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

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.

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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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$\frac{1}{2}(V - 2bc) = a(b+c)$ where did that $2bc$ come from? – Austin Broussard Jul 16 '12 at 3:05
Note that $\frac{V}{2} - bc = \frac{1}{2}(V-2bc)$ because $\frac{1}{2}\cdot2bc = bc$. I was just trying to make it easier to see the simplified answer. There are other forms of the correct expression for $a.$ – Joe Jul 16 '12 at 3:07
I made a LaTeX error, I had edited it. Sorry. Based on your work, your error is in the second to third line. Once you have $$\left(\frac{V}{2}\right)-bc=ab+ca$$, you can factor an $a$ term out on the RHS and you get my expression for $a.$ – Joe Jul 16 '12 at 3:11
I always make errors. No big deal. And I see, since it would become, $\dfrac{1}{2}\cdot \dfrac{2bc}{1}=\dfrac{2bc}{2}$ and the twos would cancel. – Austin Broussard Jul 16 '12 at 3:12
Glad to hear it. Don't worry, you'll get a hang of the LaTeX. Feel free to click the "edited _ minutes ago" to learn the code. I remember, as do others, our first time on the site trying to eagerly learn as others marked up our posts. – Joe Jul 16 '12 at 3:17

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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By backwards, do you mean multiply? I tried individually dividing $b$ and $c$. and got: $$\left(\frac{V-bc}{2cb}\right)=a$$ – Austin Broussard Jul 16 '12 at 2:43
By backwards, I mean take your expression $\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$ and multiply it by $b+c$. Do you get the expression you had before you divided by $b+c$? – David Spencer Jul 16 '12 at 2:44
No, you get; $\left(\frac{V}{2}\right)-bc=2a(b+c)$. But that doesn't help solve for $a$ because you would have to divide by $b+c$ anyway to get it unattached from the $a$. – Austin Broussard Jul 16 '12 at 2:46
Since you did that one step forwards and then backwards and got something different, you know that you did something wrong there. If you distribute your $2a$ in what you just wrote, you get $\left(\frac{V}{2}\right)-bc=2ab+2ca$. This is very close to what it should be, but the right is a bit different - there's an extra factor of $2$ on each term. That should tell you what you did wrong. – David Spencer Jul 16 '12 at 2:52
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 $\left(\frac{V}{2}\right)-bc=ab+ca$. – David Spencer Jul 16 '12 at 3:09