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

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Answer 1 · 2017-05-10T19:22:42

I got:
#x=2#
#y=3#

We can substitute the second equation into the first for #3x# and get:

#y-color(red)(2y)=-3#

so that:

#-y=-3#

and:

#y=3#

Let us use this into, say, the second equation:

#2*color(blue)(3)=3x#

giving:

#x=6/3=2#