# How do you solve -3+2abs(n-9)=1?

Aug 1, 2015

$n = 7$ or $n = 11$

#### Explanation:

First, isolate the modulus on one side of the equation by adding $3$ to both sides of the equation and dividing everything by $2$.

$- \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} + 2 | n - 9 | = 1 + 3$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} | n - 9 |}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} = \frac{4}{2}$

$| n - 9 | = 2$

Now, this equation can produce 2 distinct values for $n$ depending on which condition is true

• If $\left(n - 9\right) < 0$, then you have

$| n - 9 | = - \left(n - 9\right) = - n + 9$

This means that the equation becomes

$- n + 9 = 2 \implies n = \textcolor{g r e e n}{7}$

• If $\left(n - 9\right) > 0$, then you have

$| n - 9 | = n - 9$

This means that $n$ will be equal to

$n - 9 = 2 \implies n = \textcolor{g r e e n}{11}$

Aug 1, 2015

#### Explanation:

Starting with $- 3 + 2 \left\mid n - 9 \right\mid = 1$,

we can see that we must have:

$2 \left\mid n - 9 \right\mid = 4$, so

$\left\mid n - 9 \right\mid = 2$

On the number line, this means that the distance between $n$ and $9$ is 2.
There are two numbers that are $2$ away from $9$: one on the left of $9$ at $9 - 2$ and the other on the right at $9 + 2$.

So, we get:

$n = 9 - 2 \text{ }$ or $\text{ } n = 9 + 2$

and

$n = 7 \text{ }$ or $\text{ } n = 11$