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I'm guessing i should use substitution but I'm not sure :/
in the first equation get the y to the opposite side and the 2 to the other side. so it should be -2=1/3x-y
well since y is already solved shouldn't I just plug it in to the second equation???
oh man, lol i was going to do elimination but yeah you're right lol plug it into the 2nd equation
kk, when I did that and tried to solvee, x=10 over 3 or 10 thirds
thats weird, i went through it and distributed -6 to 1/3 x and 2 and got 2x-2x-12=16
so the 2x's would cancel so that wouldn't work,
well lets continue doing it the way i was with elimination because that seems easier... so when we put the y to the other side we will get 1/3x-y=-2 to get the y's to cancel out we would multiply that equation by -6 to get 2x+6y=-2 combine those equations and get 2x+2x+6y-6y=14 4x=14 divide 4 from both sides and get x=3.5 understand?