JAMB 1987 · UME · Q23

Find two values of $y$ which satisfy the simultaneous equations $x + y = 5$ and $x^2 - 2y^2 = 1$ .

Worked solution (try it first)

Make $x$ the subject of the linear equation: $x = 5 - y$ .
Substitute: $(5 - y)^2 - 2y^2 = 1$ .
Expanding, $25 - 10y + y^2 - 2y^2 = 1$ .
Collect terms and multiply by $-1$ : $y^2 + 10y - 24 = 0$ .
Factorise: $(y + 12)(y - 2) = 0$ , so $y = -12$ or $y = 2$ , option C.

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