, 2012

Since $ax+ay=4$, then $a(x+y)=4$.

Since $x+y=12$, then $12a=4$ or $a=\tfrac{4}{12}=\tfrac{1}{3}$.
Since the two lines are parallel, then their slopes are equal.

We re-write the given equations in the form “$y=mx+b$".

The first equation becomes $6y = -4x+5$ or $y = -\frac{4}{6}x+\frac{5}{6}$ or $y = -\frac{2}{3}x+\frac{5}{6}$.

Since the first line is not vertical, then the second line is not vertical, and so $k \neq 0$.

The second equation becomes $ky = -6x+3$ or $y = -\frac{6}{k}x+\frac{3}{k}$.

Therefore, $-\frac{2}{3} = -\frac{6}{k}$ and so $\frac{k}{6}=\frac{3}{2}$ or $k = 6\times \frac{3}{2} = 9$.
Adding the two equations, we obtain $x+x^2=2$ or $x^2+x-2=0$.

Factoring, we obtain $(x+2)(x-1)=0$, and so $x=-2$ or $x=1$.

From the first equation, $y=-x$. If $x=-2$, then $y=2$ and if $x=1$, then $y=-1$.

Therefore, the solutions are $(x,y)=(-2,2)$ and $(x,y)=(1,-1)$.

(We can check that each of these solutions satisfies both equations.)