How do you solve #abs(x + 6) = abs(x - 1) #?

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Answer 1 · 2015-05-23T21:56:34

I don't usually solve moduli this way, but let's square both sides of the equation...

LHS: #|x+6|^2 = (x+6)^2 = x^2+12x+36#

RHS: #|x-1|^2 = (x-1)^2 = x^2-2x+1#

So #x^2+12x+36 = x^2-2x+1#

Subtract #x^2# from both sides to get:

#12x+36 = -2x+1#

Add #2x# to both sides to get:

#14x+36 = 1#

Subtract #36# from both sides to get:

#14x=-35#

Divide both sides by #14# to get:

#x = -35/14 = -5/2#

Check:

#|x+6| = |-5/2+6| = |-5/2+12/2| = |7/2| = 7/2#

#|x-1| = |-5/2-1| = |-5/2-2/2| = |-7/2| = 7/2#