3 Equations 3 Unknowns | Irish Mathematics Olympiad 1992

[echandler]

Solving in terms of the variables squared first - very nice! A great *aha* moment indeed.

[peamutbubber]

literally my first thought was alpha beta gamma because edexcel love their roots of polynomials questions 😳

[m.h.]

While there are likely other, more complex solutions, the easiest solutions are found by ignoring the second formula and only focus on the first and third:
Which numbers squared add up to 9, while multiplied result in -4?
There is only one set of digits with two varieties, that fit that criteria: 1, 2, 2 with either one negative or all three negatives.
Given, that x, y and z are completely interchangeable, there is no way to assign a specific value to a specific variable.
So there are 3 basic solutions:
-1, -2, -2
-1, 2, 2
1, -2, 2 (which also fits 1, 2, -2 as order is interchangeable)

[thorinpalladino]

Technically your answer is incomplete as you did not specify the domain for the correct answers. There are several imaginary number solutions.
If the domain is integers then there are easier ways to determine the answer, like the method shown by M. H.
If the domain is real numbers then you have to do all the algebra in the video.
In any case you should list the domain of the solutions,

[bait]

nice to know you don't need the authors method to solve.

[rishavbagri]

If u are bored solve this
find range of a for all value of y lie in R as y = (ax²+3x-4)/(3x-4x²+9)
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Ans- a€(1, 7)

[dandeleanu]

There are 24 solutions all together not only 12

[ΠασχάληςΜπαμπουλης]

Can you solve; 2(x^y)=2005+y
(It's 2 times x^y and x, y are natural numbers)

[aimlessexplorer]

Did in my head in 20 seconds. You can quickly see that 33 and 9 are 1 away from numbers that have factors -2, 2, 1.