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

May 23, 2015

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

LHS: $| x + 6 {|}^{2} = {\left(x + 6\right)}^{2} = {x}^{2} + 12 x + 36$

RHS: $| x - 1 {|}^{2} = {\left(x - 1\right)}^{2} = {x}^{2} - 2 x + 1$

So ${x}^{2} + 12 x + 36 = {x}^{2} - 2 x + 1$

Subtract ${x}^{2}$ from both sides to get:

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

Add $2 x$ to both sides to get:

$14 x + 36 = 1$

Subtract $36$ from both sides to get:

$14 x = - 35$

Divide both sides by $14$ to get:

$x = - \frac{35}{14} = - \frac{5}{2}$

Check:

$| x + 6 | = | - \frac{5}{2} + 6 | = | - \frac{5}{2} + \frac{12}{2} | = | \frac{7}{2} | = \frac{7}{2}$

$| x - 1 | = | - \frac{5}{2} - 1 | = | - \frac{5}{2} - \frac{2}{2} | = | - \frac{7}{2} | = \frac{7}{2}$