How do you solve #4 + abs(r + 2 ) =7#?

1 Answer
Jun 26, 2015

First subtract #4# from both sides to get: #abs(r+2) = 3#

So either #r+2 = 3#, giving #r = 1#

or #r+2 = -3#, giving #r = -5#

Explanation:

Let's look at the definition of absolution value first:

#abs(x) = x# if #x > 0#

#abs(x) = 0# if #x = 0#

#abs(x) = -x# if #x < 0#

So if #abs(x) = a#, then #a >= 0# and #x = +-a#

Now subtract #4# from both sides of our original equation to get:

#abs(r+2) = 3#

From what we have found, #r+2 = +-3#

Subtract #2# from both sides to get:

#r = -2 +- 3#

That is #r = -5# or #r = 1#