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If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th

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Bunuel
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If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   04 Jan 2017, 02:05
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If y and x are numbers such that x + y = 17 and 2x – y = 6, what is the value of x/2?

A. 23/6
B. 11/2
C. 23/3
D. 11
E. 23
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   04 Jan 2017, 02:30
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Bunuel wrote:
If y and x are numbers such that x + y = 17 and 2x – y = 6, what is the value of x/2?

A. 23/6
B. 11/2
C. 23/3
D. 11
E. 23


x + y = 17
2x – y = 6

Adding the two equations

3x = 23

i.e. x = 23/3
i.e. x/2 = 23/6

Answer: Option A
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   04 Jan 2017, 12:01
Adding 2 equations => x=23/3
=>x/2=23/6

A.

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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   06 Jan 2017, 10:59
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Bunuel wrote:
If y and x are numbers such that x + y = 17 and 2x – y = 6, what is the value of x/2?

A. 23/6
B. 11/2
C. 23/3
D. 11
E. 23


We can add the two given equations together and we have:
(x + y = 17)
+ (2x – y = 6)

3x = 23

x = 23/3

Thus, x/2 = (23/3)/2 = 23/6.

Answer: A
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   10 Jan 2017, 09:19
x + y = 17
2x – y = 6
3x = 23
x =\(\frac{23}{3}\)
\(\frac{x}{2}\) = \(\frac{23}{6}\)
A
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   18 Jan 2017, 03:19
Why not 11/2 if we solve them simultaneously?
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   18 Jan 2017, 03:39
ahmedimran1 wrote:
Why not 11/2 if we solve them simultaneously?

Simultaneously? Can you explain a bit? :)
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Re: If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   18 Jan 2017, 03:49
Sirakri wrote:
ahmedimran1 wrote:
Why not 11/2 if we solve them simultaneously?

Simultaneously? Can you explain a bit? :)


Multiply first equation by 2 and then change the signs of the 2nd equation. 1st equation subtracted from 2nd equation y comes out to be 28/3 , x=17-28/3 which gives x as 11 then since the question asks for x/2 answer should be 11/2.
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If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink] New post   18 Jan 2017, 04:08
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ahmedimran1 wrote:
Sirakri wrote:
ahmedimran1 wrote:
Why not 11/2 if we solve them simultaneously?

Simultaneously? Can you explain a bit? :)


Multiply first equation by 2 and then change the signs of the 2nd equation. 1st equation subtracted from 2nd equation y comes out to be 28/3 , x=17-28/3 which gives x as 11 then since the question asks for x/2 answer should be 11/2.


Hey,

When you multiply the first equation by 2, you will get -
\(2x + 2y = 34\)

And when you change the sign of the second equation you will get -
\(-2x + y = -6\)

When you add the above two equation, you will get -

    \(3y = 34 - 6\)
    \(y = 28/3\)

Therefore till point you are absolutely correct.

When you substitute this value in first equation, the value of x will be -
    \(x = 17 - \frac{28}{3}\)
    \(x = \frac{51 - 28}{3}\)
    \(x = \frac{23}{3}\)

This is exactly where you have made the mistake of just take the difference of 28 and 17 and ignoring the 3 in the denominator.

Thus, the value of x/2 will be 23/6.

Thanks,
Saquib
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If y and x are numbers such that x + y = 17 and 2x – y = 6, what is th [#permalink]
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