How do you solve the following system: #y= 3/4x+3 , 3x - 3y = 2 #?

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Answer 1 · 2017-05-10T11:07:07

Instead of y in the second question, write the first equation (of the right hand side) to solve

#3x-3y=2#

Instead of y in the above equation, write (#3x/4+3#) where you see y.

It is:

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

#3x-9x/4-9=2#

#(12x-9x)/4=11#

#3x/4=11#

#x=44/3#

From here you can solve y:

#y=(3/4)*(44/3) + 3#

#y=11+3#

#y=14#