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# If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

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Joined: 02 Sep 2009
Posts: 58327
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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09 Nov 2015, 23:02
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25% (medium)

Question Stats:

73% (01:23) correct 27% (02:09) wrong based on 164 sessions

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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

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Joined: 29 Aug 2015
Posts: 11
Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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10 Nov 2015, 08:58

(1/X+1/Y)=5 canbe solved as {(x+y)/xy}=6. Substituting for 1/xy=6, we get
x+y=5/6

==> (x+y)/5= 5/(6*5)=1/6.
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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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23 Oct 2017, 10:14
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

HarrishGowtham wrote:

(1/X+1/Y)=5 canbe solved as {(x+y)/xy}=6. Substituting for 1/xy=6, we get
x+y=5/6

==> (x+y)/5= 5/(6*5)=1/6.

I am stuck trying to understand this explanation. I got the right answer, but not this way.

$$\frac{1}{x}+ \frac{1}{y} = 5$$

$$(\frac{1}{x}+ \frac{1}{y}) = \frac{x + y}{xy}$$

Aren't the two RHS equivalent?

Quote:
(1/X+1/Y)=5 can be solved as {(x+y)/xy}=6.

Can anyone explain the steps from the first part of the above sentence to the last part? If the two RHS above quote are equivalent, where does "= 6" come from?
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Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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02 Apr 2018, 16:10
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.
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Math Expert
Joined: 02 Sep 2009
Posts: 58327
Re: If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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02 Apr 2018, 20:52
msurls wrote:
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.

$$\frac{1}{x} + \frac{1}{y} = 5$$;

$$\frac{x+y}{xy} = 5$$;

Since given that $$\frac{1}{xy}= 6$$, then $$6(x+y)=5$$.

Divide by 30: $$\frac{x+y}{5}=\frac{1}{6}$$.

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If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?  [#permalink]

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10 Apr 2018, 09:03
Bunuel wrote:
msurls wrote:
Bunuel wrote:
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?

A. 1/25
B. 1/6
C. 1/5
D. 5
E. 6

Bunuel, will you please provide the solution?

I don't understand the solution above either. Thanks in advance.

$$\frac{1}{x} + \frac{1}{y} = 5$$;

$$\frac{x+y}{xy} = 5$$;

Since given that $$\frac{1}{xy}= 6$$, then $$6(x+y)=5$$.

Divide by 30: $$\frac{x+y}{5}=\frac{1}{6}$$.

My approach is pretty similar:

$$\frac{1}{x} + \frac{1}{y} = 5$$;

$$\frac{x+y}{xy} = 5$$;

$$\frac{x+y}{5} = XY$$;

$$\frac{1}{xy} = 6$$ --> XY = $$;\frac{1}{6}$$;

so

$$\frac{x+y}{5} = XY = 1/6$$;
If xy > 0, 1/x + 1/y = 5, and 1/xy = 6, then (x+y)/5 = ?   [#permalink] 10 Apr 2018, 09:03
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