How do you solve the system #y^2-2x^2=6# and #y=-2x#? Precalculus Solving Systems of Two Equations Solving by Substitution 1 Answer Noah G Oct 3, 2016 The solution set is #(sqrt(3), -2sqrt(3))# and #(-sqrt(3), 2sqrt(3))# Explanation: #y = -2x -> (-2x)^2 - 2x^2 = 6# #4x^2 - 2x^2 = 6# #2x^2 = 6# #x^2 = 3# #x = +-sqrt(3)# #y = -2x# #y = -2(sqrt(3))" AND "y = -2(-sqrt(3))# #y = -2sqrt(3)" AND "y = 2sqrt(3)# 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 1199 views around the world You can reuse this answer Creative Commons License