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If x+y=3 and x^2+y^2=7, then xy=?

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If x+y=3 and x^2+y^2=7, then xy=? [#permalink]

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New post 03 May 2017, 02:28
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  15% (low)

Question Stats:

74% (00:47) correct 26% (01:47) wrong based on 62 sessions

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If x+y=3 and \(x^2+y^2=7\), then xy=?

A. -2
B. -1
C. 0
D. 1
E. 2
[Reveal] Spoiler: OA

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Re: If x+y=3 and x^2+y^2=7, then xy=? [#permalink]

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New post 03 May 2017, 02:34
(x+y)^2=x^2+y^2+2xy
(3)^2=7+2xy
Solve to get 1

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Re: If x+y=3 and x^2+y^2=7, then xy=? [#permalink]

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New post 05 May 2017, 01:50
==> You get \((x+y)^2=x^2+y^2+2xy\), and if you substitute this, from \(3^2=7+2xy\), you get 2xy=2, and xy=1.

Therefore, the answer is D.
Answer: D
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Re: If x+y=3 and x^2+y^2=7, then xy=? [#permalink]

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New post 06 May 2017, 17:53
MathRevolution wrote:
If x+y=3 and \(x^2+y^2=7\), then xy=?

A. -2
B. -1
C. 0
D. 1
E. 2


We are given that x + y = 3 and that x^2 + y^2 = 7. If we square both sides of x + y = 3, we will have:

x^2 + y^2 + 2xy = 9

x^2 + y^2 = 9 - 2xy

Since x^2 + y^2 = 7:

9 - 2xy = 7

-2xy = -2

xy = 1

Answer: D
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Re: If x+y=3 and x^2+y^2=7, then xy=? [#permalink]

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New post 28 Aug 2017, 00:17
2xy = (x+y)^2-(x^2+y^2)

xy =9-7/2=1
Hence Option D

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Re: If x+y=3 and x^2+y^2=7, then xy=?   [#permalink] 28 Aug 2017, 00:17
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