# Simple algebra question

#### 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

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#### TD

Homework Helper
Multiply both sides with $v_w$ to get:

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

#### DB

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

i get 20 btw

#### TD

Homework Helper
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.

DB said:
i get 20 btw
Correct You can check it by plugging it in the initial equation.

#### DB

thanks TD, i see it perfectly now

#### TD

Homework Helper
No problem, glad I could help

#### DB

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

#### VietDao29

Homework Helper
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,

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