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(x – 2y)(x + 2y) = 5 (2x – y)(2x + y) = 35
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07 Apr 2019, 07:33
07 Apr 2019, 07:33
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80% (01:45) correct
20% (02:34) wrong based on 65 sessions
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\((x – 2y)(x + 2y) = 5\)
\((2x – y)(2x + y) = 35\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\((2x – y)(2x + y) = 35\)
Quantity A |
Quantity B |
\(2x^2 - y^2\) |
\(x^2 - 2y^2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: (x – 2y)(x + 2y) = 5 (2x – y)(2x + y) = 35
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07 Apr 2019, 07:47
07 Apr 2019, 07:47
2
Carcass wrote:
\((x – 2y)(x + 2y) = 5\)
\((2x – y)(2x + y) = 35\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\((2x – y)(2x + y) = 35\)
Quantity A |
Quantity B |
\(2x^2 - y^2\) |
\(x^2 - 2y^2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
GIVEN:
(x – 2y)(x + 2y) = 5
(2x – y)(2x + y) = 35
Expand and simplify each equation to get:
x² - 4y² = 5
4x² - y² = 35
ADD the two equations to get: 5x² - 5y² = 40
Divide both sides by 5 to get: x² - y² = 8
Rewrite each quantity as follows:
QUANTITY A: x² + x² - y²
QUANTITY B: x² - y² - y²
Add some brackets for clarity:
QUANTITY A: x² + (x² - y²)
QUANTITY B: (x² - y²) - y²
Replace x² - y² with 8 to get:
QUANTITY A: x² + 8
QUANTITY B: 8 - y²
Subtract 8 from both quantities to get:
QUANTITY A: x²
QUANTITY B: -y²
Since x and y cannot both equal 0, we know that x² is POSITIVE, and -y² is NEGATIVE
So, Quantity A must be greater.
Answer: A
Cheers,
Brent
Re: (x 2y)(x + 2y) = 5 (2x y)(2x + y) = 35
[#permalink]
04 Aug 2022, 10:26
04 Aug 2022, 10:26
1
My solution was to solve both the given equations by elimination,
End up with
\(x^2 = 7\)
\(y^2 = \frac{1}{2}\)
Quantity A = 13.5
Quantity B = 7 - <something positive>
So, the answer is A
End up with
\(x^2 = 7\)
\(y^2 = \frac{1}{2}\)
Quantity A = 13.5
Quantity B = 7 - <something positive>
So, the answer is A
