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,

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

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