math help please?

Question

Add or subtract as indicated. Express your result in simplest form.

4y / y^2 + 6y + 5
+
2y / y^2 - 1

Answer 1

=4/y+6y+5+2/y-1
=4/y+2/y+6y+4
=2(2/y+1/y)+2(3y+2)
=2y(3)+2y(3y+2)
=6y+6y^2+4y
=2(3y+3y^2+2y)

Answer 2

4y / y^2 + 6y + 5 + 2y / y^2 - 1 =
(4y + 2y) / y^2 + 6y + 5 - 1 =
6y/y^2 + 6y + 4 =
6/y + 6y + 4 =
2 (3/y + 3y +2)

cheers!

Answer 3

Question should read:-
4y/(y² + 6y + 5) + 2y/(y² - 1)
= 4y / (y + 5).(y + 1) + 2y / (y - 1).(y + 1)
= [4y(y - 1) + 2y(y + 5)] / [(y + 5).(y - 1).(y + 1)]
= (6y² + 6y) / (y + 5).(y - 1).(y + 1)
= 6y.(y + 1) / (y + 5).(y - 1).(y + 1)
= 6y / (y + 5).(y - 1)

Answer 4

4y / y^2 + 6y + 5 + 2y / y^2 - 1

*regrouping {4y/y^2 + 2y/y^2 = (4y + 2y) / y^2}

= (4y + 2y) / y^2 + 6y + 5 - 1

*cancellation {y/y^2=1/y)

= 6y/y^2 + 6y + 4

* 6y/y^2 = 6/y

= 6/y + 6y + 4

* GCF of numerical coefficients is 2 (num. coef. are 6,6,& 4)

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

Answer 5

(4y/y^2+6y+5)+(2y/y^2-1)

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

4y(y+5)+2y(y-1)

4y^2+20y+2y^2-2y

(6y^2+18y)/(y+1)

Answer:
6y(y+3)/y+1

The common denominator is (y+1). Then just follow as you would a normal addition problem and you should get the above answer.

Answer 6

4y/ (y^2 + 6y + 5) + 2y / (y^2 - 1)
= 4y/ [(y + 5) (y + 1)] + 2y / [(y + 1) (y - )]
= [4y (y - 1) + 2y (y + 5)] / [(y + 5) (y + 1) (y -1)]
= (4y^2 - 4y + 2y^2 + 10y) / [(y + 5) (y + 1) (y -1)]
= (6y^2 + 6y) /[(y + 5) (y + 1) (y -1)]
= 6y (y + 1) / [(y + 5) (y + 1) (y -1)]
= 6y / [(y + 5) (y -1)]
= 6y / (y^2 - 4y - 5)

Answer 7

(4y/y^2+6y+5) + (2y/y^2-1)

taking LCM

(4y+6y^3+5y^2)/y^2 + (2y-y^2)/y^2

adding those terms

(4y+6y^3+5y^2+2y-y^2)/y^2

(6y+6y^3+4y^2)/y^2

cancelling "y" from numerator and denominator

(6+6y^2+4y)/y

for above quadratic expression we get imaginary roots

so wot all i can solve is this much only

Answer 8

6/y+6y+4

Answer 9

4y/y^2+6y+5+2y/y^2-1

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

=6y/y^2+4

Answer 10

Simplifying the roots, we get:
[4y / (y+5) (y+1)] + [2y / (y-1) (y+1)]
Adding both we get:
4y(y+1)(y-1) + 2y (y+1) (y+5)
= -----------------------------------------
(y+5)(y+1)(y+1)(y-1)
Taking (y+1) commonly, we get
(y+1) {4y(y-1) + 2y(y-1)}
--------------------------------
(y+5)(y+1)(y+1)(y-1)
Cancelling (y+1), we get:
4y(y-1) + 2y(y-1)
--------------------------------
(y+5)(y+1)(y-1)
Further simplifying:
4y^2-4y+2y^2-2y
--------------------------------
(y+5)(y+1)(y-1)
and we get:
6y^2 -6y
--------------------------------
(y+5)(y+1)(y-1)
or:
6y(y-1)
--------------------------------
(y+5)(y+1)(y-1)
Cancelling (y-1), we get:
6y/(y+5)(y+1)
In other words,
6y/(y^2+6x+5)