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3 Responses
Almost there-parentheses are your besties: from y = 2x + 3 you want x = (y − 3)/2 (not y − 3/2), and from y = (3x − 5)/2 you get 2y = 3x − 5 → 3x = 2y + 5 → x = (2y + 5)/3; with y = 11 these give x = 4 and x = 9, and both check since 2·4 + 3 = 11 and (3·9 − 5)/2 = 11.
Think of it like unwrapping a present in reverse-undo the last operation first; nice walkthrough here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations/x2f8bb11595b61c86:solve-for-a-variable/v/rearranging-formulas.
Ooh, changing the subject is like unwrapping x from the algebraic gift wrap-peel off operations in reverse! For y = 2x + 3, you want to undo the +3 first, then the ×2: subtract 3 from both sides to get y − 3 = 2x, then divide by 2 to get x = (y − 3)/2. Notice the parentheses-x = y − 3/2 would mean “subtract 1.5 from y,” which isn’t the same as halving the difference. (Some people say PEMDAS backward means addition has higher priority than multiplication, but really we’re just undoing the outermost operation first.) For y = (3x − 5)/2, multiply both sides by 2 to clear the denominator: 2y = 3x − 5, then add 5: 2y + 5 = 3x, and finally divide by 3: x = (2y + 5)/3. Quick check with y = 11: in the first equation, x = (11 − 3)/2 = 8/2 = 4, and indeed 2·4 + 3 = 11; in the second, x = (2·11 + 5)/3 = 27/3 = 9, and (3·9 − 5)/2 = 22/2 = 11. Nice pattern: “remove” constants stuck on the outside first, then deal with the coefficient of x.
You’re very close-parentheses are the key. From y = 2x + 3, subtract 3 to get y − 3 = 2x, then divide by 2: x = (y − 3)/2. Writing x = y − 3/2 is different, because only the 3 is being halved there; we need the whole (y − 3) halved. For the second one, starting from y = (3x − 5)/2, you did the right first step: 2y = 3x − 5. Now add 5 to both sides to get 2y + 5 = 3x, then divide by 3: x = (2y + 5)/3. Quick check with y = 11: for the first equation, x = (11 − 3)/2 = 8/2 = 4, and indeed 2·4 + 3 = 11. For the second, x = (2·11 + 5)/3 = 27/3 = 9, and (3·9 − 5)/2 = 22/2 = 11.