How do you solve #abs(2/3x-1/4x)=abs(1/4x+8)#?

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Answer 1 · 2015-04-03T19:16:41

Answers:
#x = 48# and #x = -12#

Solution:
#|2/3x - 1/4x| = |1/4x + 8|#

#=> |(8 - 3)/12x| = |1/4x + 8|#

#=> |5/12x| = |1/4x + 8|#

Square both sides, you get

#(5/12x)^2 = (1/4x + 8)^2#

#=> (5/12x)^2 - (1/4x + 8)^2 = 0#

This is a difference of two squares, as #a^2 - b^2 = (a - b)(a + b)#

#=> (5/12x - (1/4x + 8))*(5/12x + (1/4x + 8)) = 0#

#=> ( 2/12x - 8)(8/12x + 8) = 0#

#=> 2/12x - 8 = 0 => 1/6x = 8 => x = 48#

Also,

# 8/12x + 8 = 0 => 2/3x = -8 => x = -12#