How do you find the exact solutions to the system #x^2/30+y^2/6=1# and #x=y#? Precalculus Solving Systems of Two Equations Solving by Substitution 1 Answer Shwetank Mauria · erinspoons Nov 23, 2016 #x=y=-sqrt5# or #x=y=sqrt5# Explanation: #y^2/30+y^2/6=1# #y^2/30xx30+y^2/6xx30=1xx30# #y^2+5y^2=30# #6y^2=30# #y^2=30/6=5# #y^2-5=0# #(y+sqrt5)(y-sqrt5)=0# Hence, either #y+sqrt5=0# i.e. #y=-sqrt5# or #y-sqrt5=0# i.e. #y=sqrt5# Hence, #x=y=-sqrt5# or #x=y=sqrt5# 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 1669 views around the world You can reuse this answer Creative Commons License