Dgdrxrt

This question has been asked here, but its results are inconclusive.

Suppose that $a$ is a number such that the system of equations

$$|2x| − y = 5$$

$$x − |2y + 2| = a$$

has an odd number of distinct solutions. What is the product of all
possible values of $a$?

I start with a little simplifying first.

The first equation gives $$y=pm2x-5$$

The second equation gives $$y=fracpm(x-a)-22$$

After getting these equations, I don't see any way to go further.

Therefore, I then try another approach using logic.

The first equation is a "v" shifted down $5$(hence the "-5") units and vertically stretched by $2$(hence the "2x") giving it a $22.5^circ$ angle with the $y$-axis.

As for the second one, I am not that sure how to graph. How do I graph the second equation?

The maximum number of odd intersections, that is, solutions to the two equations is obviously $3$.

Using similar logic, there is then cases where there are 1 intersections.

There are three cases with an odd number of solutions.

Case 1: 3 solutions; The second equation's "v" has to have 1 side of the "v" intersect with the first equation two times, and the second line of the "v" touch the vertex of the first equation.

Case 2: 3 solutions; The second equation has its vertex on a side of the "v" of the first equation, and has the lines of the "v" intersecting with the other line of the "v" of the first equation.

Case 3: 1 solution; The vertex of the second equation touches the side of the "v" of the first equation.

Are there any cases I have missed?

I can see that if I know that line of symmetry $y=c$(where $c$ is a constant) of the second equation, I can set equations for the two equations to find $a$ for all three cases I have found. How do I do that? Is there any way to find the line of symmetry for the second equation using its equation with a number missing?

Thanks! Your help is appreciated!

Max0815

Note: Thank you to @WW1 for hinting. He gets an upvote! :)

There are 2 absolute values. An absolute value means two solutions, so in this problem, there are $2+2=4$ cases. However, there are three cases that have odd solutions, given in the diagram:

The first absolute value is a "v" with a vertical stretch of $2$ and a vertex at $(0, -5)$. We note that the vertex occurs whenever the stuff in the absolute value signs is equal to $0$. So for the second equation, $|2y+2|=0$ and we conclude that $x=a$.

Furthermore, setting $a=0$ and graphing, we can conclude from the data that the second equation's line of symmetry is $y=-1$. We can then conclude from this that the vertex of the second equation is $(a, -1)$, and the vertex shifts with the change of $a$.

According the the diagram, we can see there are three cases. Let's start with the first case(the furthest one on the left):

We see that the vertex of the first equation is the solution to both. Plugging in the $x$ and $y$, we can solve for the $a$: I set $0=|2(-5)+2|+a$ so $a=-8$.

For the second case, we can see that there is a solution in the form of $(x, -1)$. Plugging this in the first equation gives $x=-2$. Plugging this into the second equation gives $a=-2$.

The third case has a solution at $(x, -1)$ again. Realizing that this is an absolute value equation, $x$ can also equal $2$. So we have the pair $(2, -1)$ which gives $a=2$ when I plug this into the second equation.

As a result, we have three solutions for $a$: $-8, -2$, and $2$.

Multiplying these together gives $(-8)cdot(-2)cdot2=boxed32$