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

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Answer 1 · 2015-07-15T09:13:01

$r=-2,-1$

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

Add $5$ to both sides of the equation.

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

Divide both sides by $5$ .

$abs(2r+3)=1$

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

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

Positive equation

$2r+3=1$ =

$2r=1-3$ =

$2r=-2$

Divide both sides by $2$ .

$r=-1$

Negative equation

$-(2r+3)=1$

$-2r-3=1$

$-2r=1+3$ =

$-2r=4$

Divide both sides by $-2$ .

$r=4/(-2)$ =

$r=-2$

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