Algebra - Absolute Value Equations

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Section 2.14 : Absolute Value Equations

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1. Solve the following equation.

\[\left| {4p - 7} \right| = 3\]

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Start Solution

There really isn’t all that much to this problem. All we need to do is use the formula we discussed in the notes for this section. Doing that gives,

\[4p - 7 = - 3\hspace{0.25in}{\mbox{or}}\hspace{0.25in}4p - 7 = 3\]

Do not make the common mistake of just turning every minus sign inside the absolute value bars into a plus sign. That is just not how these work. The only way for the value of the absolute value to be 3 is for the quantity inside to be either -3 or 3. In other words, we get rid of the absolute value bars by using the formula from the notes.

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At this point all we need to do is solve each of the linear equations we got in the previous step. Doing that gives,

\[\begin{align*}4p & = 4 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}4p & = 10\\ p & = 1 & \hspace{0.25in} & {\mbox{or}} & \hspace{0.25in}p & = \frac{{10}}{4} = \frac{5}{2}\end{align*}\]

The two solutions are then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{p = 1\,\,\,\,{\mbox{and }}p = \frac{5}{2}}}\) .