As with any "solve for ..." problem, you start by looking at the operations performed on the variable of interest. Here, when you evaluate this expression according to the order of operations, you
- add b1
- multiply by (h/2)
When you want to solve for b2, you undo these operations in reverse order. To undo multiplication by a fraction, you multiply by the inverse (reciprocal) of the fraction. To undo addition, you add the opposite.
Whatever you do must be done to both sides of the equation.
Here we go ...
... (2/h)A = b1 + b2 . . . . . we undid the multiply by h/2, by multiplying by 2/h
... (2/h)A - b1 = b2 . . . . . we undid the addition of b1 by adding the opposite of b1
Then your solution is
... b2 = 2A/h - b1
If you want to, you can combine these terms over a single denominator to get
... b2 = (2A -h·b1)/h
I think so but it’s kind of hard to tell.
The answer should be: Yes because the terms are all written in the order of highest degree to the lowest degree.
I might be able to help you out
