How do you solve #-7+8abs(-7x-3)=73#?

1 Answer
Feb 19, 2017

See the entire solution process below:

Explanation:

First, add #color(red)(7)# to each side of the equation to isolate the absolute value term while keeping the equation balanced:

#color(red)(7) -7 + 8abs(-7x - 3) = color(red)(7) + 73#

#0 + 8abs(-7x - 3) = 80#

#8abs(-7x - 3) = 80#

Next, divide each side of the equation by #color(red)(8)# to isolate the absolute value function while keeping the equation balanced:

#(8abs(-7x - 3))/color(red)(8) = 80/color(red)(8)#

#(color(red)(cancel(color(black)(8)))abs(-7x - 3))/cancel(color(red)(8)) = 10#

#abs(-7x - 3) = 10#

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1)

#-7x - 3 = -10#

#-7x - 3 + color(red)(3) = -10 + color(red)(3)#

#-7x - 0 = -7#

#-7x = -7#

#(-7x)/color(red)(-7) = (-7)/color(red)(-7)#

#x = 1#

Solution 2)

#-7x - 3 = 10#

#-7x - 3 + color(red)(3) = 10 + color(red)(3)#

#-7x - 0 = 13#

#-7x = 13#

#(-7x)/color(red)(-7) = 13/color(red)(-7)#

#x = -13/7#

The solution is: #x = 1# and #x = -13/7#