JAMB 2012 · UTME · Q13

Solve for $x$ and $y$ in the equations $x^2 - y^2 = 4$ and $x + y = 2$ .

Worked solution (try it first)

Factorise the difference of two squares: $x^2 - y^2 = (x + y)(x - y) = 4$ .
Since $x + y = 2$ , this gives $2(x - y) = 4$ , so $x - y = 2$ .
Add to $x + y = 2$ : $2x = 4$ , so $x = 2$ and $y = 0$ , option C.

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