How do you solve #abs(8x+8) = abs(9x-4)#?

2 Answers
Jul 11, 2015

Answer:

Split into cases for intervals #(-oo, -1)#, #[-1, 4/9)# and #(4/9, oo)#, solving the linear equations that result to find:

#x=-4/17# or #x=12#

Explanation:

Split into cases:

Case 1: #x in (-oo, -1)#

#8x+8 < 0#, so #abs(8x+8) = -(8x+8) = -8x-8#
#9x-4 < 0#, so #abs(9x-4) = -(9x-4) = -9x+4#

Equation becomes:

#-8x-8 = -9x+4#

Add #9x+8# to both sides to get:

#x = 12#

This lies outside #(-oo, -1)# so is not valid for this case.

Case 2: #x in [-1, 4/9)#

#8x+8 >= 0#, so #abs(8x+8) = 8x+8#
#9x-4 < 0#, so #abs(9x-4) = -(9x-4) = -9x+4#

Equation becomes:

#8x+8 = -9x+4#

Add #9x-8# to both sides to get:

#17x=-4#

Divide both sides by #17# to get:

#x=-4/17#

This does lie in #[-1, 4/9)# so is a valid solution.

Case 3: #x in [4/9, oo)#

#8x+8 >= 0#, so #abs(8x+8) = 8x+8#
#9x-4 >= 0#, so #abs(9x-4) = 9x-4#

Equation becomes:

#8x+8 = 9x-4#

Add #-8x + 4# to both sides to get:

#x = 12#

This does lie in #[4/9, oo)# so is a valid solution.

Jul 11, 2015

Answer:

#abs(u) = abs(v)# if and only if either #u=v# or #u=-v#.

Explanation:

Two numbers have the same absolute value when the numbers are equal or they are opposites (negatives) of each other.

#abs(8x+8) = abs(9x-4)#

if and only if:

#8x+8 = 9x-4# #color(white)"xx"# or #color(white)"xx"# #8x+8 = -(9x-4)#

Solving #8x+8 = 9x-4# , we get

#8+4 = 9x-9x# so #12=x#, that is: #x=12#

Solving #8x+8 = -(9x-4)#. we get

#8x+8 = -(9x-4)#, so #8x+8 = -9x+4#

and then #8x+9x = 4-8#, so #17x = -4#, and finally #x = -4/17#

The solutions are: #12# and #-4/17#