How do you solve the system of equations #3x + 4y = 4# and #2x - y = 2#?

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Answer 1 · 2016-12-05T16:42:17

#x = 12/11# and #y = 2/11#

Step 1) Solve the second equation for #y#:

#2x - y + y - 2 = 2 + y - 2#

#2x - 0 - 2 = 0 + y#

#2x - 2 = y#

#y = 2x - 2#

Step 2) Substitute #2x - 2# for #y# in the first equation and solve for #x#:

#3x + 4(2x - 2) = 4#

#3x + 8x - 8 = 4#

#11x - 8 = 4#

#11x - 8 + 8 = 4 + 8#

#11x - 0 = 12#

#11x = 12#

#(11x)/11 = 12/11#

#x = 12/11#

Step 3) Substitute #12/11# for #x# in the solution to the second equation in Step 1) and calculate #y#:

#y = 2(12/11) - 2#

#y = 24/11 - 2#

#y = 24/11 - (11/11)*2#

#y = 24/11 - 22/11#

#y = 2/11#