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Question #2128f

1 Answer

Feb 24, 2016

$x=2$

Explanation:

$x^2 + y^2 = 8$
$x-y =4$

$:.y = x-4$

substituting this back into the first equation gives
$x^2 + (x-4)^2 = 8$

$x^2 +x^2 - 8x +16 = 8$
$2x^2 - 8x +8 =0$
$x^2 -4x +4 = 0$
$(x-2)^2 = 0$

$:.x=2$

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