How do you solve the system #3x^2-y^2=-6# and #y=2x+1#? Precalculus Solving Systems of Two Equations Solving by Substitution 1 Answer Ratnaker Mehta Aug 29, 2016 #(x,y)=(1,3), or, (-5,-9)#. Explanation: We submit #y=2x+1# in the first eqn. and get, #3x^2-(2x+1)^2+6=0# #rArr -x^2-4x+5=0, or, x^2+4x-5=0#. #rArr (x-1)(x+5)=0# #rArr x=1, or, -5#. #rArr y=2x+1=3, or, -9# Therefore, the solns. are #(x,y)=(1,3), or, (-5,-9)#, which satisfy the set of eqns. 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 1345 views around the world You can reuse this answer Creative Commons License