# How do you solve -abs(x+1)=-2?

Mar 30, 2015

$x = 1$ and
$x = - 3$

The way to solve it is as follows.

Equal parts of an equation, left and right, can be multiplied by the same non-equal to zero multiplier getting an equivalent equation.
Let's multiply them by $- 1$:
$| x + 1 | = 2$

Now we have to remember the definition of the absolute value of a number.
If the number is positive or zero, its absolute value equals to itself:
if $Z \ge 0$ then $| Z | = Z$.
If the number is negative, its absolute value equals to its opposite (or, not very scientifically, minus this number)
if $Z < 0$ then $| Z | = - Z$.

Applying this to a problem at hand:

CASE 1
Looking for solutions in the area defined by an inequality
$x + 1 \ge 0$ (that is, $x \ge - 1$) then $| x + 1 | = x + 1$ and,
using our equation,
$x + 1 = 2$, that is, $x = 1$.
This value is within the area $x \ge - 1$ and, therefore, is the first legitimate solution.

CASE 2
Looking for solutions in the area defined by an inequality
$x + 1 < 0$ (that is, $x < - 1$) then $| x + 1 | = - \left(x + 1\right)$ and,
using our equation,
$- \left(x + 1\right) = 2$, that is, $x = - 3$.
This value is within the area $x < - 1$ and, therefore, is the second legitimate solution.

We can demonstrate it graphically.
Our equation is equivalent to
$| x + 1 | - 2 = 0$
Let's draw a graph of a function
$y = | x + 1 | - 2$
graph{|x+1|-2 [-10, 10, -5, 5]}
As you see, it intersects the X-axis (that is, equals to zero) at points $x = - 3$ and $x = 1$. This confirms our solutions.