How do you solve the system $x² + y² = 16$ and $x + y = 4$ ?

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Answer 1 · 2016-05-03T01:13:51

$y = 4 - x -> x^2 + (4 - x)^2 = 16$

$x^2 + 16 - 8x + x^2 = 16$

$2x^2 - 8x = 0$

$2x(x - 4) = 0$

$x = 0 and 4$

$0 + y = 4 or 4 + y = 4$

$y = 4 or 0$

The solution sets are ${0, 4} and {4, 0}$

Hopefully this helps!

How do you solve the system x² + y² = 16x² + y² = 16 and x + y = 4x + y = 4?