How do you solve the following system: #x+y=4 , 3x + 4y = 11 #?

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Answer 1 · 2018-07-07T19:00:41

#x=5#, #y=-1#

Given that

#x+y=4\ .........(1)#

#3x+4y=11\ .........(2)#

Multiplying (1) by #3# & subtracting from (2) as follows

#3x+4y-3(x+y)=11-3\times 4#

#y=-1#

setting #y=-1# in (1), we get

#x=4-y=4-(-1)=5#

hence, the solution is #x=5# & #y=-1#