How do you solve the system $x² + y² = 20$ and $x + y = 6$ ?

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Answer 1 · 2016-07-07T14:33:35

$y = 6 - x$

By substitution

$:.x^2 + (6 - x)^2 = 20$

$x^2 + 36 - 12x + x^2 = 20$

$2x^2 - 12x + 16 = 0$

$2(x^2 - 6x + 8) = 0$

$2(x - 4)(x - 2) = 0$

$x = 4 and 2$

$:.x + y = 6$

$4 + y = 6" and " 2 + y = 6$

$y = 2" and " y = 4$

Hence, the solution set is ${4, 2}$ and ${2, 4}$ .

Hopefully this helps!

How do you solve the system x² + y² = 20x² + y² = 20 and x + y = 6x + y = 6?