JAMB 1991 · UME · Q20

Find the two values of $y$ which satisfy the simultaneous equations $3x + y = 8$ and $x^2 + xy = 6$ .

Worked solution (try it first)

Make $y$ the subject of the linear equation: $y = 8 - 3x$ .
Substitute: $x^2 + x(8 - 3x) = 6$ , so $-2x^2 + 8x - 6 = 0$ .
Divide by $-2$ : $x^2 - 4x + 3 = 0$ .
Factorise: $(x - 1)(x - 3) = 0$ , so $x = 1$ or $x = 3$ .
Then $y = 8 - 3x$ gives $y = 5$ or $y = -1$ , option A.

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