Figure 2 shows a sketch of the graph with equation $$y = 2|x + 4| - 5$$ The vertex of the graph is at the point P, shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 13 - 2020 - Paper 2

To solve the equation

$3x + 40 = 2|x + 4| - 5,$

we first simplify the right side:

$3x + 40 = 2|x + 4| - 5$ $3x + 45 = 2|x + 4|.$

Next, we can isolate the absolute value:

$|x + 4| = \frac{3x + 45}{2}.$

We will solve this by considering two cases for the absolute value.

Case 1: x + 4 ≥ 0

In this case,

$x + 4 = \frac{3x + 45}{2}$.

Multiplying through by 2:

$2(x + 4) = 3x + 45$ $2x + 8 = 3x + 45$ $x = 45 - 8 = 37.$

Case 2: x + 4 < 0

Here,

$-(x + 4) = \frac{3x + 45}{2}$ $-x - 4 = \frac{3x + 45}{2}.$

Multiplying through by 2:

$-2x - 8 = 3x + 45$ $-8 - 45 = 5x$ $x = -\frac{53}{5}.$

Thus, the two solutions are:

$x = 37 \text{ and } x = -\frac{53}{5}.$

To find the range of possible values of a, we consider when the line

$y = ax$

intersects the given graph. For intersections to occur, the slope a must ensure that the line can touch or cross the V-shaped graph. Since the vertex is downward at P(-4, -5), the intersection occurs when

$a > 2 \text{ or } a < -2.$

Thus, the range of values for a in set notation is:

$\{ a : a > 2 \} \cup \{ a : a < -2 \}.$