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Solve for x - y

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Solve for x - y

by govind_raj_76 » Sun Jul 18, 2010 8:23 am
Solve for X - Y

X + KY = 7

KX + Y - X = 10

A. (7-10K) / (1 + K+ K^2)

B. (7K-3) / (K^2 - K -1)

C. (10+3K) / (K^2 - K -1)

D. (3K-4) / (K^2 - K -1)

E. (4-7K) / (K^2 + K +1)

Please assist.
Govind
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by aloneontheedge » Sun Jul 18, 2010 11:06 am
govind_raj_76 wrote:Solve for X - Y

X + KY = 7

KX + Y - X = 10

A. (7-10K) / (1 + K+ K^2)

B. (7K-3) / (K^2 - K -1)

C. (10+3K) / (K^2 - K -1)

D. (3K-4) / (K^2 - K -1)

E. (4-7K) / (K^2 + K +1)

Please assist.
In these kind of situations it is always helpful to plug values
lets say K= 2 then X= 13 Y = -3
x-y = 16
substitute these values in the options
we find C = 16
Pick C
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by sandysai » Mon Jul 19, 2010 5:34 pm
Hi Alone...

How did you come up with these smart numbers guess. i had troubling selecting the right numbers which satisfies both the equations.

is there any technique to do that
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by kanagasu » Wed Jul 21, 2010 8:17 am
I guess the best solution would be a general Algebric solving method.

Given two equations are :

X+KY = 7 ----- (1)

KX+Y-X = 10 ----- (2)

Since we have all the options in terms of K, lets make X and Y in terms of K.

From the first equation we have Y = (7-X)/K

Substitue this into the Second equation.

Where we get X as, X = (10K - 7)/ (K^2 -K -1) ----------- Equation A

Now again from the first equation we have X= 7-kY, lets substitute this in the second equation.

Where we get Y as, Y = (7K-17)/(K^2-K-1) ---------------Equation B

Lets subtract A and B

We get X-Y = 3K+10 /(K^2-K-1)

hence, C is the answer.

Thought it might seem lengthy, I guess it works more efficiently than plugging in some arbitrary values (In this case).

Cheers,
Suresh
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by Stuart@KaplanGMAT » Wed Jul 21, 2010 9:38 am
sandysai wrote:Hi Alone...

How did you come up with these smart numbers guess. i had troubling selecting the right numbers which satisfies both the equations.

is there any technique to do that
Hi,

Alone simply picked a nice simply value for k, then solved the two equations with k=2.

Plugging in:

x + 2y = 7

2x + y - x = 10
(simplifying the second equation)
x + y = 10

Now we have two equations and two unknowns, so we can solve. Let's subtract the second equation from the first:

x + 2y = 7
-(x + y = 10)

y = -3

and subbing y = -3 into the second equation:

x + (-3) = 10
x = 13

So:

x - y = 13 - (-3) = 16

Now we plug k=2 into the choices and look for the one that works out to 16.
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by aloneontheedge » Wed Jul 21, 2010 10:02 am
sandysai wrote:Hi Alone...

How did you come up with these smart numbers guess. i had troubling selecting the right numbers which satisfies both the equations.

is there any technique to do that
Sorry buddy did not see the post.
Stuart has already answered this.
Just pick some number and plug in.Plug the same values in answer choices and both should be equal.
Probably,at times when u pick a number,u might end up with 2 answers choices which satisfy,in such cases plug some different numbers and check for them(usually happens in number system probs which asks for even r definetely odd)
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by zareentaj » Wed Aug 11, 2010 4:36 am
Stuart Kovinsky wrote:
sandysai wrote:Hi Alone...

How did you come up with these smart numbers guess. i had troubling selecting the right numbers which satisfies both the equations.

is there any technique to do that
Hi,

Alone simply picked a nice simply value for k, then solved the two equations with k=2.

Plugging in:

x + 2y = 7

2x + y - x = 10
(simplifying the second equation)
x + y = 10

Now we have two equations and two unknowns, so we can solve. Let's subtract the second equation from the first:

x + 2y = 7
-(x + y = 10)

y = -3

and subbing y = -3 into the second equation:

x + (-3) = 10
x = 13

So:

x - y = 13 - (-3) = 16

Now we plug k=2 into the choices and look for the one that works out to 16.
Thank you very much Stuart Kovinsky. I understood clearly.
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