# Adding fractions with variables and using common denominator? Rather basic math, but here I am..

So I have these fractions I need to merge together and shorten:

$$\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}$$

I understand that I need to make the denominators the same so that I can merge these fractions together, but I have no idea how.

I assume I need to find the common denominator, which is a friend told me was $2a^2 - 64$. I don't know what to do from here, please help?

How do I use the common denominator to make these denominators the same?

EDIT: Okay so I failed miserably at this MathJax thing. What I am trying to say is:

$$\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}$$

A quick lesson in writing fractions on this forum would also be immensely appreciated.

• Formatting hint: Fractions written as \$\frac {a+b}{c+d} \$ compile to $\frac {a+b}{c+d}$
– lulu
Oct 12 '15 at 19:41
• Try this link: meta.math.stackexchange.com/questions/5020/… Oct 12 '15 at 19:42

Consider that this expression can be rewritten as:

$$\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}=\frac{1}{2(a+4)} + \frac{4}{a^2-16} + \frac{4}{a-4}$$

Now $(a+4)(a-4)=a^2-16$, so multiply fractions by clever forms of one:

$$\frac{1}{2(a+4)}\frac{(a-4)}{(a-4)} + \frac{4}{a^2-16}\frac{2}{2} + \frac{4}{a-4}\frac{2(a+4)}{2(a+4)}=\frac{(a-4)}{2(a^2-16)} + \frac{8}{2(a^2-16)} + \frac{8(a+4)}{2(a^2-16)}$$

Now, we can simply add up fractions:

\begin{align}\frac{(a-4)}{2(a^2-16)} + \frac{8}{2(a^2-16)} + \frac{8(a+4)}{2(a^2-16)}&=\frac{8a+32+8+a-4}{2(a^2-16)}\\\\ &=\frac{9a+36}{2(a^2-16)}\\\\ &=\frac{9(a+4)}{2(a+4)(a-4)}\\\\ &=\frac{9}{2(a-4)}\end{align}

• I don't understand at all what you did when you "multiplied fractions by clever forms of one"... Maybe it's the formatting that puts me off, but are you multiplying by (a-4) in the first fraction, 2 in the second fraction and (a+4) in the third fraction? While I see how this gets 2(a^2-16) in all the denominators I still don't understand how on Earth you figured that out. Does finding the common denominator only work when you're adding two fractions? I get that I need to get the denominators to be the same and then add the fractions, I just don't get HOW I get them to be the same.. :/ Oct 12 '15 at 19:57
• I think the first instinct you should have, is to look at the denominators and consider if they have common factors. You want to make them equal, so multiply by fractions equal to one, until the denominators are the same. I particularly use the fact that $a^2-b^2=(a+b)(a-b)$. Oct 12 '15 at 20:12
• Just to clarify, what do you mean by "fractions equal to one"? Oct 12 '15 at 20:23
• Fractions such that the numerator equals the denominator, else we'd be changing the value of the expression. Oct 12 '15 at 20:37
• Ah I see, that's why your format was confusing at first; you multiply with fractions with the same numerator and denominator. I'm used to just "multiplying the same number in the numerator and denominator" but it's the same thing. Another question, why is 2(a-4) in the end there better than 2a-8? I mean the latter is shorter, and the assignment is to make it as short as possible, isn't it? I find that confusing. I greatly appreciate all the help, by the way! Oct 12 '15 at 20:50