Question

Solution of $|x+2|+|2x+6|+|3x-3|=12$ is

• A. $\{-1,1\}$
• B. $\{-\frac{2}{5},\frac{6}{7}\}$
• C. $\{-\frac{5}{2},\frac{7}{6}\}$
• D. $\{\frac{5}{2},-\frac{7}{6}\}$

Answer 1

The correct option is C $\{-\frac{5}{2},\frac{7}{6}\}$

The turning points are $x=-2,x=-3,x=1$, respectively. Hence, we need to check the cases $-\infty <x<-3,-3\le x\le -2,-2<x\le 1,1<x<\infty$

Case $1:$ $-\infty <x<-3$
$-(x+2)-(2x+6)-(3x-3)=12$
$\Rightarrow x=-\frac{17}{6}$
However, $x=-\frac{17}{6}>-3$ is not within the domain $-\infty <x<-3$. Thus this solution is not valid.

Case $2:$ $-3\le x\le -2$
$-(x+2)+(2x+6)-(3x-3)=12$
$\Rightarrow x=-\frac{5}{2}$
$x=-\frac{5}{2}$ lies between $-3$ and $-2$. Thus $x=-\frac{5}{2}$ is one of the solutions.

Case $3:$ $-2<x\le 1$
$(x+2)+(2x+6)-(3x-3)=11\ne 12$
Thus there is no solution within this domain.

Case $4:$ $1<x<\infty$
$(x+2)+(2x+6)+(3x-3)=12$
$x=\frac{7}{6}$
which lies between $1$ and $\infty$. Thus $x=\frac{7}{6}$ is another solution.
In conclusion, $x=-\frac{5}{2}$ and $x=\frac{7}{6}$ are the solutions for the given equation.