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Simple algebra question

  • Thread starter DB
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DB

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(another) simple algebra question

[tex]\frac{300}{5v_w}+\frac{300}{3v_w}=8[/tex]

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

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Multiply both sides with [itex]v_w[/itex] to get:

[tex]\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 [/tex]
 

DB

501
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okay but i only multiply the numerators by v_w? otherwise i would get v_w^2 in the denominators...right?
 

DB

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i get 20 btw
 

TD

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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 :smile: You can check it by plugging it in the initial equation.
 

DB

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thanks TD, i see it perfectly now
 

TD

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No problem, glad I could help :smile:
 

DB

501
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just out of curiosity...how would i approach:
[tex]\frac{300}{5v_w+10}+\frac{300}{3v_w}=8[/tex]
?
thanks
 

VietDao29

Homework Helper
1,417
1
You can try simplify the equation a bit:
[tex]\frac{300}{5v_w + 10} + \frac{300}{3v_w} = 8[/tex]
[tex]\Leftrightarrow \frac{60}{v_w + 2} + \frac{100}{v_w} = 8[/tex]
Then, you can try multiplying both sides with vw(vw + 2) to get rid of the denominator. And you will have:
[tex]60v_w + 100(v_w + 2) = 8v_w(v_w + 2)[/tex]
You can rearrange a bit, and you will have a quadratic equation.
Can you go from here?
Viet Dao,
 

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