Solve the system by the substitution method. xy = 12 x2 + y2 = 40

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Solve the system by the substitution method. xy = 12 x2 + y2 = 40

Mathematics
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Why can't you do your own work
y dont u pop off
no thanks

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yea get off my question if u just wanna start trouble toodels
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....
All you need to do is plug in.
haha i know .. ok then y does it give me 4 quandrants to pick from ?
example {( 2, 6), ( 6, 2), ( 2, -6), ( 6, -2)}
From what I solved I got these solutions. 6, 2, -2
im just confused on plugging them in ..
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.
im going w/ b
I was thinking for along the lines of C?
i just dont understand how i allways mess that up
Because in B, if you try using ( 2, -6) and ( -2, 6) you already have a problem for the equation below:\[xy = 12\]\[(2)(-6) = -12\]\[(-2)(6)= -12\]
Its okay! Just choose the options that give you BOTH positive answers
thank u i appreciate it !! :)
You're welcome! :D

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