britbrat4290
Solve the system by the substitution method.
xy = 12
x2 + y2 = 40
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live10000000
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Why can't you do your own work
britbrat4290
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y dont u pop off
live10000000
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no thanks
britbrat4290
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yea get off my question if u just wanna start trouble toodels
ilfy214
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Thats hilarious sorry! :) But anyways you can use either equation:
\[xy = 12 \]\[x= \frac{ 12 }{ y }\] OR
\[x^2 + y^2 = 40 \]\[x + y = \sqrt{40}\]\[x = \sqrt{40} - y\] **The y and the x can be switched....
ilfy214
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All you need to do is plug in.
britbrat4290
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haha i know .. ok then y does it give me 4 quandrants to pick from ?
britbrat4290
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example {( 2, 6), ( 6, 2), ( 2, -6), ( 6, -2)}
britbrat4290
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@ilfy214
ilfy214
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From what I solved I got these solutions. 6, 2, -2
britbrat4290
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im just confused on plugging them in ..
ilfy214
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Doesn't it depend on how you solved your equation? Also, when I plugged in -2 and 6 for the first problem, I got -12 instead of 12.
britbrat4290
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im going w/ b
ilfy214
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I was thinking for along the lines of C?
britbrat4290
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i just dont understand how i allways mess that up
ilfy214
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Because in B, if you try using ( 2, -6) and ( -2, 6) you already have a problem for the equation below:\[xy = 12\]\[(2)(-6) = -12\]\[(-2)(6)= -12\]
ilfy214
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Its okay! Just choose the options that give you BOTH positive answers
britbrat4290
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thank u i appreciate it !! :)
ilfy214
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You're welcome! :D