How do you solve #y = -x+2# and #y= 3x-2#? Algebra Systems of Equations and Inequalities Linear Systems with Multiplication 1 Answer Nallasivam V · Stefan V. Aug 10, 2015 The answer is #x = 1# ; #y = 1# Explanation: Since #{(y = -x + 2), (y = 3x - 2) :}# We shall have #-x +2 = 3x - 2# #-x +2 - 3x + 2 = 0# #- 4x + 4 = 0# #- 4x = -4 implies x = (-4)/(-4) = 1# Substitute #x = 1# in the first equation to get #y = -1 + 2# #y = 1# Answer link Related questions How do you solve systems of equations by elimination using multiplication? Can any system be solved using the multiplication method? How do you find the least common number to multiply? How do you solve #4x+7y=6# and #6x+5y=20# using elimination? Are there more than one way to solve systems of equations by elimination? How do you solve the system #5x-10y=15# and #3x-2y=3# by multiplication? Which method do you use to solve #x=3y# and #x-2y=-3#? How old are John and Claire if twice John’s age plus five times Claire’s age is 204 and nine... How do you solve the system of equations #2x - 5y = 10# and #4x - 10y = 20#? How do you solve the system of equations #2x-3y=6# and #3y-2x=-6#? See all questions in Linear Systems with Multiplication Impact of this question 3589 views around the world You can reuse this answer Creative Commons License