How do you solve the system #y^2+x=2# and #3x+y=8#? Precalculus Solving Systems of Two Equations Solving by Substitution 1 Answer Noah G Jul 4, 2016 #x = 2 - y^2# #:.3(2 - y^2) + y = 8# #6 - 3y^2 + y = 8# #0 = 3y^2 - y + 2# #y = (-(-1) +- sqrt(-1^2 - 4 xx 3 xx 2))/(2 xx 3)# #y = (1 +- sqrt(-23))/6# #y = O/# #:.# This systems of equations has no real solution. 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 1188 views around the world You can reuse this answer Creative Commons License