Algebra II

$6 \div 2 = 3$, one $x$ cancels leaving $x$, and the $y$ cancels completely.

Two numbers multiplying to 15 and adding to 8 are 3 and 5. Expanding confirms it: $x^2 + 5x + 3x + 15$.

Two numbers multiplying to 12 and adding to $-7$ are $-3$ and $-4$, so $(x-3)(x-4) = 0$.

Subtract 3 from both sides to get $2x > 8$, then divide by 2. Dividing by a positive number leaves the sign unchanged.

The $y$ terms have opposite signs, so add the equations: $5x = 15$, giving $x = 3$. Then $2(3) + y = 7$, so $y = 1$.

The top is $(x-2)(x+2)$ and the bottom is $(x-3)(x+2)$. Cancelling the factor $(x+2)$ leaves $\dfrac{x-2}{x-3}$.