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If x^2+xy=100-y^2
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01 Nov 2021, 06:26
01 Nov 2021, 06:26
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Question Stats:
85% (01:21) correct
14% (02:06) wrong based on 21 sessions
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Joined: 10 Apr 2015
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Re: If x^2+xy=100-y^2
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01 Nov 2021, 08:33
01 Nov 2021, 08:33
1
Carcass wrote:
Given: x² + xy = 100 - y² - xy
Add xy and y² to both sides of the equation to get: x² + 2xy + y² = 100
Now let's find the value of \((x \sqrt{3} + y \sqrt{3})^2\)
Use FOIL to expand and simplify: 3x² + 6xy + 3y²
Factor to get: 3(x² + 2xy + y²)
Substitute to get: 3(100)
Evaluate: 300
Answer: 300
General Discussion
Re: If x^2+xy=100-y^2
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15 Feb 2023, 11:53
15 Feb 2023, 11:53
Hi Brent GreenlightTestPrep
Use FOIL to expand and simplify: 3x² + 6xy + 3y²
Divide by 3 to get: (x² + 2xy + y²)
Substitute and I got 100 as answer.
Not sure why am I getting differnet answer here? Could you clairfy? Thanks
Use FOIL to expand and simplify: 3x² + 6xy + 3y²
Divide by 3 to get: (x² + 2xy + y²)
Substitute and I got 100 as answer.
Not sure why am I getting differnet answer here? Could you clairfy? Thanks
