Question 1017312
{{{abs(x-5) = 2abs(x--7)}}}

Squaring both sides of the equation to eliminate the absolute value bars, we get

==> {{{(x-5)^2 = 4*(x+7)^2}}}

<==> {{{x^2 -10x +25 = 4(x^2+14x+49)}}}

<==> {{{x^2 -10x +25 = 4x^2+56x+196)}}}

<==> {{{0 = 3x^2 + 66x +171}}}

<==> {{{0 = (x+3)(x+19)}}}

==> x = -3 or -19