# Homework Help: Expressing as a single fraction

1. Oct 4, 2007

### Theinvoker

Hiya,

Just found the forums and loving them so far, going to spend some time tomorrow trauling through some of the intersting discusssions I've seen.
I am starting a course soon and have been going over some maths problems to prepare, however there is one that I never used to be able to figure out and turns out i still cant lol It's todo with adding fractions with more than 2 parts.

The practice question I have here is:

1/2(x+4) + 3/(x+4)^2 + 1/2

I can see without too much trouble that the common denominator is 2(x+4)^2 but the closest I have got to the correct answer is x^2+x+25 / 2(x+4)^2 and according to the answers its slightly wrong.

I am fine when it's just 2 fractions to be added but cant understand how you get the answer when its 3 or more.

If anyone could help explain it would be most appreciated :)

Regards,

Theinvoker

2. Oct 4, 2007

### Dick

For 1/(2(x+4)) (I assume that's what you mean, you should use more parentheses), multiply top and bottom by (x+4) for the second term multiply top and bottom by 2 and for the last one by (x+4)^2. Now they are all over your common denominator. What's the sum of the numerators?

3. Oct 4, 2007

### real10

$$\frac{a}{u}+\frac{b}{v}+\frac{c}{w}=\frac{v*w*a+u*w*b+u*v*c}{u*v*w}$$
follow the pattern generalizing it further to n fractions....

in ur specific case v = u^2 which leads to simplifcations but the formula above would help if in the general sense..

Last edited: Oct 4, 2007
4. Oct 4, 2007

### Dick

Forget the formula. You got the common denominator. That's half the battle. Now multiply top and bottom of each fraction by what's lacking from the common denominator. Formula? Jeez.

5. Oct 4, 2007

### real10

I thought the original poster wanted to understand how to add/sub for more than 2 fractions....

6. Oct 4, 2007

### Dick

I suspect the poster has already memorized a formula for two fractions. I wouldn't give the OP a formula for three. The procedure for adding doesn't depend on a formula. Just a pedagogical notion. Sorry if I seemed dismissive of your contribution.

7. Oct 4, 2007

### real10

the procedure works for fraction addition and makes sense,doesnt it??

8. Oct 4, 2007

### Dick

What procedure? Finding a common denominator and learning how to use it makes sense.

9. Oct 5, 2007

### HallsofIvy

The point is, do you really think it is a good thing for someone to memorize a formula for adding two fractions, a different formula for adding three, fractions, still a different formula for adding four fractions, ... ? Far better to understand that once you have changed each fraction, no matter how many there are, to the common denominator, you just add the numerators.

10. Oct 5, 2007

### Theinvoker

Thanks for all the help. I thought I had it so I went back to the question and followed what was said, but I still get the wrong answer :(

My workings:

1/2(x+4) + 3/(x+4)^2 + 1/2 =

1(x+4) + 6 + 1(x+4)^2 / 2(x+4)^2 =

x^2 + x + 26 / 2(x+4)^2

I've been over it and over it and all seems to be right, but the answer in the book is x^2 + 9x + 26 / 2(x+4)^2 and I cant see where you get 9x from for the life of me lol.

Thanks again.

11. Oct 5, 2007

### Dick

Use more parentheses. 1+2/1 can be either 3/2, (1+2)/1 or 3, 1+(2/1). You might know what you mean but somebody else might not. That said, your problem is that (x+4)^2 is not equal to x^2+4^2. (1+4)^2=5^2=25. 1^2+4^2=17. They aren't the same. What went wrong?

12. Oct 5, 2007

### Theinvoker

1(x+4)^2 = (x+4)(x+4) = x^2 + 8x +16 add that to the rest and you get the right answer!!! Yay! :)

Thank you very much! If it had been explained to me like that 10 years ago I might have got it sooner (except the last (x+4)^2 bit that as me being a plonker hehe).

Thanks again for all the help, sure Ill be back again when the course starts if not before, and once I start picking up new knowledge off it, help others out that are where i am now :)