# Verify if the difference (1-9/y^2)/(1-3/y)-3/y=1

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### 2 Answers

We have to verify (1-9/y^2)/(1-3/y)-3/y=1

(1-9/y^2)/(1-3/y)-3/y

=> ((y^2 - 9)/y^2)/((y-3)/y) - 3/y

=> ((y^2 - 9)y/y^2(y-3) - 3/y

=> ((y^2 - 9)/y(y-3) - 3/y

=> (y - 3)(y + 3)/y(y-3) - 3/y

=> (y + 3)/y - 3/y

=> y/y + 3/y - 3/y

=> y/y

=> 1

**This proves that (1-9/y^2)/(1-3/y)-3/y=1**

We'll multiply the second fraction by the term (1-3/y):

[(1-9/y^2) - (3/y)*(1-3/y)]/(1-3/y)

The difference of squares returns the product:

1-9/y^2 = (1-3/y)(1+3/y)

[(1-9/y^2) - (3/y)*(1-3/y)]/(1-3/y) = [(1-3/y)(1+3/y) - (3/y)

(1-3/y)]/(1-3/y)

We'll factorize the numerator by (1-3/y):

(1-3/y) [(1+3/y) - (3/y)]/(1-3/y)

We'll simplify by (1-3/y):

(1-3/y) [(1+3/y) - (3/y)]/(1-3/y) = [(1+3/y) - (3/y)]

We'll eliminate like terms inside brackets:

(1-9/y^2)/(1-3/y)-3/y=1

We notice that the given difference yields 1.