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 $\text{(D)}$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2002 AMC 10B Problems/Problem 13

Contents

Problem

Solution

Solution 2

See Also