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

Oct 19, 2017

See below.

#### Explanation:

The solutions for $\left\mid 2 + 3 x \right\mid = \left\mid 4 - 2 x \right\mid$ or equivalently

$\sqrt{{\left(2 + 3 x\right)}^{2}} = \sqrt{{\left(4 - 2 x\right)}^{2}}$ are included into the solutions for

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

${\left(2 + 3 x\right)}^{2} - {\left(4 - 2 x\right)}^{2} = 0$ or

$\left(2 + 3 x + 4 - 2 x\right) \left(2 + 3 x - 4 + 2 x\right) = 0$

or

$\left(x + 6\right) \left(5 x - 2\right) = 0$

or $x = \left\{- 6 , \frac{2}{5}\right\}$ and both solutions are feasible.

Oct 19, 2017

${x}_{1} = - 6$ and ${x}_{2} = \frac{2}{5}$

#### Explanation:

$\left\mid 2 + 3 x \right\mid = \left\mid 4 - 2 x \right\mid$

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

$9 {x}^{2} + 12 x + 4 = 4 {x}^{2} - 16 x + 16$

$5 {x}^{2} + 28 x - 12 = 0$

$5 {x}^{2} + 30 x - 2 x - 12 = 0$

$5 x \cdot \left(x + 6\right) - 2 \cdot \left(x + 6\right) = 0$

$\left(5 x - 2\right) \cdot \left(x + 6\right) = 0$

Hence ${x}_{1} = - 6$ and ${x}_{2} = \frac{2}{5}$