kashishh
Thankyou Bunuel,
It is asking for range.
i understand,so is in this question "|x+3|-|4-x|=|8+x| -How many solutions does the equation have?
Please can you elaborate on "four conditions given"?
pavanpuneet
If for the same question, |x+3|-|4-x|=|8+x|, we had to identify for what what value of x will the equation be valid. How would we do that?
How many solutions does \(|x+3| - |4-x| = |8+x|\) have?
Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.
If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;
If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;
If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;
If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.
So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).
Hope it's clear.