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# If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ?

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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
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VeritasPrepKarishma wrote:
zatspeed wrote:
If Y = 2 + 2K and Y(not equal to) 0 , Then 1/Y + 1/Y + 1/Y + 1/Y = ?

You should give the options.

1/Y + 1/Y + 1/Y + 1/Y
= 4/Y
= 4/(2 + 2K)
= 2/(1 + K)

Sorry, I have updated it now.
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
2
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VeritasPrepKarishma wrote:
zatspeed wrote:
If Y = 2 + 2K and Y(not equal to) 0 , Then 1/Y + 1/Y + 1/Y + 1/Y = ?

You should give the options.

1/Y + 1/Y + 1/Y + 1/Y
= 4/Y
= 4/(2 + 2K)
= 2/(1 + K)

Responding to a pm:

Quote:
For you last two steps, this involves using the conjugate of the denominator and then recognizing that 1-k2 can be expressed as a difference of squares, correct?

The question gives you that
$$Y = 2 + 2K$$

So you simply substitute that in place of Y in the denominator.

$$\frac{4}{Y} = \frac{4}{(2 + 2K)}$$

Then just take 2 common from the numerator and denominator to get:

$$\frac{4}{(2 + 2K)} = \frac{2*2}{2(1 + K)}$$

Cancel off the 2 of the numerator with the 2 of the denominator to get:

$$\frac{2*2}{2(1 + K)} = \frac{2}{(1 + K)}$$
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
y=2(1+k)
4(1/y)= 4(1/2(1+k))= 2/(1+k)
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
zatspeed wrote:
If y = 2 + 2K and $$y\neq{0}$$, then 1/y + 1/y + 1/y + 1/y = ?

A. 1/(8+8k)
B. 2/(1+k)
C. 1/(8+k)
D. 4/(8+k)
E. 4/(1+k)

y = 2+2k

1/y +1/y + 1/y+1/y = 4/y

4/(2+2k) = 2/(1+k)
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
zatspeed wrote:
If y = 2 + 2K and $$y\neq{0}$$, then 1/y + 1/y + 1/y + 1/y = ?

A. 1/(8+8k)
B. 2/(1+k)
C. 1/(8+k)
D. 4/(8+k)
E. 4/(1+k)

y= 2+ 2k

1/y+1/y+1/y+1/y= 4/y
plug 2+2k in for y
4/2+2k--> simplify to 2/1+K
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
zatspeed wrote:
If y = 2 + 2K and $$y\neq{0}$$, then 1/y + 1/y + 1/y + 1/y = ?

A. 1/(8+8k)
B. 2/(1+k)
C. 1/(8+k)
D. 4/(8+k)
E. 4/(1+k)

Adding the given fractions we have:

4/y

Since y = 2 + 2k, we have:

4/(2 + 2k) = 4/[2(1 + k)] = 2/(1 + k)

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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
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Re: If y = 2 + 2K and y#0, then 1/y + 1/y + 1/y + 1/y = ? [#permalink]
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