How do you solve $\left| 3x-2 \right|=\left| 4-x \right|$ ?

Hint: We have given to solve a linear equation in one variable consisting of two absolute value functions. We shall consider each absolute value function as a separate equation to be solved. Then, we will sketch the graphs of both the functions on the same plane and further obtain their points of intersection to compute our final solution.

Complete step by step solution:
Given that $\left| 3x-2 \right|=\left| 4-x \right|$.
Let $f\left( x \right)=\left| 3x-2 \right|$ and let $g\left( x \right)=\left| 4-x \right|$.
We shall sketch the graphs of both the functions taking each of them one-by-one.
In $f\left( x \right)=\left| 3x-2 \right|$, we can observe that this function resembles the absolute value function $\left| x \right|$ and can by modified as $f\left( x \right)=3\left| x-\dfrac{2}{3} \right|$.
We know that the graph of $\left| x \right|$ is as follows:

We will shift the graph $\dfrac{2}{3}$ points towards the right of the x-axis first. Doing so, we will get the graph for $\left| x-\dfrac{2}{3} \right|$ only. To change $f\left( x \right)$ to $\dfrac{f\left( x \right)}{3}$, we shall make the graph 3 points steeper, and thus the graph of $f\left( x \right)=3\left| x-\dfrac{2}{3} \right|$ will look as follows:


Now our second function$g\left( x \right)=\left| 4-x \right|$, can also be written as $g\left( x \right)=\left| x-4 \right|$. This is because multiplying the absolute value function with a negative sign does not make any difference to it. We will shift the graph 4 points towards the left of the x-axis first. Doing so, we will get the graph for $\left| x-4 \right|$ and thus the graph of $g\left( x \right)=\left| 4-x \right|$, will look as follows:

Finally, we shall sketch both the graphs shown above on the same plane.

From this, we can observe that the graphs intersect at two point, $\left( -1,5 \right)$ and $\left( \dfrac{3}{2},\dfrac{5}{2} \right)$.
Therefore, the solutions of the equation $\left| 3x-2 \right|=\left| 4-x \right|$ are $x=-1,\dfrac{3}{2}$.

Note: Another method of finding the solutions was by the algebraic approach instead of the graphical one. We shall open each absolute modulus function with a positive and a negative sign one-by-one. Thus, we will obtain four equations as a result of the positive and negative combinations of opening the absolute value function. We shall solve these equations thus obtained to find the solutions for the given function.