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Solve for $a$
$V=2(ab+bc+ca)$

$\left(\frac{V}{2}\right)=ab+bc+ca$ $\left(\frac{V}{2}\right)-bc=ab+ca$ $\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$ $\frac{\left(\frac{V}{2}\right)-bc}{2b+c}=a??$ I honestly do not know what to do with this problem. And I think I may have messed up when I divided by $b+c$. Any pointers. please, no answers. Only generalized hints.

2 Answers 2

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$V=2(ab+ac+bc)$

$\frac{V}{2} = (ab+ac+bc)$

Note that two terms on the RHS have an $a$ term, so we can factor them and rewrite it as:

$\frac{V}{2} = a(b+c) + bc$

$\frac{V}{2} - bc = a(b+c)$

I rewrote the LHS so it is easier to see the step when dividing through by $b+c$:

$\frac{1}{2}(V - 2bc) = a(b+c)$

Divide through by $b+c$ to get:

$a = \frac{V-2bc}{2(b+c)}$ for $b+c \ne 0$.

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    Use \pm and enclose it with dollar signs.2012-07-16
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Hint: You are on the right track, but you did make an error. Try doing the step when you divided by $b+c$ backwards.

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    Not quite; perhaps I've not explained my hint clearly. Try it this way: I see you refer to "take out the a". Do this from your (correct) expression $\left(\frac{V}{2}\right)-bc=ab+ca$.2012-07-16