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tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
The equation looks like this: dw:1363123735015:dw
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix can you help me, please?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\]Does this look accurate? :) I added the brackets so we can see which one is the fraction on top of the fraction :P
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
Aha! Yes, it's perfect. :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Let's start by multiplying by \(\large \dfrac{x}{x}\).\[\large \frac{\left(\dfrac{4}{x+3}\right)}{\dfrac{1}{x}+3}\left(\frac{x}{x}\right)\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Understand where to put the x within the brackets?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We can think of it like this, \[\large \frac{\left(\dfrac{4}{x+3}\right)x}{\left(\dfrac{1}{x}+3\right)x}\]
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
I believe so.... I'll end up with... 4x/3 for the top fraction? And 1 + 3x for the bottom fraction?
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
4x/ x^2 + 3x
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{\left(\dfrac{4x}{x\color{orangered}{+3}}\right)}{\left(1+3x\right)}\]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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Woops I did the wrong portion in orange, my bad
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
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. :)
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
I didn't put it into 'fancy' form, though. :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large \frac{\left(\dfrac{4x}{x+3}\right)}{\left(1+3x\right)} \qquad = \qquad \frac{4x}{(x+3)(3x+1)}\]This part make sense? :o
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Sorry for the delay :c I couldn't connect to the site for a bit there.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1363126766341:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1363126871122:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

danielcvalencia Group TitleBest ResponseYou've already chosen the best response.0
Welcome to OpenStudy, @tafkas77 !
 one year ago

tafkas77 Group TitleBest ResponseYou've already chosen the best response.0
@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! :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
yah looks right c: good job.
 one year ago
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