Welcome to our community

Be a part of something great, join today!

Alexander's problem via Facebook about solving a modulus equation

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Solve $\displaystyle \begin{align*} \left| x - 4 \right| - \left| x + 2 \right| = 6 \end{align*}$ for $\displaystyle \begin{align*} x \end{align*}$.
To start with, we need to realise that each modulus function will be defined differently depending on the value of $\displaystyle \begin{align*} x \end{align*}$.

Notice that

$\displaystyle \begin{align*} \left| x - 4 \right| = \begin{cases} x - 4 \textrm{ if } x \geq 4 \\ 4 - x \textrm{ if } x < 4 \end{cases} \end{align*}$

and

$\displaystyle \begin{align*} \left| x + 2 \right| = \begin{cases} x + 2 \textrm{ if } x \geq -2 \\ - x - 2 \textrm{ if } x < -2 \end{cases} \end{align*}$

Thus

$\displaystyle \begin{align*} \left| x - 4 \right| - \left| x + 2 \right| &= \begin{cases} \left( 4 - x \right) - \left( -x - 2 \right) \textrm{ if } x < -2 \\ \left( 4 - x \right) - \left( x + 2 \right) \textrm{ if } -2 \leq x < 4 \\ \left( x - 4 \right) - \left( x + 2 \right) \textrm{ if } x \geq 4 \end{cases} \\ &= \begin{cases} 6 \textrm{ if } x < -2 \\ 2 - 2\,x \textrm{ if } -2 \leq x < 4 \\ -6 \textrm{ if } x \geq 4 \end{cases} \end{align*}$

Notice that we already have $\displaystyle \begin{align*} \left| x - 4 \right| - \left| x + 2 \right| = 6 \textrm{ if } x < -2 \end{align*}$.

If we solve $\displaystyle \begin{align*} 2 - 2\,x = 6 \end{align*}$ for $\displaystyle \begin{align*} x \end{align*}$ we find

$\displaystyle \begin{align*} 2 - 2\,x &= 6 \\ 2\,x &= -4 \\ x &= -2 \end{align*}$

which satisfies the condition $\displaystyle \begin{align*} -2 \leq x < 4 \end{align*}$.

Thus the solution to $\displaystyle \begin{align*} \left| x - 4 \right| - \left| x + 2 \right| = 6 \end{align*}$ is $\displaystyle \begin{align*} x \leq -2 \end{align*}$.