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Algebra Difficulty 2.5 Junior Find the answer Canada

Suppose that $x$ and $y$ are real numbers with $-4 \leq x \leq -2$ and $2 \leq y \leq 4$ . The greatest possible value of $\dfrac{x + y}{x}$ is

$1$
$-1$
$-\frac{1}{2}$
$0$
$\frac{1}{2}$

Your answer

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

Solution

We note that $\dfrac{x+y}{x} = \dfrac{x}{x} + \dfrac{y}{x} = 1 + \dfrac{y}{x}$ .
The greatest possible value of $\dfrac{x+y}{x} = 1 + \dfrac{y}{x}$ thus occurs when $\dfrac{y}{x}$ is as great as possible.
Since $x$ is always negative and $y$ is always positive, then $\dfrac{y}{x}$ is negative.
Therefore, for $\dfrac{y}{x}$ to be as great as possible, it is as least negative as possible (i.e. closest to 0 as possible).
Since $x$ is negative and $y$ is positive, this happens when $x$ is as negative as possible and $y$ is as small as possible – that is, when $x=-4$ and $y=2$ .
Therefore, the greatest possible value of $\dfrac{x+y}{x}$ is $1 + \dfrac{2}{-4} = \dfrac{1}{2}$ .

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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.