$\displaystyle \frac{3}{x+2}+\frac{x-2}{x}$

I'm used to finding the CLM of the denominators but what do I do in this case? Why is adding fractions so complicated.

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- Jul 29th 2013, 09:30 AMuperkurkAddition with fractions
$\displaystyle \frac{3}{x+2}+\frac{x-2}{x}$

I'm used to finding the CLM of the denominators but what do I do in this case? Why is adding fractions so complicated. - Jul 29th 2013, 09:54 AMemakarovRe: Addition with fractions
What do you do with two fractions whose denominators are relatively prime, e.g., 2/3 + 3/5?

- Jul 29th 2013, 10:01 AMuperkurkRe: Addition with fractions
$\displaystyle \frac{2}{3}+\frac{3}{5} = \frac{10}{15}+\frac{9}{15}$

I simply find the LCM of each denominator. But in my original fractions there is no CLM. I have another question here.

$\displaystyle \frac{b}{b-1}-\frac{1}{2-2b}$ again there is no LCM that I can see here. Even if I expand. $\displaystyle \frac{b}{b-1}-\frac{1}{2(1-b)}$ I don't get what I have to do here - Jul 29th 2013, 10:35 AMemakarovRe: Addition with fractions
I meant that you find the LCM of two relatively prime numbers by multiplying those numbers. With polynomials the situation is similar. The greatest common divisor of x + 2 and x is 1, i.e., x + 2 and x are relatively prime. Therefore, their LCM is their product. In other words, you multiply both the numerator and the denominator of the first fraction by x and similarly multiply both parts of the second fraction by x + 2. Then the denominators will be equal and you can add. In fact, you don't have to find the

*least*common multiple to add fractions; it is enough to have*some*common multiple, e.g., the product of the two denominators.

Here you again can multiply the denominators. Note, however, that the denominators are not relatively prime: the first divides the second because 2 - 2b = (-2)(b - 1). Therefore, the LCM is 2 - 2b and it is enough to multiply both top and bottom of the first fraction by -2. - Jul 29th 2013, 10:46 AMuperkurkRe: Addition with fractions
- Jul 29th 2013, 11:00 AMemakarovRe: Addition with fractions
$\displaystyle \frac{3}{x+2}+\frac{x-2}{x} =\frac{x\cdot3}{x(x+2)}+\frac{(x+2)(x-2)}{(x+2)x}$. Now the denominators are equal and you can add.

$\displaystyle \frac{b}{b-1}-\frac{1}{2-2b} = \frac{(-2) b}{(-2)(b-1)}-\frac{1}{2-2b}$. Again, the denominators are equal. Alternatively,

$\displaystyle \frac{b}{b-1}-\frac{1}{2-2b}= \frac{b}{b-1}+\frac{1}{2b-2} = \frac{b}{b-1}+\frac{1}{2(b-1)} = \frac{2b}{2(b-1)}+\frac{1}{2(b-1)}$ - Jul 29th 2013, 11:32 AMuperkurkRe: Addition with fractions
- Jul 29th 2013, 12:43 PMuperkurkRe: Addition with fractions
This is so annoying I just can't do this... look at this one.

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

$\displaystyle \frac{2}{(a-2)(a+2)}-\frac{1}{a(a-2)}$

Now what? What do I multiply the left and right hand sides by? The only thing they have in common is $\displaystyle a-2$ so I will multiply the left by $\displaystyle a$ and the right side by $\displaystyle a+2$

$\displaystyle \frac{2}{(a-2)(a+2)}\times\frac{a}{a}} = \frac{2a}{a(a-2)(a+2)}$

Now the right hand side.

$\displaystyle \frac{1}{a(a-2)}\times\frac{a+2}{a+2} = \frac{a+2}{a(a-2)(a+2)}$

Finally we get $\displaystyle \frac{2a}{a(a-2)(a+2)} - \frac{a+2}{a(a-2)(a+2)}$ as they now both have common multiples.

Then after the calculation we arrive at $\displaystyle \frac{a-2}{a(a-2)(a+2)}$ and after cancellation $\displaystyle \frac{1}{a^2+2a}$ which I think is wrong - Aug 1st 2013, 04:33 AMmarybaloghRe: Addition with fractions
Thanks for this type of post.

- Aug 1st 2013, 09:11 AMemakarovRe: Addition with fractions
The last term should be $\displaystyle \frac{1}{a(a+2)}$ instead of $\displaystyle \frac{1}{a(a-2)}$.

What you did next was correct, and you should have arrived at $\displaystyle \frac{a+2}{a(a-2)(a+2)}=\frac{1}{a^2-2a}$. Note, however, that this last equality is true only for $\displaystyle a\ne-2$. - Aug 1st 2013, 01:13 PMuperkurkRe: Addition with fractions
I'm not going to open another topic but this is just pathetic, as soon as a question is slightly different I can't do it, I'm trying to be so patient but this is just a **** take.

How can I not solve this? It looks so so so so simple...

$\displaystyle \frac{x}{x+5}=\frac{2}{3}$

It's the whole denominator thing that throws me off EVERY single time.

I'm going to guess that I should multiply both sides by the LCM which is 15.

$\displaystyle 15\times\left(\frac{x}{x+5}\right)=\left(\frac{2}{ 3}\right)\times15$

$\displaystyle \frac{15x}{x+5}=\frac{30}{3}$

$\displaystyle \frac{15x}{x+5}=10$

now what?!

Why is this crap so difficult? Fractions in algebra is almost impossible. - Aug 1st 2013, 01:38 PMemakarovRe: Addition with fractions
The LCM of (x + 5) and 3 is not 15. Indeed, 15 is not divisible by (x + 5). The LCM is 3(x + 5), and that's by what you multiply both sides. Remember that you can always multiply both sides by the product of the denominators and not necessarily by their LCM. The worst thing you get is a fraction that can be reduced and/or larger expressions.

- Aug 1st 2013, 01:41 PMHallsofIvyRe: Addition with fractions
The two denominators are NOT 3 and 5, they are 3 and x+ 5. The LCM is 3(x+ 5).

You can't just pick numbers out of the algebraic expressions.

Quote:

tex]$\displaystyle \frac{15x}{x+5}=\frac{30}{3}$

So 3x= 2x+ 10. Can you solve that?

Quote:

$\displaystyle \frac{15x}{x+5}=10$

now what?!

Why is this crap so difficult? Fractions in algebra is almost impossible.

- Aug 1st 2013, 01:43 PMPlatoRe: Addition with fractions
- Aug 1st 2013, 01:48 PMtopsquarkRe: Addition with fractions
Rules of thumb here.

1) multiply both sides by the denominators

2) Add up the numerators

3) Factor everything in sight

4) Cancel out common factors.

So for your last problem:

$\displaystyle \frac{x}{x + 5} = \frac{2}{3}$

Denominators: x + 5 and 3, so the common denominator is 3(x + 5). Multiply both sides by this.

$\displaystyle \frac{x}{x + 5} \cdot 3(x + 5) = \frac{2}{3} \cdot 3(x + 5)$

Now cancel out the common denominators:

$\displaystyle x \cdot 3 = 2(x + 5)$

I'm sure you can finish from here.

-Dan

Whoops! Plato got there first. :)