Question

what is the simplified form of (1/x)-2/x^2+x

Answer 1

is that

\[\Large \frac{1}{x} - \frac{2}{x^2+x}\]

??

Answer 2

you know how to do it

Answer 3

yes

Answer 4

ok in order to subtract, the denominators must be the same

Answer 5

so you need to get both denominators equal to the LCD

Answer 6

in this case, the LCD is x(x+1)

so how do you get the first denominator equal to the LCD?

Answer 7

you have x as the denominator of the first fraction

you want x(x+1)

so what's missing?

Answer 8

idunno

Answer 9

You're missing an (x+1), so you multiply top and bottom of the first fraction by this to get

\[\Large \frac{1}{x} - \frac{2}{x^2+x}\]

\[\Large \frac{1(x+1)}{x(x+1)} - \frac{2}{x^2+x}\]

\[\Large \frac{x+1}{x(x+1)} - \frac{2}{x^2+x}\]

Answer 10

notice how x(x+1) distributes to x^2 + x, so we can further say

\[\Large \frac{x+1}{x(x+1)} - \frac{2}{x^2+x}\]

turns into

\[\Large \frac{x+1}{x^2+x} - \frac{2}{x^2+x}\]

Answer 11

so what am i missing

Answer 12

now you just need to subtract the numerators

Answer 13

and place that difference over the common denominator

Answer 14

ok so what is that

Answer 15

x+1 - 2 = ???

Answer 16

=0

Answer 17

no

Answer 18

ok i dunnno get it

Answer 19

1 - 2 = ???

Answer 20

-1

Answer 21

so x + 2 - 1 turns into x - 1

Answer 22

and

\[\Large \frac{x+1}{x^2+x} - \frac{2}{x^2+x}\]

turns into

\[\Large \frac{x-1}{x^2+x}\]

Answer 23

which is your final answer

Answer 24

so

\[\Large \frac{1}{x} - \frac{2}{x^2+x}\]

fully simplifies to

\[\Large \frac{x-1}{x^2+x}\]

Answer 25

thank u

Answer 26

np