UKMT Special (Problem $19$)

Solve the pair of simultaneous equations:

$x^2 + 3y = 10$

$3 + y = \frac{10}{x}$

[UKMT Hamilton Olympiad $2017$, H$6$]

Note by Yajat Shamji
8 months, 2 weeks ago

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You have until next Tuesday, $3:00$pm!

- 8 months, 2 weeks ago

Step 1) Substitute y

   Equation 2    =>    y = 10/x - 3
Equation 1     =>    x^2 + 3(10/x -3) = 10
=>    x^2 + 30/x - 9 = 10
=>    x^2 + 30/x - 19 = 0
=>    x^3 - 19x + 30 = 0


Step 2) Factor the cubic Let P(x) = x^3 - 19x + 30

   According to rational root theorem, the only rational root possible is
(x +- 30), (x +- 15), (x +- 10), (x +- 5), (x +- 3), (x +- 2), (x +- 1)

According to polynomial factor theorem, if P(a) = 0, then (x - a) is a root.
We try all the possible rational factor.
P(-5) = P(2) = P(3) = 0

We can conclude that P(x) = (x + 5)(x - 2)(x - 3)


Step 3) Finding x and y x^3 - 19x + 30 = 0 (x + 5)(x - 2)(x - 3) = 0 x = -5 or x = 2 or x = 3

 Now, find y by substituting x to equation 2.
x = -5   =>    y = -5
x = 2    =>    y = 2
x = 3    =>    y = 1/3


The solutions are (x,y) = {(-5, -5), (2, 2), (3, 1/3)}

- 8 months, 2 weeks ago