Maths Olympiad Prep

Track / Stage 4 / 214 of 340 #474 of 1964

Problem 474

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer

## Task A-2.4.

Find all pairs (x,y)(x, y) of real numbers for which

xy3=135,(x+y)y=6 x y^{3}=-135, \quad(x+y) y=-6

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solutions — 2

Solution 1

## Solution.

Let's introduce the substitution a=xy,b=y2a=xy, b=y^2.

Then the equations become ab=135,a+b=6ab=-135, a+b=-6. According to Viète's formulas, aa and bb satisfy the quadratic equation t2+6t135=0t^2 + 6t - 135 = 0.

Solving this equation, we get the solutions t=15t=-15 and t=9t=9.

We have two possibilities: y2=9,xy=15y^2=9, xy=-15 or y2=15,xy=9y^2=-15, xy=9. We discard the second possibility because the square of a real number cannot be negative.

Therefore, y2=9y^2=9, so y=3y=3 or y=3y=-3. From xy=15xy=-15, we get two solutions, (x,y)=(5,3)(x, y)=(-5,3) and (x,y)=(5,3)(x, y)=(5,-3).

1 point

Solution 2

Scoring: Introducing the substitution is worth 2 points. Deriving the quadratic equation for aa and bb is worth 1 point, and solving it is worth 1 point. Discarding the case y2=15y^{2}=-15 is worth 1 point, and stating the solution is worth 1 point. Writing the solution without the process is worth 1 point.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.