# 2002 AMC 10B Problems/Problem 13

## Problem

Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$. $\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14$

## Solution

We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$.

As $(4y + 1)(2x - 3) = 0$ must be true for all $y$, we must have $2x - 3 = 0$, hence $\boxed{x = \frac 32\ \mathrm{ (D)}}$.

## Solution 2

Since we want only the $y$-variable to be present, we move the terms only with the $y$-variable to one side, thus constructing $8xy - 12y + 2x - 3 = 0$ to $8xy + 2x = 12y + 3$. For there to be infinite solutions for $y$ and there is no $x$, we simply find a value of $x$ such that the equation is symmetrical. Therefore, $2x = 3$ $8xy = 12y$

There is only one solution, namely $x = \boxed{\dfrac{3}{2}}$ or $\text{(D)}$

## See Also

 2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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