Question

@Mertsj and @satellite73 Simplify each expression. 16. 3x + 6 over x^2 - 9 divided by 6x^2 + 12x over 4x + 12

Answer 1

\[\frac{3x+6}{x^2-9}\times \frac{4x+12}{6x^2+12x}\] is a start, then factor a bunch and cancel a bunch

Answer 2

Can you show me, because I am totally stuck on this problem

Answer 3

first step is to factor
\[\frac{3(x+2)4(x+3)}{(x+3)(x-3)6x(x+2)}\] then get rid of the common factors

Answer 4

Cause the next part confused me

Answer 5

then you are done

Answer 6

What would it look like when it's done

Answer 7

you should only be left with \(\frac{2}{x-3}\)

Answer 8

How did you get that?

Answer 9

\[\frac{3\cancel{(x+2)}4(x+3)}{(x+3)(x-3)6x\cancel{(x+2)}}\]
\[\frac{12(x+3)}{6x(x+3)(x-3)}\]
\[\frac{12\cancel{(x+3)}}{6x\cancel{(x+3)}(x-3)}\]\[\frac{12}{6x(x-3)}\]

Answer 10

then finally cancel the 6

Answer 11

so the final answer would be: 2 over x - 3

Answer 12

\[\frac{2}{x(x-3)}\]

Answer 13

That's the correct answer.

Answer 14

why is the 2 on top

Answer 15

Because 12/6 = 2

Answer 16

because 12 is the bigger number right