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Posts: 6,733
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Difficulty:
45%
(medium)
Question Stats:
73% (02:57) correct
27%
(03:06)
wrong
based on 535
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If y = x - 3, and x^2 - 5y^2 + 4xy - 15 = 0, then x + 2y =
A. 2
B. 4
C. 6
D. 8
E. 10
A. 2
B. 4
C. 6
D. 8
E. 10
I can think of 3 different solutions to this question.
Kudos for all correct solutions
A video solution can be found here: https://www.gmatprepnow.com/module/gmat ... video/1020
Kudos for all correct solutions
A video solution can be found here: https://www.gmatprepnow.com/module/gmat ... video/1020
Posts: 6,733
GMATPrepNow
Another approach:
We're also told that y = x - 3
Rearrange to get: x - y = 3
Also given: x² - 5y² + 4xy - 15 = 0
Add 15 to both sides: x² - 5y² + 4xy = 15
Rearrange to get: x² + 4xy - 5y² = 15
Factor: (x - y)(x + 5y) = 15
Replace x - y with 3 to get: 3(x + 5y) = 15
So, we can conclude that (x + 5y) = 5
We have two equations:
x + 5y = 5
x - y = 3
ADD the two above equations to get: 2x + 4y = 8
Divide both sides by 2 to get: x + 2y = 4
Answer: Here's the video solution: https://www.gmatprepnow.com/module/gmat ... video/1020
Cheers,
Brent
From given information, \(y = x-3\) we can get \((x-y)^2 = 9\)
Rewriting the other equation, we get
\((x-y)^2 + 6xy - 6y^ 2= 15 \) -> \(9 = 15 + 6y(y-x)\) -> \(-6 = -18 y\) (as \(y-x = -3\))
\(y = \frac{1}{3}\) and solving for x, x = \(\frac{10}{3}\)
Therefore, the value of x+2y is \(\frac{12}{3}\)(Option B)
Rewriting the other equation, we get
\((x-y)^2 + 6xy - 6y^ 2= 15 \) -> \(9 = 15 + 6y(y-x)\) -> \(-6 = -18 y\) (as \(y-x = -3\))
\(y = \frac{1}{3}\) and solving for x, x = \(\frac{10}{3}\)
Therefore, the value of x+2y is \(\frac{12}{3}\)(Option B)
General Discussion
Substitute the value of y in the given equation:
x^2 - 5(x - 3)^2 + 4x(x - 3) - 15 = 0
x^2 - (x - 3)(5x - 15 - 4x) - 15 = 0
x^2 - (x - 3)(x - 15) - 15 = 0
x^2 - x^2 + 18x - 45 - 15 = 0
x = 10/3 --> y = 1/3
x + 2y = 10/3 + 2(1/3) = 12/3 = 4
Answer: B
x^2 - 5(x - 3)^2 + 4x(x - 3) - 15 = 0
x^2 - (x - 3)(5x - 15 - 4x) - 15 = 0
x^2 - (x - 3)(x - 15) - 15 = 0
x^2 - x^2 + 18x - 45 - 15 = 0
x = 10/3 --> y = 1/3
x + 2y = 10/3 + 2(1/3) = 12/3 = 4
Answer: B
(y-x)^2=9
Substitute this in the other equation,
Final : y= 1/3 x=10/3. Hence, 4
Sent from my Nexus 5X using GMAT Club Forum mobile app
Substitute this in the other equation,
Final : y= 1/3 x=10/3. Hence, 4
Sent from my Nexus 5X using GMAT Club Forum mobile app
Mo2men
Joined: 26 Mar 2013
Last visit: 09 May 2023
Posts: 2,426
Concentration: Operations, Strategy
Schools: Erasmus (II)
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Schools: Erasmus (II)
Posts: 2,426
Hi Brent,
I have solved the question like above mentioned solutions. However, I'm curious to see if you have another short cut or though to solve it.
Thaknks
I have solved the question like above mentioned solutions. However, I'm curious to see if you have another short cut or though to solve it.
Thaknks
y=x-3
x^2 - 5y^2 +4xy -15 =0
x^2-5(x^2 -6x+9)+4x(x-3)-15=0
6(3x-10) = 0
x = 10/3
x+2(x-3) = 10/3 + 20/3 - 6 = 4
x^2 - 5y^2 +4xy -15 =0
x^2-5(x^2 -6x+9)+4x(x-3)-15=0
6(3x-10) = 0
x = 10/3
x+2(x-3) = 10/3 + 20/3 - 6 = 4
Posts: 227
GMATPrepNow
Hello GMATPrepNow !
Would you please give a detailed explanation about how did you factorize?
Rearrange to get: x² + 4xy - 5y² = 15
Factor: (x - y)(x + 5y) = 15
Thank you so much!
Posts: 6,733
jfranciscocuencag
You bet!
First, IGNORE the y's in the expression to get: x² + 4x - 5 = 15
We can now factor the left side to get: (x + 5)(x - 1) = 15
At this point, we need to add some y's.
Notice that the last term = 5y²
To get 5y², we need to add a y to each binomial to get: (x + 5y)(x - 1y) = 15
Does that help?
Cheers,
Brent
Tushrgupta
Joined: 25 Jun 2019
Last visit: 26 Jul 2020
Posts: 40
Location: India
Concentration: Strategy, International Business
GPA: 2.7
Posts: 40
Solved the question with conventional approach of substituting the y in the second equation.it took 3 mins to solve the question.
BrentGMATPrepNow please could you show the factorization process. I Find it a bit difficult to factor out two variables in one equation
Quote:
Posts: 6,733
Hoozan
Here's one way to look at it..
Take: x² + 4xy - 5y²
Ignore the y's for now to get: x² + 4x - 5
Factor as you regularly would: (x +5)(x - 1)
At this point, we must determine what we need to add to expression, (x + 5)(x - 1), so that it expands to get x² + 4xy - 5y².
We know that (5y)(-1y) = -5y²
So, let's add y-terms as follows to get: (x + 5y)(x - 1y), which is equivalent to (x + 5y)(x - y)
Does that help?
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