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If x^2+2xy+y^2=25
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15 Jan 2021, 06:04
15 Jan 2021, 06:04
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80% (01:11) correct
20% (02:34) wrong based on 20 sessions
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If \(x^2+2xy+y^2=25\) , \(x+y >0\) , and \(x-y=1\), then what is the value of x?
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Re: If x^2+2xy+y^2=25
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15 Jan 2021, 06:50
15 Jan 2021, 06:50
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Carcass wrote:
Take: \(x^2+2xy+y^2=25\)
Factor: \((x+y)^2=25\)
So, EITHER \(x+y=5\) OR \(x+y=-5\)
Since we're told that \(x+y >0\), it must be the case that \(x+y=5\)
We're also told that \(x-y=1\)
If we add these two equations we get: \(2x = 6\), which means \(x = 3\)
Answer: \(3\)
Re: If x^2+2xy+y^2=25
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17 Aug 2022, 19:45
17 Aug 2022, 19:45
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