How do you solve abs(2/3x-1/4x)=abs(1/4x+8)23x14x=14x+8?

1 Answer
Apr 3, 2015

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

Solution:
|2/3x - 1/4x| = |1/4x + 8|23x14x=14x+8

=> |(8 - 3)/12x| = |1/4x + 8|8312x=14x+8

=> |5/12x| = |1/4x + 8|512x=14x+8

Square both sides, you get

(5/12x)^2 = (1/4x + 8)^2(512x)2=(14x+8)2

=> (5/12x)^2 - (1/4x + 8)^2 = 0(512x)2(14x+8)2=0

This is a difference of two squares, as a^2 - b^2 = (a - b)(a + b)a2b2=(ab)(a+b)

=> (5/12x - (1/4x + 8))*(5/12x + (1/4x + 8)) = 0(512x(14x+8))(512x+(14x+8))=0

=> ( 2/12x - 8)(8/12x + 8) = 0(212x8)(812x+8)=0

=> 2/12x - 8 = 0 => 1/6x = 8 => x = 48212x8=016x=8x=48

Also,

8/12x + 8 = 0 => 2/3x = -8 => x = -12812x+8=023x=8x=12