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If y = x - 3, and x² - 5y²

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If y = x - 3, and x² - 5y²  [#permalink]

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New post 13 Oct 2016, 08:40
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Question Stats:

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If y = x - 3, and x² - 5y² + 4xy - 15 = 0, then x + 2y =

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


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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 13 Oct 2016, 10:03
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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
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 13 Oct 2016, 10:54
1
(y-x)^2=9
Substitute this in the other equation,
Final : y= 1/3 x=10/3. Hence, 4

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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 13 Oct 2016, 21:14
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From given information, y = x-3 we can get (x-y)^2 = 9
Rewriting the other eqn, we get
(x-y)^2 + 6xy - 6y^ 2= 15
9 = 15 + 6y(y-x)
-6 = -18 y (because y-x = -3)
y = 1/3. We can get x = 10/3

x+2y = 12/3(Option B)
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 14 Oct 2016, 07:16
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
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 14 Oct 2016, 07:36
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GMATPrepNow wrote:
If y = x - 3, and x² - 5y² + 4xy - 15 = 0, then x + 2y =

A) 2
B) 4
C) 6
D) 8
E) 10


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
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 24 Oct 2016, 19:39
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
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 08 Mar 2019, 16:26
GMATPrepNow wrote:
GMATPrepNow wrote:
If y = x - 3, and x² - 5y² + 4xy - 15 = 0, then x + 2y =

A) 2
B) 4
C) 6
D) 8
E) 10


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


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!
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Re: If y = x - 3, and x² - 5y²  [#permalink]

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New post 09 Mar 2019, 08:10
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jfranciscocuencag wrote:

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!


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
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Re: If y = x - 3, and x² - 5y²   [#permalink] 09 Mar 2019, 08:10
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