JAMB 2010 · UTME · Q16

Solve for $x$ and $y$ if $x - y = 2$ and $x^2 - y^2 = 8$ .

Worked solution (try it first)

Factorise the difference of two squares: $x^2 - y^2 = (x - y)(x + y)$ .
So $2(x + y) = 8$ , which gives $x + y = 4$ .
Add this to $x - y = 2$ : $2x = 6$ , so $x = 3$ .
Then $y = 1$ .
So $(x, y) = (3, 1)$ , option B.

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