Hamilton 2017 Q6
6. Solve the pair of simultaneous equations
Hamilton Olympiad 2017, Question 6. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: \(x = -5\), \(y = -5\); \(x = 2\), \(y = 2\); or \(x = 3\), \(y = \tfrac{1}{3}\)
First, let us number the two given equations, so that it is easy to refer to them.
It is possible to eliminate one of the two unknowns by substituting from equation (2) into equation (1), but this leads to a cubic equation. We present another method that avoids this.
By subtracting \(x \times\) equation (2) from equation (1), we get
so that
Hence either \(x = 3\) or \(x = y\). We deal with each of these two cases separately.
\(\boldsymbol{x = 3}\)
\(\boldsymbol{x = y}\)
By checking in the two given equations, we find that all three solutions are valid. Thus there are three solutions of the simultaneous equations, namely
or \(x = 2\) and \(y = 2\),
or \(x = 3\) and \(y = \tfrac{1}{3}\).
Solution © UK Mathematics Trust, from the Hamilton Olympiad 2017 solutions.