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: \(\displaystyle \frac{2A}{h}- B_2= (B_1+ B_2)- B_2= B_1\).\)