# How do you solve |x+2|= 4?

Feb 6, 2017

$x = - 6 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x = 2$

#### Explanation:

Consider the two possibilities:
{: (x+2 < 0,color(white)("XX")andcolor(white)("XX"),x+2 > 0), (rarr abs(x+2) = -x-2,,rarrabs(x+2)=x+2), ("so "abs(x+2)=4,,"so "abs(x+2)=4), (rarr -x-2 =4,,rarr x+2=4), (rarr -x=6,,rarrx==2), (rarr x=-6,,) :}

Feb 6, 2017

$x = 2 \text{ or } x = - 6$

#### Explanation:

Equations with an$\textcolor{b l u e}{\text{ absolute value}}$ normally have 2 solutions.

These are found by solving $x + 2 = \textcolor{red}{\pm} 4$

$\textcolor{b l u e}{\text{Solution 1}}$

$\text{solve } x + 2 = 4 \Rightarrow x = 4 - 2 = 2$

$\textcolor{b l u e}{\text{Solution 2}}$

$\text{solve } x + 2 = - 4 \Rightarrow x = - 4 - 2 = - 6$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$x = 2 \to | 2 + 2 | = | 4 | = 4 = \text{ right side}$

$x = - 6 \to | - 6 + 2 | = | - 4 | = 4 = \text{ right side}$

$\Rightarrow x = 2 \text{ or "x=-6" are the solutions}$