91Ó°ÊÓ

To evaluate any equation involving more than one operation, follow the below order operation

• First assume the case one in which the equation be a=b.
• Secondly assume the case two in which the equation be a=-b.
• Solve the equations of both the cases to find the value of algebraic expression

After finding the two values of algebraic expression, put the value in the original equation and check the solution

Case1-$a=b$

$\begin{array}{c}−12(9x+1)=144\\ −108x−12=144\\ −108x=144+12\\ x=−\frac{156}{108}\\ =−1.44\end{array}$

Case2-$a=−b$

$\begin{array}{c}−12(9x+1)=−144\\ −108x−12=−144\\ −108x=−144+12\\ x=\frac{132}{108}\\ =1.22\end{array}$

Put the value of x=1.22 in the equation $-12\left|9x+1\right|=144$

$\begin{array}{c}−12|9x+1|=−12|9×\frac{132}{108}+1|\\ =−12|9×1.22+1|\\ =−12×11.98\\ =−144\end{array}$

This is not a true solution.

Put the value of x=-1.44 in the equation $-12\left|9x+1\right|=144$

$\begin{array}{c}−12|9×−1.44+1|=−12|−11.96|\\ =−144\end{array}$

This is not a true solution.

As the modulus value of expression is negative it is not true solution. The solution is $(âˆ\dots )$.