How do you solve #3x + y = 1# and #x – y = 3# using substitution?

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Answer 1 · 2018-05-23T13:40:49

#(x,y) = (1,-2)#

From the second argument, we can come to conclusion that

#x = 3 + y#

Therefore, we have:

#3(3 + y) + y = 1#

Furthermore:

#9 + 3y + y = 1#

#4y = -8#

#y=-2#

Back to the second equation:

#x = 3 + y#

#x = 3 + (-2)#

#x = 1#

So

#(x,y) = (1,-2)#