2002 AMC 10B Problem 13

Treat $x$ as fixed and look at the expression as a function of $y$. Collecting the $y$-terms,
$$8xy - 12y + 2x - 3 = (8x - 12)\,y + (2x - 3).$$

A linear expression in $y$ equals $0$ for every $y$ only if its slope and its constant term are both zero:
$$8x - 12 = 0 \quad\text{and}\quad 2x - 3 = 0.$$
Both equations give $x = \frac32$, so this single value works (and it is the only one).

Check: at $x = \frac32$ the expression is $12y - 12y + 3 - 3 = 0$ regardless of $y$.

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

"True for all $y$" means the expression is the zero polynomial in $y$, so every coefficient must vanish. Equivalently, the expression factors as $(2x-3)(4y+1)$, and the factor $4y + 1$ is not zero for most $y$, so $2x - 3$ must be. Either view isolates the condition on $x$ alone.

Plug in two convenient $y$ values. With $y = 0$: $2x - 3 = 0$, so $x = \frac32$. With $y = 1$: $8x - 12 + 2x - 3 = 10x - 15 = 0$, again $x = \frac32$. Consistent, so $x = \frac32$ is the answer.

• Factoring to (2x - 3)(4y + 1) and then also solving 4y + 1 = 0, reporting -1/4 as if it were a value of x.
• Solving only $2x - 3 = 0$ without noticing that the $y$-coefficient must also vanish (it does here, but it is part of the argument).