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y=2-x, x(x+y)=5-3y^2 a solution would be helpful please, no idea how these things are solved

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You know that y=2-x, so plug in 2-x instead of y in the second equation and solve.
Hold on, lemme try
What's (2-x)^2? Is it the long answer or like, the one with 2 numbers?

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can't do it man, help me out here
If y = 2-x, then x = 2-y. Replace each x with 2-y
where do I go from 2-2y=5-3y^2?
Put it in the standard form for a quadratic equation and then solve by factoring (if it will factor) or by the quadratic formula if it will not.
English man, English
Rearrange the equation so that it is equal to 0. Make the y^2 term positive.
Also, it should be 4-2y not 2-2y
Actually, Mertsj is right, it's easier to replace x other than y.
could one of you dudes just give me the answer +solution? It's REALLY late
\[(2-y)(2-y+y)=5-3y^2\] \[(2-y)(2)=5-3y^2\] \[4-2y=5-3y^2\] \[3y^2-2y-1=0\] \[(3y+1)(y-1)=0\]
Can you solve that for y?
how did 3y^2-2y-1=0 become (3y+1)(y-1)=0 ?
By factoring.
what method?
You can also use discriminant if you want.
Too many cooks.
mmm, well
thanks dudes, I'll try to figure this dang thing out
Sorry Helper Can't give more than one medal

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