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Re: If y = x - 3, and x² - 5y² [#permalink]
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(y-x)^2=9
Substitute this in the other equation,
Final : y= 1/3 x=10/3. Hence, 4

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

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Re: If y = x - 3, and x² - 5y² [#permalink]
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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]
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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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]
Solved the question with conventional approach of substituting the y in the second equation.it took 3 mins to solve the question.
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Re: If y = x - 3, and x² - 5y² [#permalink]
BrentGMATPrepNow please could you show the factorization process. I Find it a bit difficult to factor out two variables in one equation

Quote:
Rearrange to get: x² + 4xy - 5y² = 15
Factor: (x - y)(x + 5y) = 15
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Re: If y = x - 3, and x² - 5y² [#permalink]
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Hoozan wrote:
BrentGMATPrepNow please could you show the factorization process. I Find it a bit difficult to factor out two variables in one equation

Quote:
Rearrange to get: x² + 4xy - 5y² = 15
Factor: (x - y)(x + 5y) = 15


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