Maclaurin 2016 Q4

For convenience, we number the equations, as follows:

By adding equation (1) to \(5 \times\) equation (2), we obtain

that is,

Thus either \(8x + 3y = 0\) or \(x + y = 0\).

In the first case, from equation (2) we obtain \(25x^2 = 9\), so that \(x = \pm\tfrac{3}{5}\). Because \(8x + 3y = 0\), if \(x = \tfrac{3}{5}\), then \(y = -\tfrac{8}{5}\), and if \(x = -\tfrac{3}{5}\), then \(y = \tfrac{8}{5}\).

Checking, we see that each of these is also a solution to equation (1).

In the second case, from equation (2) we get \(0 = 1\), so this is impossible.

Hence the solutions of the given simultaneous equations are \(x = \tfrac{3}{5}\), \(y = -\tfrac{8}{5}\); and \(x = -\tfrac{3}{5}\), \(y = \tfrac{8}{5}\).

Alternative

We may factorise the left-hand side of each of the given equations as follows:

From equation (4) it follows that \(x + y = \pm 1\).

When \(x + y = 1\), from equation (3) we get \(3x - 2y = -5\). Solving these two linear simultaneous equations, we obtain \(x = -\tfrac{3}{5}\) and \(y = \tfrac{8}{5}\).

Similarly, when \(x + y = -1\), we obtain \(x = \tfrac{3}{5}\) and \(y = -\tfrac{8}{5}\).

Checking, we see that each of these is also a solution to the original equations.

Hence the solutions of the given simultaneous equations are \(x = \tfrac{3}{5}\), \(y = -\tfrac{8}{5}\); and \(x = -\tfrac{3}{5}\), \(y = \tfrac{8}{5}\).