2003 CEMC Pascal Problems/Problem 11

Problem

If $x = 2$ and $y = -3$ satisfy the equation $2x^{2} + kxy = 4$ , then the value of $k$ is $\text{ (A) }\ \frac{2}{3} \qquad\text{ (B) }\ 0 \qquad\text{ (C) }\ \frac{4}{3} \qquad\text{ (D) }\ -\frac{2}{3} \qquad\text{ (E) }\ -2$

Solution

We can simply plug in the values of $x$ and $y$ into the equation, and then solve for $k$ :

$2x^{2} + kxy = 4$

$2 \times 2^{2} + k \times 2 \times -3 = 4$

$8 - 6k = 4$

$-6k = -4$

$k = \frac{-4}{-6} = \boxed {\textbf {(A) } \frac{2}{3}}$

~anabel.disher

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2003 CEMC Pascal Problems/Problem 11

Problem

Solution