Maths homework help~~~?

Question

Please help me simplify these as a fraction with a single denominator....

4/y + 3/(y+1) =?

2/(y-3) - 8/5y =?

6/(y-2b) + 4/(2b-y)=?

5/(y^2-2y+1) - 7/y-1=?

And

1/(y^2-49) + 6/(7-y)=?

Answer 1

1) Given fracton is 4/y + 3/(y+1)

= [4(y+1)+3(y)]/y(y+1)

To expand above fraction  using distributive property.

=(4y+4+3y)/y^2+y

= (4+7y)/y^2+y

 

2) 2/(y-3) - 8/5y

= [2(5y)-8(y-3)]/5y(y-3)

= (10y-8y+24)/5y^2-15y

= (2y+24)/5y^2-15y

 

3) 6/(y-2b) + 4/(2b-y)

= 6/(y-2b) + 4/[-(y-2b)]

= 6/(y-2b) - 4/(y-2b)

= (6-4)/(y-2b)

= 2/(y-2b)

 

4) 5/(y^2-2y+1) - 7/(y-1)

= 5/(y-1)² - 7/(y-1)

= [5 - 7(y-1)]/(y-1)²

= [5 - 7y + 7]/(y-1)²

= (12 - 7y)/(y-1)².

 

5) 1/(y^2-49) + 6/(7-y)

= 1/[-(49-y^2)] + 6/(7-y)

= -1/[(7-y)(7+y)] + 6/(7-y)

= [-1+6(7+y)] / (7-y)(7+y)

= [-1+42+6y] / (7-y)(7+y)

= (41+6y) / (7-y)(7+y).