JAMB 1993 · UME · Q14

Solve the simultaneous equations $x^2 + y - 5 = 0$ and $y - 7x + 3 = 0$ for $x$ .

Worked solution (try it first)

Make $y$ the subject of the linear equation: $y = 7x - 3$ .
Substitute into $x^2 + y - 5 = 0$ : $x^2 + 7x - 3 - 5 = 0$ , so $x^2 + 7x - 8 = 0$ .
Factorise: $(x - 1)(x + 8) = 0$ , so $x = 1$ or $x = -8$ , option D.