What is the sum of values that $a$ can take?

Question

$x$ and $y$ are negative integers.

$y+x = -5$

$\left\lvert x-y \right\rvert + a \cdot \left\lvert y-x \right\rvert = 12$

• What is the sum of values that $a$ can take?

I know that

$\left\lvert y-x \right\rvert = \left\lvert x-y \right\rvert$

Answer 1

in the Case of $$x\geq y$$ we have to solve $$(x-y)(a+1)=12$$ and $$x+y=-5$$ in the Case of $$x<y$$ we have to solve $$(-x+y)(a+1)=12$$ and $$x+y=-5$$ Can you proceed? we find $$a=3,x=-4,y=-1$$ or $$a=3,x=-1,y=-4$$ or $$a=11,x=-3,y=-2$$ or $$a=11,x=-2,y=-3$$ we Can also use that $$x-y=\frac{12}{a+1}$$ or $$-x+y=\frac{12}{a+1}$$

Answer 2

As you mention, we have $$|x-y| + a|x-y| = 12 \implies (a+1)|x-y| = 12$$ after factoring out $|x-y|$, which means that, after doing some algebra, $$a=\dfrac{12}{|x-y|}-1\text{.}$$ Now, given that $x, y$ are negative integers which sum to $-5$, you could have $$\begin{array}{|c|c|c|} \hline x & y & |x-y|\\ \hline -1 & -4 & 3 \\ -2 & -3 & 1\\ -3 & -2 & 1\\ -4 & -1 & 0\\ \hline \end{array}$$ Assuming that $x$ and $y$ are distinct (you can't divide by $0$), you may plug in $3, 1$, and $1$ into the formula for $a$ above to obtain its possible values.