📚 Solving Linear Equations with Brackets and Fractions | 解含括号和分数的线性方程
Linear equations are the foundation of algebra at KS3. This guide focuses on equations with brackets, fractions, and unknowns on both sides, matching the style of questions found in Cambridge Checkpoint-style worksheets such as p307_2.pdf. Mastering these skills will help you solve problems confidently and prepare for later topics like graphs and simultaneous equations.
线性方程是 KS3 代数的基础。本指南重点讲解含括号、分数以及两边都有未知数的方程，与 Cambridge Checkpoint 风格的练习（如 p307_2.pdf）题型一致。掌握这些技能可以帮助你自信地解决问题，并为图像和联立方程等后续内容做好准备。
1. Understanding Linear Equations | 认识线性方程
A linear equation is an equation in which the unknown appears only to the first power. Examples include 2x + 3 = 11 and 5(x − 2) = 3x + 4. The solution is the value of the unknown that makes the left side equal to the right side.
线性方程是指未知数只出现一次方幂的方程。例如 2x + 3 = 11 和 5(x − 2) = 3x + 4。方程的解就是使左边等于右边的未知数值。
Linear equations have exactly one solution when the coefficient of x is not zero. The solution can be positive, negative, a fraction, or a decimal.
当 x 的系数不为零时，线性方程只有一个解。这个解可以是正数、负数、分数或小数。
2. The Balance Method | 天平法
Think of an equation as a balance. Whatever you do to one side, you must do to the other side to keep it level. You can add, subtract, multiply, or divide both sides by the same non-zero number.
把方程想象成一个天平。你对一边做什么操作，就必须对另一边做同样的操作，才能保持平衡。你可以在两边同时加、减、乘或除以同一个非零数。
For example, to solve 5x − 2 = 18, add 2 to both sides: 5x = 20. Then divide both sides by 5: x = 4.
例如，解 5x − 2 = 18 时，先在两边加 2，得到 5x = 20。然后两边同时除以 5，得到 x = 4。
The order of inverse operations is important. Undo addition or subtraction first, then undo multiplication or division.
逆运算的顺序很重要。先处理加法或减法，再处理乘法或除法。
3. Expanding Brackets First | 先展开括号
When brackets are present, expand first. The distributive law says a(b + c) = ab + ac. For example, 4(x + 3) = 32 becomes 4x + 12 = 32.
有括号时，要先去括号。分配律指出 a(b + c) = ab + ac。例如 4(x + 3) = 32 可以化为 4x + 12 = 32。
Then subtract 12 from both sides: 4x = 20. Divide by 4: x = 5.
然后两边同时减去 12，得到 4x = 20。再除以 4，得到 x = 5。
Always multiply every term inside the bracket. In 3(x + 2) = 21, expanding gives 3x + 6 = 21, not 3x + 2 = 21.
一定要把括号内的每一项都乘以括号外的数。在 3(x + 2) = 21 中，展开后得到 3x + 6 = 21，而不是 3x + 2 = 21。
4. Dealing with Negative Terms | 处理负数项
Be careful with negative signs inside brackets. In 3(x − 2) = 15, expanding gives 3x − 6 = 15, not 3x + 6 = 15. Add 6 to both sides: 3x = 21, so x = 7.
括号内有负号时要格外小心。在 3(x − 2) = 15 中，去括号得到 3x − 6 = 15，而不是 3x + 6 = 15。两边加 6 得到 3x = 21，因此 x = 7。
For 15 − 2x = 7, avoid subtracting 15 immediately unless you keep the negative coefficient clear. A reliable method is to add 2x to both sides: 15 = 7 + 2x, then subtract 7: 8 = 2x, so x = 4.
对于 15 − 2x = 7，如果立刻减去 15，就需要清楚处理负系数。更稳妥的方法是两边同时加 2x，得到 15 = 7 + 2x，再减去 7，得到 8 = 2x，因此 x = 4。
When you divide by a negative number, remember the sign rules: negative ÷ negative is positive, and positive ÷ negative is negative.
当除以负数时，要记住符号规则：负数除以负数得正数，正数除以负数得负数。
5. Unknowns on Both Sides | 两边都含未知数
When an unknown appears on both sides, collect the variable terms on one side first. For 7x + 2 = 3x + 18, subtract 3x from both sides: 4x + 2 = 18. Then subtract 2: 4x = 16, so x = 4.
当未知数出现在方程两边时，要先把含未知数的项移到同一边。例如 7x + 2 = 3x + 18，两边同时减去 3x，得到 4x + 2 = 18。再减去 2，得到 4x = 16，所以 x = 4。
If you prefer to avoid negative coefficients, collect the x terms on the side with the larger coefficient. For 2x + 9 = 5x − 3, subtract 2x from both sides: 9 = 3x − 3, then add 3: 12 = 3x, so x = 4.
