Solving for b_1: Help Needed with 2A=h(b_{1}+b_{2})?

[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 ?

 

[courtrigrad]

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

So how would you solve for b_{1}?

 

[GLprincess02]

Wouldn't you divide both sides by h?

 

[courtrigrad]

yes you would.

 

[GLprincess02]

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

 

[courtrigrad]

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

 

[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!