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You need to bring the fractions to the right to a common denominator, such that:
`1/y - 1/(y+1) = (y + 1)*1/(y*(y + 1)) - y*1/(y*(y + 1))`
`1/y - 1/(y+1) = (y + 1 - y)/(y*(y + 1))`
Reducing duplicate terms yields:
`1/y - 1/(y+1) = 1/(y*(y + 1))`
Hence, testing if `1/(y*(y + 1)) = 1/y - 1/(y+1)` yields that the statement is valid.
We'll solve the problem starting from the idea that the fraction 1/y(y+1) is the result of addition or subtraction of 2 elementary fractions:
1/y(y+1) = A/y + B/(y+1) (1)
We'll multiply the ratio A/y by (y+1) and we'll multiply the ratio B/(y+1) by y.
1/y(y+1) = [A(y+1) + By]/y(y+1)
Since the denominators of both sides are matching, we'll write the numerators, only.
1 = A(y+1) + By
We'll remove the brackets:
1 = Ay + A + By
We'll factorize by a to the right side:
1 = y(A+B) + A
If the expressions from both sides are equivalent, the correspondent coefficients are equal.
A+B = 0
A = 1
1 + B = 0
B = -1
We'll substitute A and B into the expression (1):
1/y(y+1) = 1/y - 1/(y+1)
Since the LHS=RHS, the identity was verified.
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