How do you solve $5 | 2 r + 3 | - 5 = 0$ $5abs(2r + 3) -5 = 0$ ?

Question

1565 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-15T09:13:01

$r = - 2, - 1$ $r=-2,-1$

$5 | 2 r + 3 | - 5 = 0$ $5abs(2r+3)-5=0$

Add $5$ $5$ to both sides of the equation.

$5 | 2 r + 3 | = 5$ $5abs(2r+3)=5$

Divide both sides by $5$ $5$ .

$| 2 r + 3 | = 1$ $abs(2r+3)=1$

Separate the equation into one positive equation and one negative equation.

$2 r + 3 = 1$ $2r+3=1$ and $- (2 r + 3) = 1$ $-(2r+3)=1$

Positive equation

$2 r + 3 = 1$ $2r+3=1$ =

$2 r = 1 - 3$ $2r=1-3$ =

$2 r = - 2$ $2r=-2$

Divide both sides by $2$ $2$ .

$r = - 1$ $r=-1$

Negative equation

$- (2 r + 3) = 1$ $-(2r+3)=1$

$- 2 r - 3 = 1$ $-2r-3=1$

$- 2 r = 1 + 3$ $-2r=1+3$ =

$- 2 r = 4$ $-2r=4$

Divide both sides by $- 2$ $-2$ .

$r = \frac{4}{- 2}$ $r=4/(-2)$ =

$r = - 2$ $r=-2$

How do you solve 5|2r+3|−5=05abs(2r + 3) -5 = 0?