# Solve |2x-3|+|x-1|=|x-2| Find the values of x?

Jun 24, 2018

The solutions are $S = \left\{1 , \frac{3}{2}\right\}$

#### Explanation:

The equation is

$| 2 x - 3 | + | x - 1 | = | x - 2 |$

There are $3$ points to consider

$\left\{\begin{matrix}2 x - 3 = 0 \\ x - 1 = 0 \\ x - 2 = 0\end{matrix}\right.$

$\implies$, $\left\{\begin{matrix}x = \frac{3}{2} \\ x = 1 \\ x = 2\end{matrix}\right.$

There are $4$ intervals to consider

$\left\{\begin{matrix}- \infty & 1 \\ 1 & \frac{3}{2} \\ \frac{3}{2} & 2 \\ 2 & + \infty\end{matrix}\right.$

On the first interval $\left(- \infty , 1\right)$

$- 2 x + 3 - x + 1 = - x + 2$

$\implies$, $2 x = 2$

$\implies$, $x = 1$

$x$ fits in this interval and the solution is valid

On the second interval $\left(1 , \frac{3}{2}\right)$

$- 2 x + 3 + x - 1 = - x + 2$

$\implies$, $0 = 0$

There is no solution in this interval

On the third interval $\left(\frac{3}{2} , 2\right)$

$2 x - 3 + x - 1 = - x + 2$

$\implies$, $4 x = 6$

$\implies$, $x = \frac{6}{4} = \frac{3}{2}$

$x$ fits in this interval and the solution is valid

On the fourth interval $\left(2 , + \infty\right)$

$2 x - 3 + x - 1 = x - 2$

$\implies$, $2 x = 2$

$\implies$, $x = 1$

$x$ does not fit in this interval.

The solutions are $S = \left\{1 , \frac{3}{2}\right\}$