Adding Algebraic Fractions Help

Wats_in_a_name

Basically, I have never been able to get my head around Adding Algebraic Fractions even though I did Higher Level up until a few weeks back.

So, does anyone know of any good guides online as the book doesn't make sense to me!

LilMissCiara

Get the common denominator and add!

Like
4/5 + 3/4 = 16/20 +15/20 = 31/20

(to get 4/5 into 16/20 you divide 5 into 20 and then multiply the answer by the top!)

jumpguy

Okay, so first of all, pretty much what littlemissclara said.

The numerator is the top number, the denominator is the bottom number.

1/2 + 1/3 = ?

The lowest common denominator means what is the lowest number 2 and 3 both divide into? The answer is 6. This is your new denominator.

But then you want to know, what is the numerators?
Divide your original denominator by the lowest common denominator to get the numerator.

For 1/2 it will be (6dividedby2)/6
For 1/3 it will be (6dividedby3)/6

1/2 + 1/3
Becomes..
3/6 + 2/6 = 5/6

magicianz

That's numerical fractions not algebraic ones guys

Do you mean something like:

(x+2) 3

---

+

---

(x-4) (x+3)

What I do is multiply the bottom line on the left by the top right, plus the bottom right by the top left, then put that all over the 2 expressions on the bottom multiplied. so the above would be:

(x+2)(x+3) + 3(x-4)

---

(x-4)(x+3)

And then solve out

Sorry about the crazy writing of the sums, on m.boards :L

J_E

Algebraic fractions?

Okay, here's a simple example:

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

To get a common denominator you cross multiply, so basically

1(x+2) + 4(x+1)
(x+1)(x+2)

You can keep the bottom like that or continue on.

5x+6
(x+1)(x+2)

That's the basics of it.

Wats_in_a_name

Right still cannot do this.

Can someone do the following sum for me really slowly please!

2/3x + 5/2x + 1/4

I cant add 2 different algebraic fractions through in a third & I'm screwed!

**Timbuk2**

Well you have [latex]\frac{2}{3x} + \frac{5}{2x} + \frac{1}{4}[/latex]

It's the x's that put people off, but it's no different. What is the least common denominator - i.e. what is the smallest number that 3x, 2x and 4 all divide into. The answer is 12x. (Note: if for some reason you can't think of the LCD, then multiplying (3x)*(2x)*(4) also works which gives you [latex]24x^2[/latex])

You want to express the whole thing as being a large fraction over 12x. If each 'part' is over 12x, you can write it as a large fraction.

You want each part to be over 12x, but we can't change the value of each part so ask yourself what do you need to do to the top to make the overall thing unchanged.

So something like this:
[latex]\displaystyle\frac{(2)(4)}{12x} + \frac{(5)(6)}{12x} + \frac{(3x)}{12x}[/latex]

Those numbers were chosen because 2/(3x) is the same as (2)(4)/(12x), etc.

As everything is over 12x, we can just write it as a large fraction
[latex]\displaystyle \frac{38+3x}{12x}[/latex].

That's the answer. You could have done it slightly differently though, noting that 1/4 is independent of x (it's just a constant) and thus keeping it separate from the calculation. Therefore we consider
[latex]\frac{2}{3x} + \frac{5}{2x}[/latex].
Once you understand the reasoning behind getting the LCD (as above) there is a shortcut to doing it when there's two. You can cross multiply to get the numerator, and multiply the denominators to get the overall denominator.

For example [latex]\displaystyle \frac{(2)(2x) + (5)(3x)}{(3x)(2x)}[/latex]
Which can be written as [latex]\displaystyle \frac{4x + 15x}{6x^2}[/latex] which simplifies to [latex]\displaystyle \frac{19}{6x}[/latex].

So the answer is [latex]\displaystyle \frac{19}{6x} + \frac{1}{4}[/latex]

Note: These two answers are exactly the same even though they're written slightly differently. To see the equivalence, you can 'cross multiply' the last answer and you will get the same as the first answer. Either answer would be equally acceptable in an exam though!