Question

Please help...
x^2+2x/ 4x^2y - y/14x^2

Answer 1

this problem is

\[\Large \frac{x^2+2x}{4x^2y} - \frac{y}{14x^2}\]

correct?

Answer 2

yes.

Answer 3

the LCD is 28x^2y

do you see how I got this?

Answer 4

kind of

Answer 5

the LCM of the denominators will give you the LCD

so the LCM of 4x^2y and 14x^2 is 28x^2y

Further breakdown:

LCM of 4 and 14 is 28
LCM of x^2 and y is x^2y

Answer 6

so that's how I got the LCD of 28x^2y

Answer 7

okay i see it now

Answer 8

the goal is to get each denominator equal to the LCD

once you've achieved that goal, you can subtract the two fractions

Answer 9

so how do we get that first fraction to have a denominator of 28x^2y ?

Answer 10

so for the first fraction i would multiply it by 7?

Answer 11

good, multiply top and bottom by 7

Answer 12

then the second on by 2y?

Answer 13

\[\Large \frac{x^2+2x}{4x^2y} - \frac{y}{14x^2}\]

\[\Large \frac{7(x^2+2x)}{7*4x^2y} - \frac{y}{14x^2}\]

\[\Large \frac{7x^2+14x)}{28x^2y} - \frac{y}{14x^2}\]

Answer 14

very good

Answer 15

\[\Large \frac{7x^2+14x)}{28x^2y} - \frac{y}{14x^2}\]

\[\Large \frac{7x^2+14x)}{28x^2y} - \frac{2y*y}{2y*14x^2}\]

\[\Large \frac{7x^2+14x)}{28x^2y} - \frac{2y^2}{28x^2y}\]

now you can subtract the numerators and you place the result over the LCD

Answer 16

that will give you

\[\Large \frac{7x^2+14x-2y^2}{28x^2y}\]

Answer 17

so from there do you simplify the problem to get it to simplest form?

Answer 18

yes that's only true if you could simplify the numerator, but you can't

Answer 19

so you just leave it as \[\Large \frac{7x^2+14x-2y^2}{28x^2y}\]

Answer 20

okay:) thank you very much!

Answer 21

you're welcome