(a) Determine the point of intersection of the lines with equations $y=4 x-32$ and $y=-6 x+8$.

(b) Suppose that $a$ is an integer. Determine the point of intersection of the lines with equations $y=-x+3$ and $y=2 x-3 a^{2}$. (The coordinates of this point will be in terms of $a$.)

(c) Suppose that $c$ is an integer. Show that the lines with equations $y=-c^{2} x+3$ and $y=x-3 c^{2}$ intersect at a point with integer coordinates.

(d) Determine the four integers $d$ for which the lines with equations $y=d x+4$ and $y=2 d x+2$ intersect at a point with integer coordinates.
(a) To determine the point of intersection, we equate the two expressions for $y$ to successively obtain:

$$
\begin{aligned}
4 x-32 & =-6 x+8 \\
10 x & =40 \\
x & =4
\end{aligned}
$$

When $x=4$, using the equation $y=4 x-32$, we obtain $y=4 \cdot 4-32=-16$. Therefore, the lines intersect at $(4,-16)$.

(b) To determine the point of intersection, we equate the two expressions for $y$ to successively obtain:

$$
\begin{aligned}
-x+3 & =2 x-3 a^{2} \\
3+3 a^{2} & =3 x \\
x & =1+a^{2}
\end{aligned}
$$

When $x=1+a^{2}$, using the equation $y=-x+3$, we obtain $y=-\left(1+a^{2}\right)+3=2-a^{2}$. Therefore, the lines intersect at $\left(1+a^{2}, 2-a^{2}\right)$.

(c) Since $c$ is an integer, then $-c^{2}$ is a integer that is less than or equal to 0 .

The two lines have slopes $-c^{2}$ and 1 . Since $-c^{2} \leq 0$, these slopes are different, which means that the lines are not parallel, which means that they intersect.

To determine the point of intersection, we equate the two expressions for $y$ to successively obtain:

$$
\begin{aligned}
-c^{2} x+3 & =x-3 c^{2} \\
3+3 c^{2} & =x+c^{2} x \\
3+3 c^{2} & =x\left(1+c^{2}\right)
\end{aligned}
$$

Since $c^{2} \geq 0$, then $1+c^{2} \geq 1$, which means that we can divide both sides by $1+c^{2}$ to obtain $x=\frac{3+3 c^{2}}{1+c^{2}}=3$.

In particular, this means that the $x$-coordinate of the point of intersection is an integer. When $x=3$, using the equation $y=-c^{2} x+3$, we obtain $y=-c^{2} \cdot 3+3=3-3 c^{2}$. Since $c$ is an integer, then $y=3-3 c^{2}$ is an integer.

Therefore, the lines intersect at a point whose coordinates are integers.

(d) To determine the point of intersection in terms of $d$, we equate the two expressions for $y$ to successively obtain:

$$
\begin{aligned}
d x+4 & =2 d x+2 \\
2 & =d x
\end{aligned}
$$

For the value of $x$ to be an integer, we need $d \neq 0$ and $\frac{2}{d}$ to be an integer.

Since $d$ is itself an integer, then $d$ is a divisor of 2 , which means that $d$ equals one of 1 , $-1,2,-2$.

We still need to confirm that, for each of these values of $d$, the coordinate $y$ is also an integer.

When $x=\frac{2}{d}$, using the equation $y=d x+4$, we obtain $y=d \cdot\left(\frac{2}{d}\right)+4=2+4=6$.

Therefore, when $d=1,-1,2,-2$, the lines intersect at a point with integer coordinates.

We can verify this in each case:

- $d=1$ : The lines with equations $y=x+4$ and $y=2 x+2$ intersect at $(2,6)$.
- $d=-1$ : The lines with equations $y=-x+4$ and $y=-2 x+2$ intersect at $(-2,6)$.
- $d=2$ : The lines with equations $y=2 x+4$ and $y=4 x+2$ intersect at $(1,6)$.
- $d=-2$ : The lines with equations $y=-2 x+4$ and $y=-4 x+2$ intersect at $(-1,6)$.