Is x^2 > 5^2 ? (1) |x + 5| = 3|x - 5| (2) |x| > 3

|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.

another way can be
|x+5|=3|x-5|
square both
x2+10x +25 = 9x2-90x+225
2x2+25x+50=0
x=2.5orx=10