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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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Re: If 1/y − 1/(y+1) = 1/(y+4), what is the sum of all solutions for y? [#permalink]
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]
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]
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.