Algebra help please?

Question

how do i perform the indicated operations, and reduce answer to lowest terms

2 y 2y-1
____ + ______ - ____
6y-2 9y^2 -1 1-3y

thank you

Answer 1

2 y 2y-1
____ + ______ - ____
6y-2 9y^2 -1 1-3y

restated :

2y/(6y-2) + 2y/(9y^2 -1) - 1/(1-3y) = ....

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

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

etc ...

grmpffff

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

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

pffffff

[ { (y+1)(3y+1) + 2y} / (3y+1) ] / (3y-1) =


[ { (y+1)(3y+1) + 2y} / (3y+1) ] / (3y-1) =

i give up


as final answer i would give the expression just before I started to complain ... its good enough for me , tell that your teacher gjmb think its good enough

Answer 2

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^one million + 3*a^one million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^one million + 3*x^one million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 3

Would you mind restating the question. . .and formatting your question better. I don't understand what you are asking.

Hint: Use the preview button; it works wonders when trying to post fractions in the form that you attempted to post them in.

Answer 4

(2/(6y - 2)) + (y/(9y^2 - 1)) - ((2y - 1)/(1 - 3y))
(2/(2(3y - 1))) + (y/((3y - 1)(3y + 1))) - ((2y - 1)/(-3y + 1))
(1/(3y - 1)) + (y/((3y - 1)(3y + 1))) - ((2y - 1)/(-(3y - 1)))
(1/(3y - 1)) + (y/((3y - 1)(3y + 1))) + ((2y - 1)/(3y - 1))

Multiply everything by (3y - 1)(3y + 1)

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

since nothing can be reduced

ANS : (6y^2 + 3y)/(9y^2 - 1)