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If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y?

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Bunuel
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If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   16 Jun 2017, 04:01
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If \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\), what is the sum of all solutions for y?

A. -2
B. -1
C. 0
D. 1
E. 2
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C
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BrentGMATPrepNow
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If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   Updated on: 28 Dec 2019, 07:33
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Bunuel wrote:
If \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\), what is the sum of all solutions for y?

A. -2
B. -1
C. 0
D. 1
E. 2


Given: \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\)

To eliminate the fractions, we'll multiply both sides of the equation by the least common multiple of y, y+1, and y+4

So, multiply both sides by (y)(y+1)(y+4) to get: (y+1)(y+4) - (y)(y+4) = (y)(y+1)
Expand to get: [y² + 5y + 4] - [y² + 4y] = y² + y
Simplify: y + 4 = y² + y
Subtract y from both sides to get: 4 = y²
So, y = 2 or y = -2

The SUM of the solutions = 2 + (-2) = 0

Answer: C

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 16 Jun 2017, 07:12.
Last edited by BrentGMATPrepNow on 28 Dec 2019, 07:33, edited 1 time in total.
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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   16 Jun 2017, 05:09
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Bunuel wrote:
If \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\), what is the sum of all solutions for y?

A. -2
B. -1
C. 0
D. 1
E. 2


\(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\)

\(\frac{1}{y(y+1)} = \frac{1}{(y+4)}\)

i.e. y(y+1)=(y+4)

i.e. y^2 = 4

i.e. y = +2 and -2

Sum of values = 0

Answer: Option C
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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   17 Jun 2017, 02:13
Bunuel wrote:
If \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\), what is the sum of all solutions for y?

A. -2
B. -1
C. 0
D. 1
E. 2



1/y−1/(y+1) = 1/(y+4)
=>1/y(y+1) = 1/(y+4)
=> y(y+1) = (y+4)
=> y^2 +y = y +4
=> y^2-4=0
=>y = +2 or -2
Ans C
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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   21 Jun 2017, 01:43
solving:

(y + 1 - y)/y(y+1) = 1/y+4

y + 4 = y^2 + y

(y-2)(y+2) = 0

y = 2 or - 2

sum of roots is 0
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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink] New post   09 Jun 2019, 19:21
Expert Reply
Bunuel wrote:
If \(\frac{1}{y} − \frac{1}{(y+1)} = \frac{1}{(y+4)}\), what is the sum of all solutions for y?

A. -2
B. -1
C. 0
D. 1
E. 2



We start by combining the two fractions on the left hand side, obtaining:

(y + 1 - y)/[y(y + 1)] = 1/(y + 4)

1/[y^2 + y)] = 1/(y + 4)

We now cross-multiply to obtain:

y^2 + y = y + 4

y^2 = 4

So y = 2 or y = -2, and therefore, the sum of the solutions is 0.

Answer: C
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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink]
09 Jun 2019, 19:21
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