Why am I getting two solutions for this absolute value equation?

Note that $|12-5x|=\begin{cases}12-5x,&12-5x\ge0\\5x-12,&12-5x<0\end{cases}$

When $12-5x\ge0$, you get $12-5x=3-2x\implies x=3,12-5x=-3<0$, which is inconsistent with the initial assumption that $12-5x\ge0$.

When $12-5x<0$, you get $12-5x=2x-3\implies x=15/7,12-5x=9/7>0$, which is inconsistent with the initial assumption that $12-5x<0$.


You can divide the study into two cases:

Case 1

\begin{cases} 12-5x\ge0 \\[4px] 12-5x=-2x+3 \end{cases} that becomes \begin{cases} x\le 12/5 \\[4px] x=3 \end{cases} No solution.

Case 2

\begin{cases} 12-5x<0 \\[4px] 5x-12=-2x+3 \end{cases} that becomes \begin{cases} x>12/5 \\[4px] x=15/7 \end{cases} No solution.

Where did you go wrong?

In order that $|x|=y$ holds, it's necessary that $y\ge0$. For $x=15/7$, you have $$ -2x+3=-\frac{30}{7}+3=-\frac{9}{7}<0 $$

Tags:

Absolute Value

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