Simple algebra question

1. Sep 29, 2005

DB

(another) simple algebra question

$$\frac{300}{5v_w}+\frac{300}{3v_w}=8$$

can some1 give me a hand on the algebra here? thanks

Last edited: Sep 29, 2005
2. Sep 29, 2005

TD

Multiply both sides with $v_w$ to get:

$$\frac{{300}} {{5v_w }} + \frac{{300}} {{3v_w }} = 8 \Leftrightarrow v_w \left( {\frac{{300}} {{5v_w }} + \frac{{300}} {{3v_w }}} \right) = 8v_w \Leftrightarrow \frac{{300}} {5} + \frac{{300}} {3} = 8v_w$$

3. Sep 29, 2005

DB

okay but i only multiply the numerators by v_w? otherwise i would get v_w^2 in the denominators...right?

4. Sep 29, 2005

DB

i get 20 btw

5. Sep 29, 2005

TD

Yes, that wouldn't help anything. You just multiply both sides with it, and when you multiply a fraction with a factor, it's always in the nominator. c * a/b = ca/b.

You do this to get rid of the v_w in the denominators since you can cancel them out when they're also in the numerators.

Correct You can check it by plugging it in the initial equation.

6. Sep 29, 2005

DB

thanks TD, i see it perfectly now

7. Sep 29, 2005

TD

No problem, glad I could help

8. Sep 29, 2005

DB

just out of curiosity...how would i approach:
$$\frac{300}{5v_w+10}+\frac{300}{3v_w}=8$$
?
thanks

9. Sep 30, 2005

VietDao29

You can try simplify the equation a bit:
$$\frac{300}{5v_w + 10} + \frac{300}{3v_w} = 8$$
$$\Leftrightarrow \frac{60}{v_w + 2} + \frac{100}{v_w} = 8$$
Then, you can try multiplying both sides with vw(vw + 2) to get rid of the denominator. And you will have:
$$60v_w + 100(v_w + 2) = 8v_w(v_w + 2)$$
You can rearrange a bit, and you will have a quadratic equation.
Can you go from here?
Viet Dao,