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### Section 2.14 : Absolute Value Equations

5. Solve the following equation.

$\left| {\frac{1}{2}z + 4} \right| = \left| {4z - 6} \right|$

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Hint : This problem works the same as all the others in this section do.
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This problem works identically to all the problems in this section. The only way the two absolute values can be equal is if the quantities inside them are the same value or the same value except for opposite signs. Doing this gives,

$\frac{1}{2}z + 4 = - \left( {4z - 6} \right) = 6 - 4z\hspace{0.25in}{\mbox{or}}\hspace{0.25in}\frac{1}{2}z + 4 = 4z - 6$

In other words, we can use the formula discussed in this section to do this problem!

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Now solving each these linear equations gives,

\begin{align*}\frac{1}{2}z + 4 & = 6 - 4z & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}\frac{1}{2}z + 4 & = 4z - 6\\ \frac{9}{2}z & = 2 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in} - \frac{7}{2}z & = - 10\\ z & = \frac{4}{9} & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}z & = \frac{{20}}{7}\end{align*}

Now, because both sides of the equation have absolute values, we know that regardless of the value of $$x$$ we plug into the original equation the absolute value will guarantee that the result will be positive and so we donâ€™t need to verify either of these solutions.

Therefore, the two solutions to the original equation are then : $$\require{bbox} \bbox[2pt,border:1px solid black]{{z = \frac{4}{9}\,\,\,\,\,{\mbox{and}}\,\,\,\,\,z = \frac{{20}}{7}}}$$ .