Hi, I'm new here and studying for a placement test, and I needed help with this particular question (please do not just give me the answer; I want an explanation on how to solve the question).

If x is not equal to 0, then u/x + 5u/x - u/5x =
a. 7x/5u
b. 5u/7x
c. 29u/5x
d. 31u/5x

Also, what would you call this type of problem? I would like to know so that I can brush up on it, further.

Thanks in advance for the help! ^.^

(ALSO THIS IS BEING REPOSTED BECAUSE A LOVELY PERSON MENTIONED TO ME THAT I ACCIDENTALLY POSTED IN THE FEEDBACK SECTION)

Question

Hi, I'm new here and studying for a placement test, and I needed help with this particular question (please do not just give me the answer; I want an explanation on how to solve the question).

If x is not equal to 0, then u/x + 5u/x - u/5x =
a. 7x/5u
b. 5u/7x
c. 29u/5x
d. 31u/5x

Also, what would you call this type of problem? I would like to know so that I can brush up on it, further.

Thanks in advance for the help! ^.^

(ALSO THIS IS BEING REPOSTED BECAUSE A LOVELY PERSON MENTIONED TO ME THAT I ACCIDENTALLY POSTED IN THE FEEDBACK SECTION)

skullpatrol · Answer

This type of problem is called the Algebra of polynomials.

katnisssullen · Answer

Really? If not one of those answers, then what is the correct answer?

skullpatrol · Answer

Does  u/x + 5u/x - u/5x mean? $$\Huge\dfrac{u}{x}+ \dfrac{5u}{x}- \dfrac{u}{5x}$$

katnisssullen · Answer

Yes, that's what it looks like. Sorry if it wasn't clear enough.

skullpatrol · Answer

Do you recall the distributive property?

katnisssullen · Answer

Yes, I do. Would I be distributing the 5?

skullpatrol · Answer

Nope.  ba + ca = ( b + c )a

skullpatrol · Answer

Factor out the 1/x.

katnisssullen · Answer

1/x? How exactly would I go about doing that?

skullpatrol · Answer

$$\Huge\dfrac{u}{x}+ \dfrac{5u}{x}- \dfrac{u}{5x}= (u + 5u - \dfrac{1}{5}u)\dfrac{1}{x}$$

skullpatrol · Answer

Do you see how I did that?

katnisssullen · Answer

Mostly... so the x's are cancelled out? And how did you get 1/5u? I assume that 1/5u came from u/5x... Sorry if I'm a bit of a slow learner, by the way... :P

skullpatrol · Answer

You can also factor out the u and then simplify: 

[ 1 + 5 – (1/5) ] ( u/x)

skullpatrol · Answer

= [ 29/5 ] ( u/x ) c.

usukidoll · Answer

or you could just use lcd.. the lcd is like 5x and the 5 missing on two fractions so multiply 5/5 on those 2 fractions and just add up all the numerators. faster

katnisssullen · Answer

Thank you, skullpatrol.

So UsukiDoll, if I were to use that method, then does that just ignore the u variables in the equation and treat them as ones, instead?

usukidoll · Answer

$$\Large\dfrac{u}{x}\cdot \frac{5}{5}+ \dfrac{5u}{x}\cdot \frac{5}{5}- \dfrac{u}{5x} \rightarrow \frac{5u+25u-u}{5x}=\frac{29u}{5x}$$
lcd is 5x or is it gcf 5x anyway you need the denomiators to be the same. look how many I steps I took vs. that other method

katnisssullen · Answer

I see... that is helpful :D And I'm glad that there's more than one way to do it so that I can try both methods, in the future... thank you both so much ^.^ I really appreciate it.