anonymous
  • anonymous
y=2-x, x(x+y)=5-3y^2 a solution would be helpful please, no idea how these things are solved
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
You know that y=2-x, so plug in 2-x instead of y in the second equation and solve.
anonymous
  • anonymous
Hold on, lemme try
anonymous
  • anonymous
What's (2-x)^2? Is it the long answer or like, the one with 2 numbers?

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

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