How do you solve #abs(a-5)/8=5#?

Question

2076 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-02T16:50:42

#|a - 5| = 8 xx 5#

We need to consider two separate situations.

a) The absolute value is positive
b) The absolute value is negative

Starting with the positive case:

#a - 5 = 40#

#a = 45#

Now for the negative case:

#-(a - 5) = 40#

#-a + 5 = 40#

#-a = 35#

#a = -35#

Hence, the solution set is #{-35, 45}#.

Hopefully this helps!

Answer 2 · 2016-08-02T17:03:02

#a= 45" "# or #" "a = -35#

Absolute value can be negative inside or positive inside. When the value comes out of the absolute value sign it is always positive.

First, multiply both sides by #8#, the opposite of dividing by #8#.

#(a- 5)/ 8 xx 8 = 5 xx 8#

Since #8/8= 1#, you have

# a -5 = 40#

#a-5# has two possible values. Solve for both values.

#(a-5) = + 40" "# or #" " a-5 = -40#

#a-5 = + 40 -># add #5# to both sides

#a- cancel(5) + cancel(5) = + 40 + 5#

Since #-5+ 5 = 0# and #40+5 = +45#, you have

#a= + 45#

#a-5 = -40 -># add #5# to both sides

#a - cancel(5) + cancel(5) = -40 +5#

Once again, #-5 + 5 = 0# and #-40 + 5 = -35#

#a = -35#