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

Answer 1

Savannah Anderson

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#

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Answer 2

Savannah Anderson

To solve the equation ( | \frac{2}{3}x - \frac{1}{4}x | = | \frac{1}{4}x + 8 | ):

Split the equation into two cases: Case 1: (\frac{2}{3}x - \frac{1}{4}x = \frac{1}{4}x + 8) Case 2: (\frac{2}{3}x - \frac{1}{4}x = -(\frac{1}{4}x + 8))
Solve each case separately for (x).
Once you find the solutions for each case, check if they satisfy the original equation.
If they do, those are the solutions. If not, there are no solutions.
Graph the solutions on a number line to represent the solution set.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.