Question

Simplifying complex fractions.... Help!

Answer 1

The equation looks like this:

Answer 2

@zepdrix can you help me, please?

Answer 3

\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\]Does this look accurate? :)
I added the brackets so we can see which one is the fraction on top of the fraction :P

Answer 4

Aha! Yes, it's perfect. :)

Answer 5

Let's start by multiplying by \(\large \dfrac{x}{x}\).\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\left(\frac{x}{x}\right)\]

Answer 6

Understand where to put the x within the brackets?

Answer 7

We can think of it like this,
\[\large \frac{\left(\dfrac{4}{x+3}\right)x}{\left(\dfrac{1}{x}+3\right)x}\]

Answer 8

I believe so.... I'll end up with... 4x/3 for the top fraction? And 1 + 3x for the bottom fraction?

Answer 9

wait...

Answer 10

4x/ x^2 + 3x

Answer 11

\[\large \frac{\left(\dfrac{4x}{x\color{orangered}{+3}}\right)}{\left(1+3x\right)}\]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.

Answer 12

Woops I did the wrong portion in orange, my bad

Answer 13

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. :)

Answer 14

I didn't put it into 'fancy' form, though. :)

Answer 15

\[\large \frac{\left(\dfrac{4x}{x+3}\right)}{\left(1+3x\right)} \qquad = \qquad \frac{4x}{(x+3)(3x+1)}\]This part make sense? :o

Answer 16

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?

Answer 17

Sorry for the delay :c I couldn't connect to the site for a bit there.