Question

help me please!

Answer 1

@jada1701877 thanks but why'd you give me a medal? lol

Answer 2

factor and cancel

Answer 3

\[\frac{(x+3)(x-3)}{x(x-3)}\] is a start

Answer 4

Because that looks confusing and at least you asked for help. Tbh some people don't even ask for help. They just fail

Answer 5

how do i factor that? like what do i do first? @satellite73

Answer 6

thanks! :) @jada1701877

Answer 7

i factored it for you
the top factors because it is the difference of two squares
\[a^2-b^2=(a+b)(a-b)\] so
\[x^2-9=(x+3)(x-3)\]

Answer 8

the denominator factors because each term has a common factor of \(x\) making
\[x^2-3x=x(x-3)\]

Answer 9

then we subtract x^2 ?

Answer 10

no

Answer 11

\[\frac{x^2-9}{x^2-3x}=\frac{(x+3)(x-3)}{x(x-3)}\]

Answer 12

then cancel the common factor of \(x-3\) top and bottom

Answer 13

x^2-9 / x^2-3x = (x+3)/x

Answer 14

yes

Answer 15

Thank you! @satellite73 :) :) :)

Answer 16

yw