JAMB 1985 · UME · Q23

Solve for $(x, y)$ : $2x + y = 4$ and $x^2 + xy = -12$ .

Worked solution (try it first)

Make $y$ the subject of the linear equation: $y = 4 - 2x$ .
Substitute into the second: $x^2 + x(4 - 2x) = -12$ , which is $-x^2 + 4x = -12$ .
Rearrange and factorise: $x^2 - 4x - 12 = 0$ , so $(x - 6)(x + 2) = 0$ and $x = 6$ or $x = -2$ .
Then $y = 4 - 2x$ gives $y = -8$ and $y = 8$ .
So the solutions are $(6, -8)$ and $(-2, 8)$ , option A.