|x-2|+|x-4|=3 equation

For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x - 4 \geq 0$$
$$x - 2 \geq 0$$
or
$$4 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 4\right) + \left(x - 2\right) - 3 = 0$$
after simplifying we get
$$2 x - 9 = 0$$
the solution in this interval:
$$x_{1} = \frac{9}{2}$$

2.
$$x - 4 \geq 0$$
$$x - 2 < 0$$
The inequality system has no solutions, see the next condition

3.
$$x - 4 < 0$$
$$x - 2 \geq 0$$
or
$$2 \leq x \wedge x < 4$$
we get the equation
$$\left(4 - x\right) + \left(x - 2\right) - 3 = 0$$
after simplifying we get
incorrect
the solution in this interval:

4.
$$x - 4 < 0$$
$$x - 2 < 0$$
or
$$-\infty < x \wedge x < 2$$
we get the equation
$$\left(2 - x\right) + \left(4 - x\right) - 3 = 0$$
after simplifying we get
$$3 - 2 x = 0$$
the solution in this interval:
$$x_{2} = \frac{3}{2}$$


The final answer:
$$x_{1} = \frac{9}{2}$$
$$x_{2} = \frac{3}{2}$$