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

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Bunuel
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If 1/c - 1/(c + 3) = 1/(c + 5), then c could be [#permalink] New post   22 Aug 2019, 01:30
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A
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C
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E

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95% (01:57) correct 5% (01:44) wrong based on 40 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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A
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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be [#permalink] New post   22 Aug 2019, 01:50
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\(\frac{1}{c} - \frac{1}{(c + 3)}\) = \(\frac{1}{(c + 5)}\)

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

\(3c + 15 = c^2 + 3c\)

c = \(\sqrt{15}\)

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

Answer: A
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If 1/c - 1/(c + 3) = 1/(c + 5), then c could be [#permalink] New post   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!)

Answer is (A)

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Re: If 1/c - 1/(c + 3) = 1/(c + 5), then c could be [#permalink] New post   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]
22 Aug 2019, 04:48
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