How do you solve #abs(4x + 7) - 2 = 29#?

1 Answer
Aug 27, 2015

#x = 6" "# or #" "x = -19/2#

Explanation:

First, isolate the modulus on one side of the equation by adding #2# to both sides

#|4x+7| - color(red)(cancel(color(black)(2))) + color(red)(cancel(color(black)(2))) = 29 + 2#

#|4x+7| = 31#

By definition, the absolute value of a real number will always be a positive number.

This means that you need to take into account the fact thatthe expression that's inside the modulus can be positive or negative, since both cases would produce the same value, #31#.

This means that you have

  • #4x+7>=0 implies |4x+7| = 4x+7#

The equation will become

#4x+7 = 31#

#4x = 24 implies x = 24/4 = color(green)(6)#

  • #4x+7<0 implies |4x+7| = -(4x+7)#

This time you have

#-(4x+7) = 31#

#-4x - 7 = 31#

#x = (38)/((-4)) = color(green)(-19/2)#

Your original equation will thus have two possible solutions

#x=6" "#, for which #4x+7 = 3#

and

#x = -19/2" "#, for which #4x+7 = -31#