Question

find the missing numerator. 1/x-1/x+3=??/x(x+3) I will give a MEDAL!

Answer 1

\[\frac{ 1 }{ x - 1 }/{x + 3} = \frac{ ? }{ x(x + 3) }\]
Is that it?

Answer 2

no, the last is correct. but the first part is 1/x - 1/x+3

Answer 3

oh, sorry. then this:
\[\frac{ 1 }{ x } - \frac{ 1 }{ x } + 3 = \frac{ ? }{ x(x + 3) }\]
?

Answer 4

but the x+3 is in parenthesis on the denominator of the second one

Answer 5

lol no

Answer 6

the second one is 1/ (x+3)

Answer 7

sorry.
\[\frac{ 1 }{ x } - \frac{ 1 }{ x + 3 } = \frac{ ? }{x(x + 3) }\]
Okay, this equation makes much more sense. lol

Answer 8

yeahhh
lol

Answer 9

Okay, so first step. You have to change the fractions so that they have the same denominator. Do you know how to do that with variables?

Answer 10

no, i've basically forgotten all of this, and i'm taking my final now ):

Answer 11

thats why i came here lol to help me figure out how to do it

Answer 12

@StudyGurl14

Answer 13

\[\frac{ 1(x + 3) }{ x(x + 3) } - \frac{ 1(x) }{ (x + 3)x} = \frac{ ? }{ x(x + 3) }\]
Now you can add the numerators because the denominators are the same.

Answer 14

ohh okay

Answer 15

can you do the rest?

Answer 16

yes, its three correct?

Answer 17

yes, good! So ? = 3