如果你希望避免负系数，就把 x 项移到系数较大的一边。例如 2x + 9 = 5x − 3，两边同时减去 2x，得到 9 = 3x − 3，再加 3，得到 12 = 3x，所以 x = 4。
Always keep the equation balanced: if you move a term from one side to the other, its sign changes.
始终保持方程平衡：如果把一项从一边移到另一边，它的符号要改变。
6. Equations with Fractions | 含分数的方程
Equations with fractions are easier if you clear the denominator. To solve x/4 + 3 = 7, subtract 3 first: x/4 = 4. Then multiply both sides by 4: x = 16.
含分数的方程可以先消去分母，这样会更容易。解 x/4 + 3 = 7 时，先减去 3，得到 x/4 = 4。然后两边同时乘以 4，得到 x = 16。
For ⅓(x + 2) = 5, multiply both sides by 3 first: x + 2 = 15. Then subtract 2: x = 13.
对于 ⅓(x + 2) = 5，先把两边同时乘以 3，得到 x + 2 = 15。再减去 2，得到 x = 13。
For 2x/3 + 1 = x/2 + 4, multiply every term by the lowest common multiple of 3 and 2, which is 6. This gives 4x + 6 = 3x + 24. Subtract 3x: x + 6 = 24, so x = 18.
对于 2x/3 + 1 = x/2 + 4，把每一项都乘以 3 和 2 的最小公倍数 6，得到 4x + 6 = 3x + 24。再减去 3x，得到 x + 6 = 24，所以 x = 18。
When clearing fractions, multiply every term by the same denominator, including whole numbers.
消去分母时，每一项都要乘以同一个分母，包括整数项。
7. Checking Your Solution by Substitution | 用代入法检验解
Always substitute your answer back into the original equation. If the left side equals the right side, the solution is correct. This is especially important in tests such as checkpoint papers.
一定要把得到的答案代回原方程检验。如果左边等于右边，就说明解是正确的。在 checkpoint 试卷等考试中，这一步尤其重要。
Example: For 2(3x − 1) = 22, the solution x = 4 gives 2(3 × 4 − 1) = 2(12 − 1) = 2 × 11 = 22, which matches.
例如，对于 2(3x − 1) = 22，解 x = 4 代入后得到 2(3 × 4 − 1) = 2(12 − 1) = 2 × 11 = 22，与原方程一致。
Checking does more than confirm a correct answer; it also helps you catch sign and arithmetic errors before you move on.
检验不仅能确认答案正确，还能帮助你在继续做题前发现符号和计算错误。
8. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Common mistakes include forgetting to multiply every term inside the bracket, losing a negative sign, and applying operations in the wrong order. Use inverse operations in reverse BIDMAS order: undo addition or subtraction before multiplication or division.
常见错误包括忘记把括号内的每一项都相乘、丢失负号以及运算顺序错误。要按照 BIDMAS 的逆顺序使用逆运算：先处理加减，再处理乘除。
When solving 4(x + 5) = 28, the correct expansion is 4x + 20 = 28, not 4x + 5 = 28. Subtract 20 and divide by 4 to get x = 2.
解 4(x + 5) = 28 时，正确的展开是 4x + 20 = 28，而不是 4x + 5 = 28。减去 20 再除以 4，得到 x = 2。
Another common error is dividing before clearing addition or subtraction. In 3x + 9 = 27, subtract 9 first: 3x = 18, then x = 6. Dividing 27 by 3 first would give the wrong next step.
另一个常见错误是在处理加减之前就进行除法。在 3x + 9 = 27 中，要先减 9，得到 3x = 18，再得到 x = 6。如果先除以 3，就会导致下一步错误。
9. Word Problems Leading to Equations | 由文字题建立方程
Word problems often ask you to build an equation from a sentence. Change ‘a number increased by 5 is 12’ into x + 5 = 12. Change ‘three times a number subtracted from 20 gives 8’ into 20 − 3x = 8.
文字题通常要求你根据句子建立方程。把 ‘一个数加 5 等于 12′ 写成 x + 5 = 12。把 ’20 减去一个数的三倍等于 8’ 写成 20 − 3x = 8。
Example: The perimeter of a rectangle is 38 cm. If the width is x cm and the length is x + 5 cm, write 2(x + x + 5) = 38. This simplifies to 4x + 10 = 38, so x = 7. The width is 7 cm and the length is 12 cm.
例题：一个长方形的周长是 38 cm。如果宽为 x cm，长为 x + 5 cm，则可以写成
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