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

Solve for x in the following equation.

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



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]

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

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]

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) }

which can be rewritten as

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

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)

\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]

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

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

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

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

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}

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.

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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Author: Nancy Marcus

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