Complex fraction simplification

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

Complex fraction simplification

Hi, so I'm working through some problems and see two apparently contradicting rules for simplifying a 'fraction over a fraction'. Sorry I cannot figure out how to write proper expressions here... I'm just going to type them out as best as I can.

First, there's x/5 over 2. My solution was to move the 2 up to the top and get 2x/5. Book says no, answer is x/10. Apparently, instead of multiplying by 2/2, I must multiply by 1/2. Ok, this makes sense if you plug in x=1, then .2/2 is .1, not .4. And dividing a number by 5 then also by 2 would be the same as dividing by 10. Damn, don't know where I got that idea from.

Along comes another problem, where we have 1/(x+1)/x. Solution? x/(x+1). Wtf? Thought I couldn't do that? But this makes sense by multiplying by x/x. Now, in the first problem, the 'small fraction' is in the top portion of the 'big fraction'. In the this one, it's in the bottom portion. But last I checked PEMDAS, this shouldn't make a freaking difference

Could someone explain the difference here? I'm clearly missing something...

Thank you
chillywings

Posts: 2
Joined: Thu Mar 06, 2014 6:43 pm

Re: Complex fraction simplification

chillywings wrote:First, there's x/5 over 2.

Which of the following do you mean?

. . . . .$\mbox{a. }\, \frac{x}{\left(\frac{5}{2}\right)}$
. . . . .$\mbox{b. }\, \frac{\left(\frac{x}{5}\right)}{2}$

chillywings wrote:we have 1/(x+1)/x.

Which of the following do you mean?

. . . . .$\mbox{c. }\, \frac{1}{\left(\frac{x\, +\, 1}{x}\right)}$
. . . . .$\mbox{d. }\, \frac{\left(\frac{1}{x\, +\, 1}\right)}{x}$

stapel_eliz

Posts: 1720
Joined: Mon Dec 08, 2008 4:22 pm

Re: Complex fraction simplification

b and c. Thanks for the conversion. Do I merely have to operate as though those parenthesis are there all the time?
chillywings

Posts: 2
Joined: Thu Mar 06, 2014 6:43 pm

Re: Complex fraction simplification

stapel_eliz wrote:. . . . .$\mbox{b. }\, \frac{\left(\frac{x}{5}\right)}{2}$

. . . . .$\mbox{c. }\, \frac{1}{\left(\frac{x\, +\, 1}{x}\right)}$
chillywings wrote:b and c. Thanks for the conversion. Do I merely have to operate as though those parenthesis are there all the time?

If theres no parentheses then you cant tell what they mean. You have to respect the parens!