Simplest form? 3y/(y^2+5y+4) + 2y/(y^2-1)

[dkarolasz]

Add...

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

The next step what I did was....

3y/[(y+4)(y+1)] + 2y/(y^2-1)

I believe since nothing could be cancelled then I'll add across,
my answer being

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

 

[morson]

You were doing fine until that last step! We have:

\(\displaystyle \L\ \frac{3y}{y^2 + 5y + 4}\ + \frac{2y}{y^2 - 1}\ = \frac{3y}{(y + 4)(y + 1)}\ + \frac{2y}{(y - 1)(y + 1)}\\), \(\displaystyle y \not=\ 1, -1, -4\)

Get a common denominator.

 

[dkarolasz]

The common denominator would be (y+4)(y+1)(y-1)

is that right?

so then I add the top across, and get 5y/(y+4)(y+1)(y-1)

i'm not sure what the y = with the line through = 1,-1,-4????

 

[tkhunny]

so then I add the top across, and get 5y/(y+4)(y+1)(y-1)

What? You have to get them to the common denominator, first. Each term must be considered separately untill all have a common denominator. THEN you can add the resulting numerators.

 

[dkarolasz]

ok since the common denominator is (y+4)(y+1)(y-1)
next step I did;

3y(y+4)(y+1)(y-1) + 2y(Y+4)(y+1)(y-1) = 15y^2 + 18 so my answer would be

(15y^2 +18)/(y+4)(y+1)(y-1)

Is that right...

 

[dkarolasz]

help????

(15y^2 +20)/(y+4)(y+1)(y-1)


I added wrong it's not 18 it's 20

 

[skeeter]

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

3y/[(y+4)(y+1)] + 2y/[(y+1)(y-1)]

3y(y-1)/[(y+4)(y+1)(y-1)] + 2y(y+4)/[(y+4)(y+1)(y-1)]

(3y^2 - 3y + 2y^2 + 8y)/[(y+4)(y+1)(y-1)]

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

5y(y+1)/[(y+4)(y+1)(y-1)]

5y/[(y+4)(y-1)]