Let $h$ be the number of hours. The total cost is modeled by $4+2h=18$. Subtract $4$ from both sides to get $2h=14$, then divide by $2$: $h=7$. So the car was parked for $7$ hours.

Let $c$ represent the number of classes. The total cost is modeled by $12+8c=44$. Subtract $12$ from both sides to get $8c=32$. Then divide by $8$: $c=4$. So $4$ classes were taken.

Distribute the $5$ across the parentheses: $5(x-2)=5x-10$. Then add $3$: $5x-10+3=5x-7$. So the equivalent expression is $5x-7$.

Both given points have the same $y$-coordinate, $3$. That means the line is horizontal, so its equation is $y=3$.

Let $p$ be the number of pens and $n$ be the number of notebooks. The situation gives the system $p+n=7$ and $2p+5n=23$. Substitute $p=7-n$ into the cost equation: $2(7-n)+5n=23$. Then $14-2n+5n=23$, so $3n=9$ and $n=3$. The student bought $3$ notebooks.