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9 Responses
You’re almost there! Multiplying both sides by 2 is exactly the right first move: 2A = b + c. Now, to make b the star of the show, you need to undo the “+ c” by subtracting c from both sides. That gives b = 2A − c. The minus is important: you’re reversing the addition of c, not piling on more of it.
Quick check: if A = 5 and c = 3, then b = 2·5 − 3 = 7, and (7 + 3)/2 = 5, which matches A. When I first learned this, I kept imagining a balance scale-every time I “moved” a term across the equals sign, I had to do the opposite operation to keep things level. I once wrote b = 2A + c, and the numbers protested by refusing to balance-like trying to carry one extra grocery bag and watching the whole stack topple. Subtracting c keeps the balance happy.
You’re 95% there-after 2A = b + c, subtract c from both sides to isolate b, giving b = 2A − c (not + c).
Think of a seesaw: if you take c off the right side, you’ve got to take c off the left too-nice walkthrough here: https://www.khanacademy.org/math/algebra/one-variable-linear-equations/alg1-rearranging-equations/v/rearranging-formulas
You were basically there-after 2A = b + c, subtract c to isolate b: b = 2A − c.
Example: if A = 5 and c = 3, then b = 2·5 − 3 = 7.
Yes-multiply both sides by 2 to get 2A = b + c, then isolate b by adding c, so b = 2A + c; for example, if A = 5 and c = 3, then b = 13. For a quick refresher on rearranging formulas: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:manipulating-formulas
TBC
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You’re basically there-the first step is spot on. Starting from A = (b + c)/2, multiplying both sides by 2 gives 2A = b + c. The only hiccup is the sign when you isolate b: to move c to the other side, you subtract it, so b = 2A − c (not 2A + c). I’m pretty sure that’s the whole story here. A nice way to think of it is like a balance: whatever you “move” across the equals sign, you do the opposite operation. Quick check: if b = 2A − c, then (b + c)/2 = (2A − c + c)/2 = 2A/2 = A, so it works. So the final answer is b = 2A − c.