# How do you solve  |x+2| + |2x-4| = |x-3| ?

Jun 28, 2016

The equation has no solution.

#### Explanation:

$f \left(x\right) = \left\mid x + 2 \right\mid + \left\mid 2 x - 4 \right\mid - \left\mid x - 3 \right\mid = 0$

Making $\left\mid x + 2 \right\mid = y = \left\mid x - 3 \right\mid - \left\mid 2 x - 4 \right\mid = 0$
we have $x = \pm y - 2$.

I) Choosing $x = y - 2$

$y = \left\mid y - 5 \right\mid - \left\mid 2 y - 8 \right\mid$
$y - 5 = \left\mid y - 5 \right\mid - \left\mid 2 y - 8 \right\mid - 5$
$1 = \pm 1 - \frac{\left\mid 2 y - 8 \right\mid + 5}{y - 5}$ for $y \ne 5$

Here we have two possibilities:

I-1)
$0 = \left\mid 2 y - 8 \right\mid + 5$ having two outcomes
I-1-a)
$2 y - 8 = - 5 \to y = \frac{3}{2}$
I-1-b)
$2 y - 8 = 5 \to y = \frac{13}{2}$

I-2)
$2 = - \frac{\left\mid 2 y - 8 \right\mid + 5}{y - 5}$
$2 \left(y - 5\right) = - \left\mid 2 y - 8 \right\mid - 5$
$2 y - 10 = - \left\mid 2 y - 8 \right\mid - 5$
$2 y - 8 = - \left\mid 2 y - 8 \right\mid - 3$
$1 = \pm 1 - \frac{3}{2 y - 8}$
with the only posibility
$2 = - \frac{3}{2 y - 8} \to y = \frac{13}{4}$

II) Choosing $x = - \left(y + 2\right)$

$- y + \left\mid 2 y + 8 \right\mid - \left\mid y + 5 \right\mid = 0$
$- y + 2 \left\mid y + 4 \right\mid - \left\mid y + 5 \right\mid = 0$
$- y - 4 + 2 \left\mid y + 4 \right\mid - \left\mid y + 5 \right\mid + 4 = 0$
$- \left(y + 4\right) + 2 \left\mid y + 4 \right\mid - \left\mid y + 5 \right\mid + 4 = 0$
$1 \pm 2 + \frac{\left\mid y + 5 \right\mid - 4}{y + 4} = 0$ with two outcomes
II-1)
$3 + \frac{\left\mid y + 5 \right\mid - 4}{y + 4} = 0$
$3 \left(y + 4\right) + \left\mid y + 5 \right\mid - 4 = 0$
$3 \left(y + 5\right) + \left\mid y + 5 \right\mid - 4 - 3 = 0$
$3 \pm 1 - \frac{7}{y + 5} = 0$ with two outcomes
II-1-a)
$4 \left(y + 5\right) = 7 \to y = - \frac{13}{4}$
II-1-b)
$2 \left(y + 5\right) = 7 \to y = - \frac{3}{2}$

II-2)
$- 1 + \frac{\left\mid y + 5 \right\mid - 4}{y + 4} = 0$
$- \left(y + 4\right) + \left\mid y + 5 \right\mid - 4 = 0$
$- \left(y + 5\right) + \left\mid y + 5 \right\mid - 4 + 1 = 0$
$1 \pm 1 + \frac{3}{y + 5} = 0$
with the only outcome
$- 2 \left(y + 5\right) + 3 = 0 \to y = - \frac{7}{2}$

Collecting the results obtained for $x = y - 2$

$y = \left\{\frac{3}{2} , \frac{13}{2} , \frac{13}{4}\right\} \to x = \left\{- \frac{1}{2} , \frac{9}{2} , \frac{5}{4}\right\}$

and for $x = - \left(y + 2\right)$

$y = \left\{- \frac{13}{4} , - \frac{3}{2} , - \frac{7}{2}\right\} \to x = \left\{\frac{5}{4} , - \frac{1}{2} , \frac{3}{2}\right\}$

Substituting in $f \left(x\right)$ none of them is solution.