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I have a simple part of a question to solve. The problem is my answer is different to the solution in my textbook.

The equation is: $$\frac{5v}{6} = \frac{(\frac{1}{2}a+b+\frac{1}{2} c)v}{a+b+c}$$

I am supposed to get $$\frac{2}{3}(a+b+c) = b$$

But I simply get: $$b=2a +2c$$

I get my answer by cross multiplying. I then use my answer to get $\frac{b}{a+b+c}$ as some fraction. I have not worked onto this stage as I am unsure about the above work.

What am I doing wrong here?

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$b=2a+2c$

Adding $2b$ on both sides gives

$b + 2b =2a+2b+2c$

Or better

$3b =2a+2b+2c$

And if you throw the $3$ to the other side you get: $b =\frac{2}{3}(a+b+c)$

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  • $\begingroup$ Thank you, bot answers were good, this one had a little more information. $\endgroup$ – user Nov 10 '13 at 15:35
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What you get is equivalent to what you're supposed to get.

$ b= 2a+2c \implies 3b=2a+2c+2b \implies b= 2(a+b+c)/3$

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