very EASY and SIMPLE question! check~

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very EASY and SIMPLE question! check~

Mathematics
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\[\frac{ 2x }{ x+4 } \times \frac{ 3x }{ 3x} = \frac{ 6x }{ 3x^2 + 12x }\] Why wouldn't this be right, I forgot?
you can simplify that remember it's 3x over 3x = ?
no sorry I should have added more, 3 over 3 is my common denominator i'm using

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my bad about the confusion.
post the original question first plz
alright one moment
Here we are
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but i was told the answer was this? {7x^2 + 3x - 4}/{3x^2 + 12x}
only 3x isn't the common denominator both fractions hve different denominator so multiply them that would be the common denominator
im so confused now
This is what I was shown to do
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Answer
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i'll give you an example \[\frac{ a }{ b } +\frac{ c }{ d }\] b d both are common denominator
ahh i see so they divided into 2 step
\(\color{blue}{\text{Originally Posted by}}\) @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
\[\frac{ 6x }{ 3x^2 + 12 } \times \frac{ x + 4 }{ x + 4 } = \]
like that?
yes right so 3x and x+4 both are common denominator
or one step \[\huge\rm \frac{ 2x(3x)-(x-1)(x+4) }{ 3x(x+4)}\]

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