# I was trying to solve this inequality with two absolute values: | 2 x &#x2212;<!-- − -

I was trying to solve this inequality with two absolute values:
$|2x-3|+7\le 3x-3|x-7|$
I've got an empty set of solutions, but it's not correct.
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Braylon Perez
The fastest way for me is to write down LHS-RHS as a piecewise linear function.
$f\left(x\right)=|2x-3|+7-3x+3|x-7|\phantom{\rule{0ex}{0ex}}=\left\{\begin{array}{ll}3-2x+7-3x+3\left(7-x\right)=31-8x,& x⩽\frac{3}{2}\\ 2x-3+7-3x+3\left(7-x\right)=25-4x,& \frac{3}{2}⩽x⩽7\\ 2x-3+7-3x+3\left(x-7\right)=2x-17,& x⩾7\end{array}$
Note that $f\left(\frac{3}{2}\right)=19,f\left(7\right)=-3$
When $x⩽\frac{3}{2},f$ is decreasing so $f\left(x\right)>0$;
When $x⩾7,2x-17⩽0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x⩽\frac{17}{2}.$
Therefore $\frac{25}{4}⩽x⩽\frac{17}{2}.$