displaystyle \frac{ 2x+2}{15x+5}\right] \end{eqnarray*}$$

which is equivalent to

$$\begin{eqnarray*}\displaystyle \frac{\left( 8x-1\right) \left( 7x+6\right) \left... ...15x+5\right) }{1} \left[ \displaystyle \frac{2x+2}{15x+5}\right] \end{eqnarray*}$$

which can be rewritten as

$$\begin{eqnarray*}\displaystyle \frac{\left( 8x-1\right) \left( 7x+6\right) \left... ...) \left( 15x+5\right) \left( 2x+2\right) }{\left( 15x+5\right) } \end{eqnarray*}$$

which can be rewritten as

$$\begin{eqnarray*}\displaystyle \frac{\left( 8x-1\right) }{\left( 8x-1\right) }\c... ...ac{\left( 8x-1\right) \left( 7x+6\right) \left( 2x+2\right) }{1} \end{eqnarray*}$$

which can be simplified to

$$\begin{eqnarray*}1\cdot \displaystyle \frac{\left( 7x+6\right) \left( 15x+5\righ... ...ight) +\left( 8x-1\right) \left( 7x+6\right) \left( 2x+2\right) \end{eqnarray*}$$

$$\begin{eqnarray*}\left( 7x+6\right) \left[ \left( 15x+5\right) \left( 3x+7\right... ...ight] -1\left[ 45x^{2}-5\right] +8x\left[ 14x^{2}+26x+12\right] \end{eqnarray*}$$

$$\begin{eqnarray*}&&-1\left[ 14x^{2}+26x+12\right] \\ && \\ && \\ 315x^{3}+840... ...40x-45x^{2}+5+112x^{3}+208x^{2}+96x \\ && \\ &&-14x^{2}-26x-12 \end{eqnarray*}$$

$$\begin{eqnarray*}315x^{3}+1110x^{2}+965x+210 &=&472x^{3}+149x^{2}+30x-7 \\ && \\ && \\ -157x^{3}+961x^{2}+935x+217 &=&0 \end{eqnarray*}$$

$$\begin{eqnarray*}157x^{3}-961x^{2}-935x-217 &=&0 \\ && \\ && \\ \left( x-7\right) \left( 157x^{2}+138x+31\right) &=&0 \end{eqnarray*}$$

The only way the product of two expressions equals zero if at least one of the expressions is zero.

$$\begin{eqnarray*}If x-7 &=&0,\quad then\quad x=7 \end{eqnarray*}$$

To determine what values of x will cause the expression $\left( 157x^{2}+138x+31\right)$ to equal zero, we use the Quadratic Formula

$$\begin{eqnarray*}157x^{2}+138x+31 &=&0 \\ && \\ && \\ x &=&\displaystyle \fra... ...&& \\ && \\ x &=&\displaystyle \frac{-138\pm \sqrt{-424}}{314} \end{eqnarray*}$$

Since you cannot compute the square root of a negative number in the real number system, these two solutions are not real.

The only real solution is $x=7.\bigskip\bigskip\medskip$

Check the solution x=7 by substituting 7 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: $\qquad \displaystyle \frac{3x+7}{8x-1}=\displaystyle \frac{3\left( 7\right) +7}{8\left( 7\right) -1}=\displaystyle \frac{28}{55}\approx 0.509091$

  
  

  
  

• Right Side: $\qquad \displaystyle \frac{3x-1}{7x+6}+\displaystyle \frac{2x+2}{15x+5}=\displa... ...e \frac{16}{110}= \displaystyle \frac{56}{110}\approx 0.509091\bigskip\bigskip$

Since the left side of the equation equals the right side of the equation after the substitution, we have verified that x=7, 7 is a solution.

You can also check your answer by graphing $\quad f(x)=\displaystyle \frac{3x+7}{8x-1}- \displaystyle \frac{3x-1}{7x+6}-\displaystyle \frac{2x+2}{15x+5}\bf {\bigskip\bigskip }.\quad$(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 at one place: x=7.

This verifies our solution graphically.

We have verified our solution both algebraically and graphically.

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