Maclaurin 2017 Q5
5. Solve the pair of simultaneous equations
Maclaurin Olympiad 2017, Question 5. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: \((a, b) = \bigl(\tfrac{3}{2}, \tfrac{1}{2}\bigr)\) or \((a, b) = \bigl(-\tfrac{1}{2}, -\tfrac{3}{2}\bigr)\)
Since \(a^2 - b^2 = (a - b)(a + b)\), we may write the first equation as \((a + b)^2(a - b) = 4\).
Note that \(a - b \ne 0\), since this would make the left-hand side of both equations zero, which is false. Hence we can divide the first equation by the second and cancel the term \(a - b\) to produce
Multiplying each side by \(5\bigl(a^2 + b^2\bigr)\), we get
When we multiply this out and collect like terms, we obtain
so either \(a = 3b\) or \(b = 3a\).
We substitute each of these in turn back into the first equation.
\(\boldsymbol{a = 3b}\)
Then \(4b \times 8b^2 = 4\), so that \(b = \tfrac{1}{2}\) and \(a = \tfrac{3}{2}\).
\(\boldsymbol{b = 3a}\)
Then \(4a \times \bigl(-8a^2\bigr) = 4\), so that \(a = -\tfrac{1}{2}\) and \(b = -\tfrac{3}{2}\).
Hence we have two solutions \((a, b) = \bigl(\tfrac{3}{2}, \tfrac{1}{2}\bigr)\) or \((a, b) = \bigl(-\tfrac{1}{2}, -\tfrac{3}{2}\bigr)\). These solutions should be checked by substituting back into the second equation.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2017 solutions.