# How do you graph and solve |3x+1| + |3-2x| =11?

Jan 7, 2017

$x = - \frac{9}{5}$ and $x = \frac{13}{5}$

#### Explanation:

$\left\mid 3 x + 1 \right\mid + \left\mid 2 x - 3 \right\mid = 11 = 3 x + 1 - 3 x + 10$ so
$\frac{\left\mid 3 x + 1 \right\mid}{3 x + 1} + \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 1 + \frac{10 - 3 x}{3 x + 1}$ or
$\pm 1 + \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 1 + \frac{10 - 3 x}{3 x + 1}$
Now we have two options:

a)
$1 + \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 1 + \frac{10 - 3 x}{3 x + 1} \to \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = \frac{10 - 3 x}{3 x + 1}$

b)
$- 1 + \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 1 + \frac{10 - 3 x}{3 x + 1} \to \frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 2 + \frac{10 - 3 x}{3 x + 1}$

Following with a) we have

a-1)
$\left\mid 2 x - 3 \right\mid = 10 - 3 x = 2 x - 3 - 5 x + 13$ or
$\frac{\left\mid 2 x - 3 \right\mid}{2 x - 3} = 1 + \frac{13 - 5 x}{2 x - 3}$ or
$\pm 1 = 1 + \frac{13 - 5 x}{2 x - 3}$ with two options:

a-1-1)
$1 = 1 + \frac{13 - 5 x}{2 x - 3} \to 13 - 5 x = 0 \to \textcolor{red}{x = \frac{13}{5}}$

a-1-2)
$- 1 = 1 + \frac{13 - 5 x}{2 x - 3} \to 0 = 2 + \frac{13 - 5 x}{2 x - 3} \to \textcolor{red}{x = 7}$

Now following with b)

$\frac{\left\mid 2 x - 3 \right\mid}{3 x + 1} = 2 + \frac{10 - 3 x}{3 x + 1} \to \left\mid 2 x - 3 \right\mid = 6 x + 2 + 10 - 3 x$ so
$\left\mid 2 x - 3 \right\mid = 3 x + 12$ then

b-1)
$\left\mid 2 x - 3 \right\mid = 2 x - 3 + x + 15 \to \frac{\left\mid 2 x - 3 \right\mid}{2 x - 3} = 1 + \frac{x + 15}{2 x - 3}$ or
$\pm 1 = 1 + \frac{x + 15}{2 x - 3}$ with two options

b-1-1)
$1 = 1 + \frac{x + 15}{2 x - 3} \to \textcolor{red}{x = - 15}$

b-1-2)
$- 1 = 1 + \frac{x + 15}{2 x - 3} \to 0 = 2 + \frac{x + 15}{2 x - 3} \to \textcolor{red}{x = - \frac{9}{5}}$

After checking the found values, we pick the feasible values $x = - \frac{9}{5}$ and $x = \frac{13}{5}$