|x+2|+|x-3|=7 equation

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

1.
$$x - 3 \geq 0$$
$$x + 2 \geq 0$$
or
$$3 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 3\right) + \left(x + 2\right) - 7 = 0$$
after simplifying we get
$$2 x - 8 = 0$$
the solution in this interval:
$$x_{1} = 4$$

2.
$$x - 3 \geq 0$$
$$x + 2 < 0$$
The inequality system has no solutions, see the next condition

3.
$$x - 3 < 0$$
$$x + 2 \geq 0$$
or
$$-2 \leq x \wedge x < 3$$
we get the equation
$$\left(3 - x\right) + \left(x + 2\right) - 7 = 0$$
after simplifying we get
incorrect
the solution in this interval:

4.
$$x - 3 < 0$$
$$x + 2 < 0$$
or
$$-\infty < x \wedge x < -2$$
we get the equation
$$\left(3 - x\right) + \left(- x - 2\right) - 7 = 0$$
after simplifying we get
$$- 2 x - 6 = 0$$
the solution in this interval:
$$x_{2} = -3$$


The final answer:
$$x_{1} = 4$$
$$x_{2} = -3$$