madn800
Then, how did you conclude that they have different signs?? Please explain it to me because I didn't understand that part.
|x+5| and 3|x - 5| can be expanded only in two ways, with the same sign and with different signs.
Here is more conventional approach for you:(1) |x+5| = 3|x - 5|. Critical points are -5 and 5.
If \(x\leq{-5}\), then \(x+5\leq{0}\), and \(x-5<0\), hence \(|x+5|=-(x+5)\) and \(x - 5=-(x-5)\). Thus for this range the equation becomes: \(-(x+5)=-3(x-5)\) --> \(x=10\). Discard because this solution is out of the range.
If \(-5<x<5\), then \(x+5>0\), and \(x-5<0\), hence \(|x+5|=x+5\) and \(x - 5=-(x-5)\). Thus for this range the equation becomes: \(x+5=-3(x-5)\) --> \(x=2.5\). Valid solution because it falls into the range.
If \(x\geq{5}\), then \(x+5>0\), and \(x-5\geq{0}\), hence \(|x+5|=x+5\) and \(x - 5=x-5\). Thus for this range the equation becomes: \(x+5=3(x-5)\) --> \(x=10\). Valid solution because it falls into the range.
So, we have two solutions: \(x=2.5\) and \(x=10\).
Hope it helps.