Matt71
<Simultaneous equation>
y=2x, x(x+y)=53y^2
a solution would be helpful please, no idea how these things are solved



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Helper93
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You know that y=2x, so plug in 2x instead of y in the second equation and solve.

Matt71
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Hold on, lemme try

Matt71
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What's (2x)^2?
Is it the long answer or like, the one with 2 numbers?

Matt71
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can't do it man, help me out here

Mertsj
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If y = 2x, then x = 2y. Replace each x with 2y

Matt71
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where do I go from 22y=53y^2?

Mertsj
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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.

Matt71
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English man, English

Mertsj
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Rearrange the equation so that it is equal to 0. Make the y^2 term positive.

Mertsj
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Also, it should be 42y not 22y

Matt71
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Oh

Helper93
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Actually, Mertsj is right, it's easier to replace x other than y.

Matt71
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Eh?

Matt71
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could one of you dudes just give me the answer +solution? It's REALLY late

Mertsj
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\[(2y)(2y+y)=53y^2\]
\[(2y)(2)=53y^2\]
\[42y=53y^2\]
\[3y^22y1=0\]
\[(3y+1)(y1)=0\]

Mertsj
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Can you solve that for y?

Matt71
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waaaait

Matt71
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how did 3y^22y1=0 become (3y+1)(y1)=0
?

Mertsj
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By factoring.

Matt71
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what method?

Helper93
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You can also use discriminant if you want.

Mertsj
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Too many cooks.

Matt71
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mmm, well

Matt71
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thanks dudes, I'll try to figure this dang thing out

Matt71
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Sorry Helper
Can't give more than one medal