anonymous
  • anonymous
Simplifying complex fractions.... Help!
Algebra
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Aha! Yes, it's perfect. :)
zepdrix
  • zepdrix
Let's start by multiplying by \(\large \dfrac{x}{x}\).\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\left(\frac{x}{x}\right)\]
zepdrix
  • zepdrix
Understand where to put the x within the brackets?
zepdrix
  • zepdrix
We can think of it like this, \[\large \frac{\left(\dfrac{4}{x+3}\right)x}{\left(\dfrac{1}{x}+3\right)x}\]
anonymous
  • anonymous
I believe so.... I'll end up with... 4x/3 for the top fraction? And 1 + 3x for the bottom fraction?
anonymous
  • anonymous
wait...
anonymous
  • anonymous
4x/ x^2 + 3x
zepdrix
  • zepdrix
\[\large \frac{\left(\dfrac{4x}{x\color{orangered}{+3}}\right)}{\left(1+3x\right)}\]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.
zepdrix
  • zepdrix
Woops I did the wrong portion in orange, my bad
anonymous
  • anonymous
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. :)
anonymous
  • anonymous
I didn't put it into 'fancy' form, though. :)
zepdrix
  • zepdrix
\[\large \frac{\left(\dfrac{4x}{x+3}\right)}{\left(1+3x\right)} \qquad = \qquad \frac{4x}{(x+3)(3x+1)}\]This part make sense? :o
anonymous
  • anonymous
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?
zepdrix
  • zepdrix
Sorry for the delay :c I couldn't connect to the site for a bit there.
zepdrix
  • zepdrix
|dw:1363126766341:dw|
zepdrix
  • zepdrix
|dw:1363126871122:dw|
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
Welcome to OpenStudy, @tafkas77 !
anonymous
  • anonymous
@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! :)
zepdrix
  • zepdrix
yah looks right c: good job.

Looking for something else?

Not the answer you are looking for? Search for more explanations.