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Algebra Difficulty 1.6 Junior Find the answer Canada

Suppose that $x$ and $y$ are integers satisfying $-4\le x\le 11$ and $9\le y\le 17$ . Which of the following
statements is correct for all such values of $x$ and $y$ ?

$-2\le x-y\le 21$
$6\le x-y\le 13$
$-13\le x-y\le -6$
$5\le x-y\le 28$
$-21\le x-y\le 2$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The least possible value of $x-y$ is when $x$ is minimized and $y$ is maximized.

The least possible value of $x$ is
$-4$ and the greatest possible value
of $y$ is $17$ , in which case we get $x-y = -4-17=-21$ .

Attaining $x-y = -21$ rules out
options (A) $\$-2\leq x-y\leq$
21 $, (B)$ $6\leq x-y\leq$ 13 $, (C)$ $-13\leq x-y\leq$ -6$, and
(D) $\$5\leq x-y\leq$
28 $since$ -21$ is
less than $-2$ , $6$ , $-13$ , and $5$ .

Note that the greatest possible value of $x-y$ is when $x$ is maximized and $y$ is minimized. This happens when $x=11$ and $y=9$ , giving $x-y=11-9=2$ , so indeed $x-y\leq 2$ (as in (E)) is true for all
such $x$ and $y$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.