2002 AMC 10B Problem 13

"True for all $y$" means: pick $x$ first, and then the expression must be $0$ whatever $y$ you choose.

1. Gather the terms that contain $y$:
$$8xy - 12y + 2x - 3 = (8x - 12)\,y + (2x - 3).$$
2. For fixed $x$ this is "a number times $y$, plus another number." Think about what happens as $y$ changes. If the number in front of $y$ were not zero, the expression would change as $y$ changes, so it could not always be $0$.
3. So we need $8x - 12 = 0$. And then, with the $y$ part gone, the leftover $2x - 3$ must also be $0$.
4. Both give $x = \frac32$.
5. Check: at $x = \frac32$ the expression is $12y - 12y + 3 - 3 = 0$ for every $y$.

The answer is $\boxed{\textbf{(D)}\ \frac32}$.

An expression that must be zero for every value of a variable cannot depend on that variable at all, so everything multiplying the variable must be zero, and so must what is left. The same idea shows up when an equation is supposed to hold "for all $x$": match up the pieces and set each one to zero.

Test two easy values of $y$. With $y = 0$ the expression is $2x - 3$, which must be $0$, so $x = \frac32$. With $y = 1$ it is $8x - 12 + 2x - 3 = 10x - 15$, which again needs $x = \frac32$. Both agree. (You can also factor: the expression equals $(2x - 3)(4y + 1)$, and since $4y + 1$ is usually not zero, $2x - 3$ has to be.)