Solve the equation $|2x-1|+|3x+2|+|x|=3$

Question

Solve the equation $$|2x-1|+|3x+2|+|x|=3$$ My question is regarding the case when $x\le-\dfrac23$. According to the answer of the problem, $x=-1$ is NOT a solution, but I simply don't see why.

The answer is $x\in\left[-\dfrac23;0\right]$

In this case we have $$-(2x-1)-(3x+2)+x=3\\-4x=4\\x=-1,$$ which is indeed less than $-\dfrac{2}{3}$. What's wrong?

Answer 1

first, the expression in the module must be equated to zero

$2x−1 = 0$

$3x+2 = 0$

$x = 0$

We have this: $x = \frac{1}2; x = -\frac{2}3; x = 0$.

and now you need solve 3 equation: if $x \le -\frac{2}3$:

$−(2x−1)−(3x+2)-x=3$

$x = \frac{2}3$

it is false answer because we interested only in $x \le -\frac{2}3$

if $-\frac{2}3 < x \le 0$:

$−(2x−1)+(3x+2)-x=3$

$3 = 3$, it is mean, all x in $-\frac{2}3 \le x \le 0$ can be solution.

if $0 < x \le \frac{1}2$:

$−(2x−1)+(3x+2)+x=3$

$x = 0$

we have this answer already.

if $ x > \frac{1}2$:

$+(2x−1)+(3x+2)-x=3$

$x = \frac{1}2$, our interval do not contain this.

so answer it is $-\frac{2}3 \le x \le 0$.

Answer 2

As an alternative, you can use the triangle inequality: $$\begin{align}|2x-1|+|3x+2|+|x|&=|1-2x|+|3x+2|+|-x| \\ &\ge|(1-2x)+(3x+2)+(-x)|=3 \end{align}$$ so the inequality must be an equality. This occurs if and only if the expressions $(1-2x),(3x+2),(-x)$ have the same signs. You can make a table with the signs of those expressions in each of the intervals $(-\infty,-\frac 23],[-\frac 23, 0],[0,\frac 12],[\frac 12,\infty)$ and find that this happens only when $x\in[-\frac 23,0]$ (where they are all non-negative).

Answer 3

$|2x-1|+|3x+2|+|x|=3$, we have $2x-1$ is negative when $x<\frac{1}{2}$, $3x+2$ is negative for $x<\frac{-2}{3}$ and $x$ is negative for $x<0$.

Now for $x<\frac{-2}{3}\Rightarrow 2x-1 -(3x+2)+x=3\Rightarrow -1=3$, as such for $x<\frac{-2}{3}$ we have no solution. If $x=\frac{-2}{3}$, then we have $\left|\frac{-4}{3}-1\right|+0+\left|\frac{-2}{3}\right| = \frac{7}{3}+\frac{2}{3} = 3$, so $\boxed{x=\frac{-2}{3}}$ is a solution.

For $\frac{-2}{3}<x<0, 2x-1+3x+2-x=4x-1=3\Rightarrow x=1$, but $1$ is not in that interval as such it is not a solution.

For $0\leq x < \frac{1}{2}, 1-2x+3x+2+x = 2x+3=3\Rightarrow \boxed{x=0}$.

For $x\geq \frac{1}{2}, 2x-1+3x+2+x = 6x + 1 = 3\Rightarrow 6x=2\Rightarrow \boxed{x=\frac{2}{3}}$