91Ó°ÊÓ

Absolute value describes the distance from zero that a number is on the number line, without considering direction. The absolute value of a number is never negative.

Consider the expression $3|p-5|=2p$.

Divide both sides of the equation by 3 as follows:

$\begin{array}{l}3|p−5|=2p\\ |p−5|=\frac{2}{3}p\end{array}$

Apply the absolute rule as follows:

Case 1. When $p-5=-\frac{2}{3}p$.

$\begin{array}{l}p−5=−\frac{2}{3}p\\ p+\frac{2}{3}p−5=0\\ \frac{3p+2p}{3}−5=0\\ \frac{5p}{3}−5=0\\ \frac{5p}{3}=5\\ 5p=15\\ p=3\end{array}$

Case 2. When $p-5=\frac{2}{3}p$

$\begin{array}{l}p−5=\frac{2}{3}p\\ p−\frac{2}{3}p−5=0\\ \frac{3p−2p}{3}−5=0\\ \frac{p}{3}−5=0\\ \frac{p}{3}=5\\ p=15\end{array}$

Therefore, the value is $p=3,15$.

Substitute $p=3,15$in the equation $3|p-5|=2p$and simplify as follows:

$\begin{array}{l}3|p−5|=2p\\ 3|3−5|=2\left(3\right)\\ 3|−2|=6\\ 3\left|2\right|=6\\ 6=6\end{array}$

$\begin{array}{l}3|p−5|=2p\\ 3|15−5|=2\left(15\right)\\ 3\left|10\right|=30\\ 3\left(10\right)=30\\ 30=30\end{array}$

Since the left-hand side and right-hand side is equal in both the cases thus the values are $p=3,15$.