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Algebra Difficulty 4.7 AIME Find the answer Canada

The real numbers $x$ , $y$ and $z$ satisfy the three equations $\begin{aligned} x+y &= 7\\ xz&=-180\\ (x+y+z)^2&=4\end{aligned}$ If $S$ is the sum of the two possible values of $y$ , then $-S$ equals

$56$
$14$
$36$
$34$
$42$

Your answer

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

Solution

Since $x+y=7$ , then $x+y+z=7+z$ .

Thus, the equation $(x+y+z)^2 = 4$ becomes $(7+z)^2 = 4$ .

Since the square of $7+z$ equals 4, then $7+z=2$ or $7+z=-2$ .

If $7+z=2$ , then $z=-5$ .

In this case, since $xz = -180$ , we get $x = \dfrac{-180}{-5} = 36$ which gives $y = 7 - x = -29$ .

If $7+z=-2$ , then $z = -9$ .

In this case, since $xz = -180$ , we get $x = \dfrac{-180}{-9} = 20$ which gives $y = 7 - x = -13$ .

We can check by direct substitution that $(x,y,z)=(36,-29,-5)$ and $(x,y,z)=(20,-13,-9)$ are both solutions to the original system of equations.

Since $S$ is the sum of the possible values of $y$ , we get $S = (-29)+(-13) = -42$ and so $-S=42$ .

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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.