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!