Modulus Functions 1

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Submitted by roamingfree on Fri, 02/05/2021 - 11:38
To complete this question, you need to be familiar with the modulus function and how to eliminate it to solve equations.
Solve the equation below, giving your answer in terms of $a$ $$ |4a - 12x| = 2a $$
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The answers are $x = \frac{1}{2}a$ and $x = \frac{1}{6}a$
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To solve the equation $|4a - 12x| =2a$, we need to consider two equations
  1. The equation $+(4a -12x) = 2a$
  2. The equation $-(4a - 12x) = 2a$

Consider the first equation $$\begin{array}{rcl} (4a -12x) &=& 2a \\ 2a &=& 12x \\ x &=& \frac{1}{6}a \end{array} $$ The second question yields $$\begin{array}{rcl} -(4a -12x) &=& 2a \\ -4a + 12x &=& 2a \\ 12x &=& 6a \\ x &=& \frac{1}{2}a \end{array} $$ Note that these distinct solutions exist only if $a\gt 0$. If $a=0$, then we have only the solution $x=0$. If $a\lt 0$ then the equation has no solutions.