What would the LCM be for this?

Question

rezper · Answer

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rezper · Answer

@mathmale Aye can you help me again? xD

mathmale · Answer

What can we learn here from our previous discussion?  In the previous problem, as here, we looked for a way to simplify a complicated fraction.

Look at (a/b) + c in the original problem.  You need to multiply this by the LCD to clear fractions; you must also multiply (a/b)-c by that same LCD.  Then look for a way to reduce the resulting new fraction.  Show all your work, please.

rezper · Answer

Well I don't see how to reduce a variable if it's already at its lowest form

rezper · Answer

I can't reduce c to b and b to a so wouldnt it already be at its lowest form?

rezper · Answer

@mathmale You still there?

mathmale · Answer

Resper, you're not following my instructions.  I wanted you to go back and review your previous post.  We cleared fractions in that problem by multiplying numerator and denominator by 4.  The problem at hand is similar; if you want to reduce this complicated fraction, you must multiply numerator and denominator by the same quantity.  What is that quantity?

rezper · Answer

Alright so I guess multiply the fractions by b?

rezper · Answer

It doesn't look similar... This time it's just variables no numbers.

mathmale · Answer

"I can't reduce c to b and b to a" is an inappropriate use of the word "reduce."
You are reducing fractions; you are not reducing variables such as a, b, c.

Actually, the priniciple is exactly the same.  You're in an algebra course and need to become used to using variables such as a, b and c instead of numbers. 

As you said, mult. numerator and den. by b, simplifying the result as much as you can.

mathmale · Answer

"Alright so I guess multiply the fractions by b?"  yes.

rezper · Answer

I don't know what I would do with b though. Do I combine b into something?

mathmale · Answer

Please read what I typed immediately above.  

You want to mult the numerator (a/b) + c by b, and also mult the denom (a/b)-c by b.
Show your work, please.

rezper · Answer

I see that but I dont know how to multiply b into A/b + c and A/b - c

rezper · Answer

Would the final form be: a+bc/a-bc

rezper · Answer

$$\frac{ a+bc }{ a-bc }$$

mathmale · Answer

Again I ask you to review what we did in the previous problem.  You may need to take notes on our discussions.  You did this work successfully before and can do it again.  
Yes, you have arrived at the correct answer, 
EXCEPT that you must enclose that (a-bc within parentheses unless you use Equation Editor.  Your Equation Editor result is fine, perfect.

rezper · Answer

Well I came to that but I still don't know what the LCM is

rezper · Answer

I guess its B?

rezper · Answer

Since we multiplied b into the other fractions to get something different

mathmale · Answer

You'll need to look that up.  That's the kind of thing you could easily learn thru Google.  
But yes, b is probably the LCM here...the smallest quantity with which you could mult. the numerator and denom. of the original expression to eliminate the fractions in each.

rezper · Answer

Sigh.. This stuff is so hard. I wish it came natural like it does to you.

mathmale · Answer

You're learning.  I advise that you drop phrases such as "this stuff is so hard" and "confusing" and instead make lists of things you have learned and can apply again later.

rezper · Answer

Well it's just that when I watch the videos on this stuff I get so lost so fast, but I appreciate the help.

mathmale · Answer

I will be offline for a while.  perhaps we can continue later.