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Simplifying complex fractions.... Help!

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The equation looks like this: |dw:1363123735015:dw|
@zepdrix can you help me, please?
\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\]Does this look accurate? :) I added the brackets so we can see which one is the fraction on top of the fraction :P

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Other answers:

Aha! Yes, it's perfect. :)
Let's start by multiplying by \(\large \dfrac{x}{x}\).\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\left(\frac{x}{x}\right)\]
Understand where to put the x within the brackets?
We can think of it like this, \[\large \frac{\left(\dfrac{4}{x+3}\right)x}{\left(\dfrac{1}{x}+3\right)x}\]
I believe so.... I'll end up with... 4x/3 for the top fraction? And 1 + 3x for the bottom fraction?
4x/ x^2 + 3x
\[\large \frac{\left(\dfrac{4x}{x\color{orangered}{+3}}\right)}{\left(1+3x\right)}\]Hmm you were on the right track with your first guess :) you just missed this little orange part. See how the top x is multiplying the 4, and the bottom one multiplied both terms.
Woops I did the wrong portion in orange, my bad
I see my mistake... :) So was my second guess right? I changed my answer to \[\frac{4x }{ x^2 + 3x } \] while you were typing, sorry. :)
I didn't put it into 'fancy' form, though. :)
\[\large \frac{\left(\dfrac{4x}{x+3}\right)}{\left(1+3x\right)} \qquad = \qquad \frac{4x}{(x+3)(3x+1)}\]This part make sense? :o
yes; I see what you did. But how come the denominator of the first fraction isn't x^2? Why is it only "x"... we multiplied the x's, right?
Sorry for the delay :c I couldn't connect to the site for a bit there.
The top multiplication looks like this,\[\large \left(\frac{4}{x+3}\right)x\]Which we can think of as,\[\large \left(\frac{4}{x+3}\right)\frac{x}{1}\]Understand why it doesn't produce an x^2 in the bottom? :O We're not multiplying the x+3 by anything.
okay, I understand. Thank you for explaining that to me! :D I'm sorry it took me so long - the site wasn't working! Okay, so after I do all that, I got this as my answer: \[\frac{ 4x }{ 3x^2 + 10x + 3 }\] Is that right?
Welcome to OpenStudy, @tafkas77 !
@danielcvalencia Hee hee, thanks! :D But I am not new to OS! :) I have a 65 SmartScore - I've been here for a little while. This is just my first question in Algebra, not my first question ever. But I appreciate your kindness very much! :)
yah looks right c: good job.

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