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2000 CEMC Pascal Problems/Problem 3

Revision as of 13:47, 2 July 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== If <math>x = 4</math> and <math>y = -3</math>, then the value of <math>\frac{x - 2y}{x + y}</math> is <math> \text{ (A) }\ -\frac{1}{2} \qquad\text{ (B) }\ -2 \qq...")
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Problem

If $x = 4$ and $y = -3$, then the value of $\frac{x - 2y}{x + y}$ is

$\text{ (A) }\ -\frac{1}{2} \qquad\text{ (B) }\ -2 \qquad\text{ (C) }\ \frac{10}{7} \qquad\text{ (D) }\ -\frac{2}{7} \qquad\text{ (E) }\ 10$

Solution

Plugging in $x = 4$ and $y = -3$ into the expression, we get:

$\frac{x - 2y}{x + y} = \frac{4 - 2 \times -3}{4 + (-3)}$

$=\frac{4 + 6}{4 - 3} = \frac{10}{1} = \boxed {\textbf {(E) } 10}$

~anabel.disher

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