Question 211557
There is no solution to {{{abs(x)=-a}}} where {{{a>0}}} since {{{abs(x)}}} is ALWAYS nonnegative (ie not negative).



In order to make {{{abs(x)=-a}}} have the solutions of {{{x=-a}}} or {{{x=a}}}, you need to force 'a' to be negative. So {{{abs(x)=-a}}} has the solutions of {{{x=-a}}} or {{{x=a}}} when {{{a<0}}}.




Ex: {{{abs(x)=-(-5)}}} has the solutions {{{x=-(-5)}}} and {{{x=-5}}}. If you plug in either solution, you get


{{{abs(-(-5))=-(-5)}}} ---> {{{abs(5)=5}}} ---> {{{5=5}}}


{{{abs(-5)=-(-5)}}} ---> {{{abs(-5)=5}}} ---> {{{5=5}}}