Okay, so I'm having a difficult time understanding how to solve problems such as

Solve for B1: A=1/2(B1+B2)h

Help please!?

Thanks.

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- Jul 11th 2010, 06:48 AMLizziebethV4Solving for Variables
Okay, so I'm having a difficult time understanding how to solve problems such as

Solve for B1: A=1/2(B1+B2)h

Help please!?

Thanks. - Jul 11th 2010, 06:58 AM1005
Your first step should be to make B1 not be in the denominator of a fraction. Either take the reciprocal of each side or multiply both sides by that denominator. You could multiply both sides just by (B1 + B2) also. In the end, you'll end up with the same answer regardless.

edit: I'm assuming 2(b1 + b2)h are all in the denominator, though it's a little ambiguous the way you've written it. - Jul 11th 2010, 07:12 AMHallsofIvy
It may help to think of "undoing" what has been done to the variable.

Here, the equation you have is, I suspect, $\displaystyle A= (\frac{1}{2})(B_1+ B_2)h$. That is NOT what 1005 assumed- as he said, what you wrote is ambiguous. I am assuming only the "2" is in the denominator because that looks to me like the formula for the area of a trapezoid.

Suppose you were**given**a value of $\displaystyle B_1$ and the other things on the right and wanted to find A, what would you do? The formula**tells**you what do do- first add $\displaystyle B_1$ and $\displaystyle B_2$ (because they are in parentheses) then multiply by h and then multiply by 1/2 (the order of those two is not really important). To**solve**for $\displaystyle B_1$, you do the opposite of each step**and**in the opposite order! That is first,**divide**by 1/2 (which is the same as multiply by 2) and, of course, do the same thing to both sides of the equation: multiplying both sides of $\displaystyle A= (\frac{1}{2})(B_1+ B_2)h$ by 2 gives $\displaystyle 2A= 2(\frac{1}{2})(B_1+ B_2)h= (B_1+ B_2)h$. Now divide both sides by h: [tex]\frac{2A}{h}= (B_1+ B_2)h/h= B_1+ B_2. Finally, since the first thing you woud do in calculating A would be to**add**$\displaystyle B_1$ and $\displaystyle B_2$, the last thing you do in solving for $\displaystyle B_1$ is**subtract**$\displaystyle B_2$ from both sides: $\displaystyle \frac{2A}{h}- B_2= (B_1+ B_2)- B_2= B_1$. - Jul 12th 2010, 08:11 AMLizziebethV4
Thank you all! I understand now :D