MHB Finding Solutions for a Modulus Equation

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The discussion focuses on solving the modulus equation |x - 4| - |x + 2| = 6 by analyzing the behavior of the modulus functions based on the value of x. It identifies three cases for x: when x < -2, -2 ≤ x < 4, and x ≥ 4, leading to different expressions for the equation. The solution reveals that for x < -2, the equation holds true, while solving the equation for the middle case yields x = -2, which is valid within the specified range. Ultimately, the solution concludes that all values of x less than or equal to -2 satisfy the original equation. This highlights an interesting aspect of modulus equations where the solution set can be an interval rather than a single value.
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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 realize 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 $\displaystyle \begin{align*} x \leq -2 \end{align*}$.
 
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Very interesting. At first glance I thought your answer couldn't be right because it was not a specific value but I see that it IS right in that all values <= -2 work. Very cool.
 
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It's more obvious if we let ##x = y +4## and look for$$|y| = | y + 6|+6$$Especially if we look at that graphically.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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