91Ó°ÊÓ

For any real numbers $a,b$, where $bâ‰\yen 0$, if $\left|a\right|=b$then $a=b$or $a=-b$.

Consider the equation $|x+11|=42$.

Use the concept of absolute value equation as follows:

$x+11=42$ or $x+11=-42$.

Case 1. Simplify $x+11=42$.

Subtract 42 from both sides of the equation and simplify as follows:

$\begin{array}{l}x+11=42\\ x+11−42=42−42=0\\ x−31=0\\ x=31\end{array}$

Case 2. Simplify $x+11=−42$.

Add 42 on both sides of the equation and simplify as follows:

$\begin{array}{l}x+11=−42\\ x+11+42=−42+42\\ x+53=0\\ x=−53\end{array}$

Therefore, the solution is $x=31,-53$.

Case 1. Substitute $x=31$in the equation $|x+11|=42$and simplify as follows:

$\begin{array}{l}|x+11|=42\\ |31+11|=42\\ \left|42\right|=42\\ 42=42\end{array}$

Case 2. Substitute $x=-53$ in the equation $|x+11|=42$and simplify as follows:

$\begin{array}{l}|x+11|=42\\ |−53+11|=42\\ |−42|=42\\ 42=42\end{array}$

Since the left-hand side of the equation is equal to the right-hand side of the equation in both the cases therefore, the solution is $x=31,-53$.