How do you solve the system #x^2+y^2=41# and #y=-x-1#? Precalculus Solving Systems of Two Equations Solving by Substitution 1 Answer Noah G Sep 22, 2016 #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! Answer link Related questions What is a system of equations? What does it mean to solve a system of equations by substitution? How do I use substitution to find the solution of the system of equations #c+3d=8# and #c=4d-6#? How do you write a system of linear equations in two variables? How does a system of linear equations have no solution? How many solutions can a system of linear equations have? What is the final step of completing a solve by substitution problem? How do I use substitution to find the solution of the system of equations #4x+3y=7# and #3x+5y=8#? How do I use substitution to find the solution of the system of equations #y=2x+1# and #2y=4x+2#? How do I use substitution to find the solution of the system of equations #y=1/3x+7/3# and... See all questions in Solving by Substitution Impact of this question 2048 views around the world You can reuse this answer Creative Commons License