Q:

Precalculus equation solving with inequalities

For which a $\in \mathbb{R}$ does the equation
$$|x-1| + 2|x-2| = a$$
have two solutions?
I divided this into three cases: 
Case 1
$x \geq 2$:
$$x-1 +2(x-2) = a$$
$$x-1+2x-4 = a$$
$$3x-5 = a$$
$$x = \frac{a +5}{3}$$
This has a solution if $a \geq 1$ to match the requirement $x \geq 2$.
Case 2
$1 < x < 2$:
$$-2 (x-2)+x-1 = a$$
$$-x+3 = a$$
$$x = -a + 3$$
This has a solution if $1 < a < 2$.
Case 3
$x \leq 1$
$$-2(x-2)-x+1 = a$$
$$-2x+4-x+1 = a$$
$$-3x+5 = a$$
$$x = \frac{a-5}{-3}$$
This has a solution if $a \geq 2$.
However, I do not quite understand how to make a judgement about what values for a this equation has two solutions, since there are no square terms?

A:

You have seen that if $a\geq 1$ then this has a solution $x_0$ where $x_0\geq 2$, and you have also seen that if $a\geq 2$, then this has a solution $x_1 \leq 1$. Combining these two, you can conclude that if $a\geq 2$ then your equation has two solutions, one less than or equal to $1$ and one greater than or equal to $2$.
$\textbf{Note}:$ For better understanding I suggest you draw the graph of $|x+1|+2|x-2|$.

