# How do you solve 3|p-5|=2p?

Jul 18, 2017

See a solution process below:

#### Explanation:

First, divide each side of the equation by $\textcolor{red}{3}$ to isolate the absolute value function while keeping the equation balanced:

$\frac{3 \left\mid p - 5 \right\mid}{\textcolor{red}{3}} = \frac{2 p}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left\mid p - 5 \right\mid}{\cancel{\textcolor{red}{3}}} = \frac{2 p}{3}$

$\left\mid p - 5 \right\mid = \frac{2 p}{3}$

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1

$p - 5 = - \frac{2 p}{3}$

$- \textcolor{red}{p} + p - 5 = - \textcolor{red}{p} - \frac{2 p}{3}$

$0 - 5 = - \left(\frac{3}{3} \times \textcolor{red}{p}\right) - \frac{2 p}{3}$

$- 5 = - \frac{3 p}{3} - \frac{2 p}{3}$

$- 5 = \frac{- 5 p}{3}$

$- 5 \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 5}} = \frac{- 5 p}{3} \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 5}}$

$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 5}}} \times \frac{\textcolor{red}{3}}{\cancel{\textcolor{b l u e}{- 5}}} = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 5}}} p}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \times \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{- 5}}}$

$3 = p$

$p = 3$

Solution 2

$p - 5 = \frac{2 p}{3}$

$- \textcolor{red}{p} + p - 5 = - \textcolor{red}{p} + \frac{2 p}{3}$

$0 - 5 = - \left(\frac{3}{3} \times \textcolor{red}{p}\right) + \frac{2 p}{3}$

$- 5 = - \frac{3 p}{3} + \frac{2 p}{3}$

$- 5 = \frac{- 1 p}{3}$

$- 5 \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 1}} = \frac{- 1 p}{3} \times \frac{\textcolor{red}{3}}{\textcolor{b l u e}{- 1}}$

$\frac{- 15}{-} 1 = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 1}}} p}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \times \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{- 1}}}$

$15 = p$

$p = 15$

The Solutions Are: $p = 3$ and $p = 15$