|x+3| - |4-x| = |8+x|. How many solutions does the equation

Responding to pm.

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

STEP BY STEP SOLUTION:

We have three transition points for \(|x+3| - |4-x| = |8+x|\): -8, -3, and 4 (transition point is the value of x for which an expression in the modulus equals to zero). Thus we have four ranges to check:

1. \(x<-8\);
2. \(-8\leq{x}\leq{-3}\);
3. \(-3<x<4\)
4. \(x\geq{4}\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<-8\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is negative. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=-(8+x)\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =-(8+x)\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x<-8\)).

2. When \(-8\leq{x}\leq{-3}\), then \(x+3\) is negative, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=-(x+3)\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(-(x+3) - (4-x) =8+x\): --> \(x=-15\). This solution is NOT OK, since \(x=-15\) is NOT in the range we consider (\(-8\leq{x}\leq{-3}\)).

3. When \(-3<x<4\), then \(x+3\) is positive, \(4-x\) is positive and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=4-x\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (4-x) =8+x\): --> \(x=9\). This solution is NOT OK, since \(x=9\) is NOT in the range we consider (\(-3<x<4\)).

4. When \(x\geq{4}\), then \(x+3\) is positive, \(4-x\) is negative and \(8+x\) is positive. Thus \(|x+3|=x+3\), \(|4-x|=-(4-x)=x-4\) and \(|8+x|=8+x\).

Therefore for this range \(|x+3| - |4-x| = |8+x|\): transforms to \(x+3 - (x-4) =8+x\): --> \(x=-1\). This solution is NOT OK, since \(x=-1\) is NOT in the range we consider (\(x\geq{4}\)).

Thus no value of x satisfies \(|x+3| - |4-x| = |8+x|\).

Answer: A.

Hope it's clear.