How do you solve the system #3x + 4y = -2# and #x = -2y#? Algebra Systems of Equations and Inequalities Systems Using Substitution 1 Answer Massimiliano Mar 24, 2015 We can substitue #x=-2y# in the first equation: #3(-2y)+4y=-2rArr-6y+4y=-2rArr-2y=-2rArr# #y=1# and than #x=-2#. Answer link Related questions How do you solve systems of equations using the substitution method? How do you check your solutions to a systems of equations using the substitution method? When is the substitution method easier to use? How do you know if a solution is "no solution" or "infinite" when using the substitution method? How do you solve #y=-6x-3# and #y=3# using the substitution method? How do you solve #12y-3x=-1# and #x-4y=1# using the substitution method? Which method do you use to solve the system of equations #y=1/4x-14# and #y=19/8x+7#? What are the 2 numbers if the sum is 70 and they differ by 11? How do you solve #x+y=5# and #3x+y=15# using the substitution method? What is the point of intersection of the lines #x+2y=4# and #-x-3y=-7#? See all questions in Systems Using Substitution Impact of this question 1489 views around the world You can reuse this answer Creative Commons License