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3 Responses
Negatives in front of brackets are the sock-gremlins of algebra-mischievous, but tameable. Two reliable rules: (1) read any coefficient with its sign, so -2(3x − 5) means “distribute −2 to both terms,” and (2) a lone minus in front of parentheses is really −1 times the bracket, so −(x − 3) = −x + 3 (it flips every sign inside). A safe sequence is: simplify inside any brackets if needed, distribute carefully (including the sign), combine like terms, then move variable terms to one side and constants to the other. If signs keep tripping you, another smart trick is to move whole brackets before expanding. For example, add (x − 3) to both sides of an equation that has −(x − 3); that way you avoid a sign flip until later.
Let’s do your examples. For −2(3x − 5) + 4 = 10: distribute −2 to get −6x + 10 + 4 = 10, so −6x + 14 = 10, then −6x = −4, so x = 2/3. Quick check: 3x = 2, so 3x − 5 = −3; then −2(−3) = 6; 6 + 4 = 10-works. For 3(x − 4) = 2(2x + 1) − (x − 3): expand each bit carefully. Left: 3x − 12. Right: 4x + 2 − x + 3 = 3x + 5 (because −(x − 3) = −x + 3). So 3x − 12 = 3x + 5. Subtract 3x from both sides and you get −12 = 5, which is impossible-so there is no solution. Tiny worked reminder: −(a − b) = −a + b, and a(b − c) = ab − ac; keep the sign glued to the number you distribute.
If you like a visual cue, draw two little arrows from the coefficient to each term in the bracket and say the product out loud with the sign: “negative two times three x is negative six x; negative two times negative five is positive ten.” More practice and explanations here: https://www.khanacademy.org/math/algebra/one-variable-linear-equations/distributive-property-equations.
I like a fixed sequence: expand every bracket by distributing the multiplier (including its sign) to each term-carry the sign across, so −(a−b)=−a+ b-then combine like terms and isolate x. For example, −2(3x−5)+4=10 -> −6x+10+4=10 -> x=2/3, and 3(x−4)=2(2x+1)−(x−3) -> 3x−12=4x+2−x−3 -> 3x−12=3x−1 so x=11.
Great question! Here’s a reliable sequence I use: (1) simplify inside each bracket if possible, (2) distribute carefully, (3) combine like terms, (4) move all x-terms to one side and constants to the other, (5) solve, and (6) quick check. Rule-of-thumb for signs: a − (b + c) = a − b − c and a − (b − c) = a − b + c-in other words, a negative in front of a bracket flips every sign inside; also think of subtraction as “add the opposite,” so 3x − 5 is 3x + (−5). Example 1: −2(3x − 5) + 4 = 10 → distribute to get −6x + 10 + 4 = 10 → −6x + 14 = 10 → −6x = −4 → x = 2/3 (check: 3(2/3) − 5 = −3, then −2(−3) + 4 = 10, works). Example 2: 3(x − 4) = 2(2x + 1) − (x − 3) → left is 3x − 12; right is 4x + 2 − x + 3 = 3x + 5 → 3x − 12 = 3x + 5 → subtract 3x to get −12 = 5, which is impossible, so there’s no solution. About “expand everything first”: if there’s a minus in front of a bracket, I almost always expand right away to avoid sign slips; otherwise, I sometimes combine terms inside a bracket first (fewer things to distribute means fewer chances for errors). Little trick that saved me in high school: I’d rewrite every subtraction as “+ (−something)” and then distribute; saying “negative times negative is positive” out loud felt silly, but it stopped the classic lost-minus mistake. If you want a clean refresher with more examples, this Khan Academy page on the distributive property is solid: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:expanding-expressions/x2f8bb11595b61c86:distributive-property/a/distributive-property-review