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3 Responses
A reliable trick is to turn every subtraction into “add a negative,” then distribute with the sign attached to the number. For example, 3(2x−5)−4(x+1) becomes 3(2x + (−5)) + (−4)(x + 1). Now distribute term by term: 3·2x + 3·(−5) + (−4)·x + (−4)·1 = 6x − 15 − 4x − 4. Combine like terms: (6x − 4x) + (−15 − 4) = 2x − 19. Thinking of the outside number as carrying its own sign (+3 and −4 here) usually prevents random sign slips.
For a quick mental check, plug in easy values. At x = 0, the original gives 3(−5) − 4(1) = −15 − 4 = −19, so the constant term must be −19. At x = 1, you get 3(2−5) − 4(2) = −9 − 8 = −17; your expanded form should match (2·1 − 19 = −17). Different people prefer different cues, but treating subtraction as “+ negative” and testing x = 0 (and maybe x = 1) catches most sign mistakes fast.
I always mess these up too, so I treat every minus as “plus a negative”: rewrite 3(2x−5)−4(x+1) as 3(2x+(-5)) + (-4)(x+1), then distribute to get 6x + (−15) + (−4x) + (−4) = 2x − 19-keeping the sign glued to the number really helps, I think. For a quick check, plug in x=0 (you should get −19) or x=1, and here’s a clear refresher: https://www.purplemath.com/modules/distprop.htm
I like to treat every subtraction as “add the opposite,” so the sign rides with the number: 3(2x−5) + (−4)(x+1) ⇒ 6x−15−4x−4 = 2x−19, and as a quick check I plug in x=0 or x=1 in both forms to see they match. Would carrying the sign with the coefficient (like −4) help you keep things straight, or do you prefer the plug-in check?