# Rewriting again

1. Oct 4, 2006

### GLprincess02

I know you all are probably sick of me posting these kind of "easy" problems, but rewriting equations is one of those things that I just seem to struggle with. So that being said, can I get help with $$2A=h(b_{1}+b_{2})$$? I need to solve for $$b_{1}$$. Maybe if someone just gave me the first step, I could get the rest ???

Last edited: Oct 4, 2006
2. Oct 4, 2006

$$2A=h(b_{1}+b_{2})$$.
$$2A = hb_{1}+hb_{2}$$.
$$2A-hb_{2} = hb_{1}$$.

So how would you solve for $$b_{1}$$?

3. Oct 4, 2006

### GLprincess02

Wouldn't you divide both sides by h?

4. Oct 4, 2006

yes you would.

5. Oct 4, 2006

### GLprincess02

So the final answer would be $$\frac{2A-b_{2}}{h}=b_{1}$$ ?

6. Oct 4, 2006

no it would be $$b_{1} = \frac{2A-hb_{2}}{h}$$ or $$b_{1} = \frac{2A}{h} - b_{2}$$

Last edited: Oct 4, 2006
7. Oct 4, 2006

### GLprincess02

Oh you're right, I just forgot the 2nd h in my equation.

And thanks for all your help. I think I'm starting to get this more, so hopefully this will be my last rewriting problem!