Help with adding and subtracting rational expressions?

(5y + 2).. 4.....(2x - 4)... y

---------- * --- + --------- * ---...............get like denominators 4xy^2 is the LCD

..xy^2......4........4xy.......y

20y + 8 + 2xy - 4y

------------------------

.......4xy^2

2xy + 16y + 8

------------------

.......4xy^2

xy + 8y + 4

---------------

...2xy^2

Your denominator COULD be (xy^2)(4xy), but (xy^2) is your lowest common denominator, meaning it will be simpler to use because you will have fewer extra factors running about. ( 4xy^2 would be the lowest common denominator, but notice how 4/(4xy)=1/xy.)

You have to multiply each term by whatever will give you a denominator with xy^2. For example:

5y( xy^2/ xy^2) = (5xy^3)/xy^2

(2/xy^2) ( 1/1) = 2/xy^2

2x( xy^2/ xy^2) = (2x^2y^2)/ xy^2

1/(xy)( y/y) = y/(xy^2) Make sure that you wrote the equation correctly on your computer or else the parenthesis (or lack of them) will completely mess up your equation.

Then add!

( (5xy^3) + 2 + (2x^2y^2) + y)/(xy^2)

4xy^2

if im not mistaken, that is the lcd or least common denominator (but im sure it's 4xy^2)..

you'll use that to divide it with the denominators of your terms and multiply the quotient to the numerators..like this:

4xy^2/xy^2 = 4

4xy^2/4xy = y

then:

[4(5y+2) + y(2x-4)] / 4xy^2 and you finish solving it ^_^

if you need help finding the lcd, let me know