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1/x+1/y=1 xy=6, then what is the value of x^2+y^2?

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Splendidgirl666
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1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   19 Jan 2012, 13:21
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Hi guys,

I am struggling with this one, can anyone help pls?

1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24

Thanks!
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E
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Bunuel
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Re: Algebra [#permalink] New post   19 Jan 2012, 13:29
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Splendidgirl666 wrote:
Hi guys,

I am struggling with this one, can anyone help pls?

1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24

Thanks!


\(\frac{1}{x}+\frac{1}{y}=1\) --> \(\frac{x+y}{xy}=1\) --> \(x+y=xy\). Since \(xy=6\) then \(x+y=6\).

Now, \(x^2+y^2=(x+y)^2-2xy=6^2-2*6=24\).

Answer: E.
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Splendidgirl666
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   19 Jan 2012, 13:34
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Got it! I forgot that a^2 + b^2 = (a+b)^2 - 2ab!

Thank you
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   14 Jul 2014, 21:33
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\(\frac{1}{x} + \frac{1}{y} = 1\)

x+y = xy

Squaring both sides

\(x^2 + 2(xy) + y^2 = (xy)^2\)

\(x^2 + y^2 = (xy)^2 - 2(xy)\)

Placing value of xy = 6

\(x^2 + y^2 = 36 - 12 = 24\)

Answer = 24
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   22 Jul 2015, 00:05
1/x + 1/y = 1; xy = 6.

1/x + 1/y = 1
x+y/xy = 1
x+y/6 = 1 (since xy = 6)
so, x+y = 6.

x^2+y^2 = (x+y)^2 - 2xy = (6)^2 - 2(6) = 36 - 12 = 24.

Ans (E).
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1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   Updated on: 14 Mar 2016, 04:06
Hi everyone!

How come that x^2 + y^2 equals (x+y)^2 - 2xy ?

Is it like an algebraic identity formula? I haven't seen this one anywhere.

Thanks!

Originally posted by iliavko on 14 Mar 2016, 04:02.
Last edited by iliavko on 14 Mar 2016, 04:06, edited 1 time in total.
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   14 Mar 2016, 04:06
iliavko wrote:
Hi everyone!

How come that x^2 + y^2 equals (x+y)^2 - 2xy ?

Is it like a difference of squares formula? I haven't seen this one anywhere.

Thanks!


It comes from rearranging terms in the formula: \((a+b)^2 = a^2+2ab+b^2\) ---> \(a^2+b^2 = (a+b)^2-2ab\)

Hope this helps.
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1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   14 Mar 2016, 04:07
Nice!!.. The stuff you discover at this forum :)
Thanks you so much!

One more question, maybe it's silly, but.. What about difference of squares? a^2 - b^2 = (a+b)(a-b) what is the rationale behind it? Any formula that manipulated leads to it?

Thank you!
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   14 Mar 2016, 04:22
Splendidgirl666 wrote:
Hi guys,

I am struggling with this one, can anyone help pls?

1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24

Thanks!

x*y=6
x+y=6
(x+y)^2=x^2+y^2+2*x*y
36=()+12
Therefore, the answer is 24
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   15 Mar 2016, 00:49
Splendidgirl666 wrote:
Hi guys,

I am struggling with this one, can anyone help pls?

1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24

Thanks!


See here => 1/x+1/y=1
so taking the lcm we get => x+y=xy
so (x+y)^2 =36
x^2+y^2=36-2xy=36-12 = 24
hence E
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1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   25 Apr 2018, 03:45
This is a cheeky little exercise.

The concepts that are tested: manipulating algebraic fractions, and square of a sum.


These are the steps I recommend to take, with the tested concepts mentioned in each step.

1/x + 1/y = 1. (we remember that with fractions with a different denominator, we first have to find the least common denominator)

y+x / xy =1

y+x = xy

y+x = 6

y^2+x^2 = 6^2

y^2+x^2 = 36.

(y+x)^2 = 36. (remember that (y+x)^2 → x^2+2xy+y^2

Now we can set up the equation with what we already have:

So x^2 + 2xy + y^2 = 36. (Hence we also know that xy = 6. Let’s plug it in.

So x^2 + 2(6) + y^2 = 36 → x^2+12+y^2 = 36 → x^2 + y^2 = 36-12 = 24. Answer choice E.
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   26 Apr 2018, 15:38
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Quote:

1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24


We can multiply the first equation by xy and we have:

y + x = xy

Since xy = 6 we have:

y + x = 6

Squaring both sides we have:

x^2 + y^2+ 2xy = 36

x^2 + y^2 + 12 = 36

x^2 + y^2 = 24

Answer: E
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   25 Apr 2020, 11:39
1/x + 1/y =1 xy=6, then what is the value of x^2 + y^2?

1. 12
2. 16
3. 18
4. 20
5. 24 --> correct: xy=6 & 1/x + 1/y =1 => x+y=xy=6=> (x+y)^2=6^2=>x^2+y^2+2xy=36=> x^2+y^2=24
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   03 May 2020, 12:25
given, 1/x + 1/y = 1; xy = 6.

1/x + 1/y = 1
or,(x+y)/xy = 1
or,(x+y)/6 = 1 (since xy = 6)
so, x+y = 6.

now,x^2+y^2 = (x+y)^2 - 2xy = (6)^2 - 2(6) = 36 - 12 = 24.

correct answer E
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink] New post   06 Apr 2022, 13:39
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Re: 1/x+1/y=1 xy=6, then what is the value of x^2+y^2? [#permalink]
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