# How do you solve abs((2x-5)/3)=abs((3x+4)/2)?

Nov 9, 2017

$x = - \frac{22}{5}$

#### Explanation:

$\left\mid \frac{2 x - 5}{3} \right\mid = \left\mid \frac{3 x + 4}{2} \right\mid$

After squaring both sides,

${\left(2 x - 5\right)}^{2} / 9 = {\left(3 x + 4\right)}^{2} / 4$

$\frac{4 {x}^{2} - 20 x + 25}{9} = \frac{9 {x}^{2} + 24 x + 16}{4}$

$4 \cdot \left(4 {x}^{2} - 20 x + 25\right) = 9 \cdot \left(9 {x}^{2} + 24 x + 16\right)$

$16 {x}^{2} - 80 x + 100 = 81 {x}^{2} + 216 x + 144$

$65 {x}^{2} + 296 x + 44 = 0$

$65 {x}^{2} + 276 x + 20 x + 44 = 0$

$13 x \cdot \left(5 x + 22\right) + 4 \cdot \left(5 x + 22\right) = 0$

$\left(13 x + 4\right) \cdot \left(5 x + 22\right) = 0$

Hence ${x}_{1} = - \frac{22}{5}$ and ${x}_{2} = - \frac{4}{13}$. But $x = - \frac{4}{13}$ doesn't provide solution for original problem. Thus solution of it is $x = - \frac{22}{5}$