# Solving for Variables

#### LizziebethV4

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

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

Thanks.

#### 1005

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.

#### HallsofIvy

MHF Helper
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: $$\displaystyle \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$$.\)

• LizziebethV4

#### LizziebethV4

Thank you all! I understand now 