Mastering Simple Algebra with Helpful Tips and Examples

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Homework Help Overview

The discussion revolves around solving algebraic equations, specifically focusing on manipulating fractions and variables in the context of simple algebra problems.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of multiplying both sides of an equation to eliminate denominators and question the implications of multiplying only numerators. There is also exploration of how to approach a more complex equation involving additional terms in the denominator.

Discussion Status

Some participants have provided guidance on how to manipulate the equations, while others are exploring different interpretations of the algebraic steps involved. There is a sense of collaborative problem-solving, but no explicit consensus on the final approach to the more complex equation has been reached.

Contextual Notes

Participants are working within the constraints of homework rules, focusing on algebraic manipulation without providing complete solutions. The discussion includes assumptions about the operations on fractions and the structure of the equations presented.

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

[tex]\frac{{300}}<br /> {{5v_w }} + \frac{{300}}<br /> {{3v_w }} = 8 \Leftrightarrow v_w \left( {\frac{{300}}<br /> {{5v_w }} + \frac{{300}}<br /> {{3v_w }}} \right) = 8v_w \Leftrightarrow \frac{{300}}<br /> {5} + \frac{{300}}<br /> {3} = 8v_w[/tex]
 
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
 
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.
 
thanks TD, i see it perfectly now
 
No problem, glad I could help :smile:
 
just out of curiosity...how would i approach:
[tex]\frac{300}{5v_w+10}+\frac{300}{3v_w}=8[/tex]
?
thanks
 
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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