# How do you solve |- 3- 3y | = | 2y + 1|?

Apr 6, 2017

Solution : $y = - 2 \mathmr{and} y = - \frac{4}{5}$

#### Explanation:

$| - 3 - 3 y | = | 2 y + 1 |$ squaring both sides we get

${\left(- 3 - 3 y\right)}^{2} = {\left(2 y + 1\right)}^{2} \mathmr{and} 9 {\left(1 + y\right)}^{2} = 4 {y}^{2} + 4 y + 1$ or

$9 \left({y}^{2} + 2 y + 1\right) = 4 {y}^{2} + 4 y + 1 \mathmr{and} 5 {y}^{2} + 14 y + 8 = 0$ or

$5 {y}^{2} + 10 y + 4 y + 8 = 0 \mathmr{and} 5 y \left(y + 2\right) + 4 \left(y + 2\right) = 0$ or

$\left(y + 2\right) \left(5 y + 4\right) = 0 \therefore$Either $y + 2 = 0 \therefore y = - 2 \mathmr{and} 5 y + 4 = 0 \therefore y = - \frac{4}{5}$

Solution : $y = - 2 \mathmr{and} y = - \frac{4}{5}$ [Ans]

Apr 6, 2017

$y = - 2 \text{ or } y = - \frac{4}{5}$

#### Explanation:

There are 2 possible solutions.

$- 3 - 3 y = \textcolor{red}{\pm} \left(2 y + 1\right)$

$\textcolor{b l u e}{\text{First solution}}$

$- 3 - 3 y = 2 y + 1$

$\Rightarrow - 5 y = 4 \Rightarrow y = - \frac{4}{5}$

$\textcolor{b l u e}{\text{Second solution}}$

$- 3 - 3 y = - 2 y - 1$

$\Rightarrow - y = 2 \Rightarrow y = - 2$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the equation and if left side is equal to right side then they are the solutions.

$\text{left side } = | - 3 + \frac{12}{5} | = | - \frac{3}{5} | = \frac{3}{5}$

$\text{right side } = | - \frac{8}{5} + 1 | = | - \frac{3}{5} | = \frac{3}{5}$

$\text{left side } = | - 3 + 6 \models | 3 | = 3$

$\text{right side } = | - 4 + 1 | = | - 3 | = 3$