How do you solve the system $x^2+y^2=41$ and $y=-x-1$ ?

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Answer 1 · 2016-09-22T00:58:25

$x^2 + (-x - 1)^2 = 41$

$x^2 + x^2 + 2x + 1 = 41$

$2x^2 + 2x - 40 = 0$

$2(x^2 + x - 20) = 0$

$(x + 5)(x - 4) = 0$

$x = -5 and 4$

$:.y = -(-5) - 1" AND " y = -4 - 1$

$y = 4" AND "y = -5$

Our solution set is thus ${-5, 4}$ and ${4, -5}$ .

Hopefully this helps!

How do you solve the system x^2+y^2=41x^2+y^2=41 and y=-x-1y=-x-1?