Question

very EASY and SIMPLE question! check~

Answer 1

\[\frac{ 2x }{ x+4 } \times \frac{ 3x }{ 3x} = \frac{ 6x }{ 3x^2 + 12x }\]
Why wouldn't this be right, I forgot?

Answer 2

you can simplify that
remember it's 3x over 3x = ?

Answer 3

no sorry I should have added more, 3 over 3 is my common denominator i'm using

Answer 4

my bad about the confusion.

Answer 5

post the original question first plz

Answer 6

alright one moment

Answer 7

Here we are

Answer 8

but i was told the answer was this?

{7x^2 + 3x - 4}/{3x^2 + 12x}

Answer 9

only 3x isn't the common denominator
both fractions hve different denominator so multiply them that would be the common denominator

Answer 10

im so confused now

Answer 11

This is what I was shown to do

Answer 12

Answer

Answer 13

i'll give you an example \[\frac{ a }{ b } +\frac{ c }{ d }\]
b d both are common denominator

Answer 14

ahh i see so they divided into 2 step

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @kaylaprincess
\[\frac{ 2x }{ x+4 } \times \frac{ 3x }{ 3x} = \frac{ 6x }{ 3x^2 + 12x }\]
Why wouldn't this be right, I forgot?
\(\color{blue}{\text{End of Quote}}\)
now multiply the numerator and denominator by x+4
so basically they separatd one step into 2

Answer 16

\[\frac{ 6x }{ 3x^2 + 12 } \times \frac{ x + 4 }{ x + 4 } = \]

Answer 17

like that?

Answer 18

yes right so
3x and x+4 both are common denominator

Answer 19

or one step \[\huge\rm \frac{ 2x(3x)-(x-1)(x+4) }{ 3x(x+4)}\]