# How do you graph and solve |3y+2|=|2y-5|?

Nov 27, 2015

$y \in \left\{\frac{3}{5} , - 7\right\}$

#### Explanation:

or $3 y + 2 = 2 y - 5$ (i) or $3 y + 2 = - 2 y + 5$ (ii)

(i) $R i g h t a r r o w y = - 7$

(ii) $R i g h t a r r o w 5 y = 3$

Jan 2, 2016

$y = - 7 , \frac{3}{5}$

#### Explanation:

Find the points when the term inside the absolute value switches sign.

$\left\mid 3 y + 2 \right\mid$

$3 y + 2 = 0$
$y = - \frac{2}{3}$
$3 y + 2 < 0$ when $y < - \frac{2}{3}$, $> 0$ when $y > - \frac{2}{3}$

$\left\mid 2 y - 5 \right\mid$

$2 y - 5 = 0$
$y = \frac{5}{2}$
$2 y - 5 < 0$ when $y < \frac{5}{2}$, $> 0$ when $y > \frac{5}{2}$

From this, we have three distinct ranges of numbers: $\left(- \infty , - \frac{2}{3}\right) , \left(- \frac{2}{3} , \frac{5}{2}\right)$, and $\left(\frac{5}{2} , + \infty\right)$.

color(blue)((-oo,-2/3)

In this set, both of the terms inside the absolute value functions will be negative. Take the negative versions of each of the absolute value expressions.

$- \left(3 y + 2\right) = - \left(2 y - 5\right)$

Solve. The answer is only valid if $y < - \frac{2}{3}$.

$3 y + 2 = 2 y - 5$
$y = - 7$

color(blue)((-2/3,5/2)

Here, the $3 y + 2$ term will be positive but $2 y - 5$ will be negative. Take the opposite version of only the $2 y - 5$ term and solve.

$3 y + 2 = - \left(2 y - 5\right)$
$3 y + 2 = - 2 y + 5$
$y = \frac{3}{5}$

This is also a valid answer, since $- \frac{2}{3} < \frac{3}{5} < \frac{5}{2}$.

color(blue)((5/2,+oo)

From the first set, we know this will result in an answer of $- 7$. Though it is invalid for this range, it was valid for $\left(- \infty , - \frac{2}{3}\right)$.

Thus, $y = - 7 , \frac{3}{5}$.

graph{abs(3x+2)-abs(2x-5) [-19.8, 20.75, -8.48, 11.79]}