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2003 CEMC Pascal Problems/Problem 14

Revision as of 17:41, 21 June 2025 by Anabel.disher (talk | contribs) (Created page with "==Problem== If <math>x</math> and <math>y</math> are positive integers and <math>x + y = 5</math>, then a possible value for <math>2x - y</math> is <math> \text{ (A) }\ 3 \qq...")

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Contents

1 Problem
2 Solution 1
3 Solution 2
4 Solution 2.5

Problem

If $x$ and $y$ are positive integers and $x + y = 5$ , then a possible value for $2x - y$ is

$\text{ (A) }\ 3 \qquad\text{ (B) }\ -3 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ -2 \qquad\text{ (E) }\ 0$

Solution 1

We can list out the possible values for $x$ and $y$ :

$x = 1$ and $y = 4$

$x = 2$ and $y = 3$

$x = 3$ and $y = 2$

$x = 4$ and $y = 1$

We can now try these values out in the expression to see if they match any of the answer choices:

$2 \times 1 - 4 = -2$

This happens to be $\boxed {\textbf {(D) } -2}$ .

~anabel.disher

Solution 2

From $x + y = 5$ , we can see that:

$x = 5 - y$

Plugging this into the second equation, we get:

$2x - y = 2(5 - y) - y = 10 - 3y$

We can now try to plug in positive values for $y$ until we get one of the answers:

$10 - 3 \times 1 = 7$

$10 - 3 \times 2 = 4$

$10 - 3 \times 3 = 1$

$10 - 3 \times 4 = -2$

This happens to be $\boxed {\textbf {(D) } -2}$ .

~anabel.disher

Solution 2.5

We can do something similar with solution 2 with the same process, but use $y = 5 - x$ instead. This gives us:

$2x - y = 2x - (5 - x) = 2x - 5 + x = 3x - 5$

We can now plug in positive values of $x$ until we get one of the answers:

$3 \times 1 - 5 = -2$

This happens to be $\boxed {\textbf {(D) } -2}$ .

~anabel.disher

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