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# If 1/c - 1/(c + 3) = 1/(c + 5), then c could be

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Joined: 02 Sep 2009
Posts: 58098
If 1/c - 1/(c + 3) = 1/(c + 5), then c could be  [#permalink]

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22 Aug 2019, 01:30
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Difficulty:

15% (low)

Question Stats:

96% (01:56) correct 4% (02:09) wrong based on 27 sessions

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If 1/c - 1/(c + 3) = 1/(c + 5), then c could be

A. $$\sqrt{15}$$
B. 1
C. 0
D. -1
E. -7

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If 1/c - 1/(c + 3) = 1/(c + 5), then c could be  [#permalink]

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22 Aug 2019, 01:37
1/c - 1/(c + 3) = 1/(c + 5)

try c value given in each option and put that value into the above equation. We easily find that Options (B) to (E) are wrong.
A. √15 --> skip
B. 1 --> 1 - 1/4 = 1/6 (No!)
C. 0 --> 1/0 - ... (No!)
D. -1 --> - 1 - 1/2 = 1/4 (No!)
E. -7 --> -1/7 + 1/4 = -1/2 (No!)

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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be  [#permalink]

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22 Aug 2019, 01:50
1
$$\frac{1}{c} - \frac{1}{(c + 3)}$$ = $$\frac{1}{(c + 5)}$$

$$\frac{3}{c(c + 3)}$$ = $$\frac{1}{(c + 5)}$$

$$3c + 15 = c^2 + 3c$$

c = $$\sqrt{15}$$

OPTION: A
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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be  [#permalink]

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22 Aug 2019, 04:48
c+3-c /c^2+3c = 1/c+5
c^2 + 3c=3c+15
c^2=15
c=√15
Option A

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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be  [#permalink]

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24 Aug 2019, 05:27
1
Bunuel wrote:
If 1/c - 1/(c + 3) = 1/(c + 5), then c could be

A. $$\sqrt{15}$$
B. 1
C. 0
D. -1
E. -7

We clear all fractions in the equation by multiplying the entire equation by c(c+3)(c+5), obtaining:

(c+3)(c+5) - c(c+5) = c(c+3)

c^2 + 8c + 15 - c^2 - 5c = c^2 + 3c

c^2 = 15

c = ±√15

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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be   [#permalink] 24 Aug 2019, 05:27
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