📚 Solving Linear Equations | 解线性方程
Linear equations are the foundation of algebra, and mastering them at KS3 is essential for success in Cambridge Mathematics. This comprehensive guide, inspired by the practice sets on p204 of a typical KS3 workbook, will walk you through every step—from one-step equations to tackling word problems—ensuring you build fluency and confidence.
线性方程是代数的基础，在 KS3 阶段掌握它们对于剑桥数学的成功至关重要。这篇综合指南，灵感来自典型 KS3 练习册第 204 页的习题集，将带你完整梳理每一步——从一元一次方程到解决文字应用题——帮助你建立熟练度和信心。
1. Understanding Linear Equations | 理解线性方程
kes both sides of the equation equal. For example, x + 5 = 12 is a simple linear equation.
线性方程是一个代数陈述，其中变量（通常是 x）的指数为 1，绘制图像时形成一条直线。它的目标是找出使等式两边相等的未知数的值。例如，x + 5 = 12 就是一个简单的线性方程。
x + 5 = 12
To solve any linear equation, remember the golden rule: whatever operation you perform on one side, you must perform on the other to maintain balance. Think of an equation as a pair of balanced scales.
要解任何线性方程，记住黄金法则：无论你在等式的一边进行什么运算，都必须在另一边进行相同的运算以保持平衡。可以把等式想象成一对平衡的天平。
2. One-step Equations | 一元一次方程
One-step equations require only one inverse operation to isolate the variable. If the equation involves addition, subtract the same number from both sides. If it involves multiplication, divide both sides by the same number.
一元一次方程只需要一次逆运算就能分离出变量。如果方程包含加法，就从两边减去同一个数。如果包含乘法，就把两边除以同一个数。
Example: x + 3 = 9 → subtract 3 from both sides → x = 6.
示例：x + 3 = 9 → 两边同时减去 3 → x = 6。
Example: 4x = 20 → divide both sides by 4 → x = 5.
示例：4x = 20 → 两边同时除以 4 → x = 5。
Always check your answer by substituting it back into the original equation to ensure it holds true.
始终将答案代回原方程进行检验，确保等式成立。
3. Two-step Equations | 两步方程
Two-step equations involve two operations, such as multiplication and addition. The strategy is to undo the addition or subtraction first (using inverse operations), then undo the multiplication or division. This is often remembered as 'SAMDEB' in reverse: undo addition/subtraction before multiplication/division.
两步方程包含两种运算，比如乘法和加法。策略是首先消去加法或减法（使用逆运算），然后消去乘法或除法。这常被记作逆向“SAMDEB”：先消加减，再消乘除。
Solve: 2x + 7 = 19. First subtract 7: 2x = 12, then divide by 2: x = 6.
解方程：2x + 7 = 19。首先减 7：2x = 12，然后除以 2：x = 6。
2x + 7 = 19 → 2x = 12 → x = 6
Another example: x/3 - 4 = 2. Add 4 to both sides: x/3 = 6, then multiply by 3: x = 18.
另一个示例：x/3 - 4 = 2。两边加 4：x/3 = 6，然后乘以 3：x = 18。
4. Equations with Brackets | 带括号的方程
When brackets appear, always expand them first using the distributive property. Multiply the term outside the brackets by each term inside. Then proceed to solve the simplified equation as usual.
当出现括号时，始终先使用分配律展开括号。将括号外的项乘以括号内的每一项。然后照常解简化后的方程。
Example: 5(x - 2) = 20. Expand: 5x - 10 = 20. Add 10: 5x = 30, then divide by 5: x = 6.
示例：5(x - 2) = 20。展开：5x - 10 = 20。加 10：5x = 30，然后除以 5：x = 6。
5(x - 2) = 20 → 5x - 10 = 20 → 5x = 30 → x = 6
= 7, expand carefully: -3x - 1 = 7, then add 1 and divide by -3, giving x = -8/3 or -2²/₃.
如果括号前有负号，例如 -(3x + 1) = 7，要小心展开：-3x - 1 = 7，然后加 1 再除以 -3，得到 x = -8/3 或 -2²/₃。
5. Equations with Variables on Both Sides | 变量在两侧的方程
These equations have the unknown on both the left and right sides. The goal is to collect all variable terms on one side and all constant terms on the other. Choose the side that makes the coefficient positive to reduce errors.
这类方程的未知数同时出现在左右两边。目标是把所有含变量的项集中到一边，所有常数项集中到另一边。选择能使系数为正的一边，以减少错误。
3 = 3x + 15. Subtract 3x from both sides: 4x + 3 = 15. Then subtract 3: 4x = 12, so x = 3.
解方程：7x + 3 = 3x + 15。两边同时减去 3x：4x + 3 = 15。然后减去 3：4x = 12，得出 x = 3。
7x + 3 = 3x + 15 → 4x = 12 → x = 3
back: left = 7(3)+3=24, right = 3(3)+15=24. Both sides equal, so the solution is correct.
始终将解代回检验：左边 = 7(3)+3=24，右边 = 3(3)+15=24。两边相等，所以解正确。
6. Equations involving Fractions | 带分数的方程
Fractions can make equations look intimidating, but they are manageable by multiplying every term by the least common denominator (LCD) to clear the fractions. This transforms the equation into a familiar integer form.
分数可能会让方程看起来令人生畏，但通过将每一项都乘以最小公分母（LCD）来去分母，就可以轻松处理。这会将方程转化为熟悉的整数形式。
2 and 4 is 4. Multiply every term by 4: 2x + 12 = x + 20. Solve: 2x - x = 20 - 12 → x = 8.
示例：x/2 + 3 = x/4 + 5。2 和 4 的 LCD 为 4。每项乘以 4：2x + 12 = x + 20。求解：2x - x = 20 - 12 → x = 8。
x/2 + 3 = x/4 + 5 → 2x + 12 = x + 20 → x = 8
, e.g., 3/(x+1) = 2/5, cross-multiply: 3 × 5 = 2(x + 1) → 15 = 2x + 2 → 2x = 13 → x = 6.5.
另一种技巧是将分数方程视为比例，例如 3/(x+1) = 2/5，交叉相乘：3 × 5 = 2(x + 1) → 15 = 2x + 2 → 2x = 13 → x = 6.5。
7. Forming Equations from Word Problems | 从文字题形成方程
Real-life problems require you to translate a situation into an algebraic equation. Identify the unknown and assign it a variable, then build expressions step by step based on the relationships described.
实际问题需要你将情境转换为代数方程。识别未知数并为其分配变量，然后根据描述的关系逐步建立表达式。
Example: 'Three more than twice a number is 19.' number be n. Twice the number is 2n, three more is 2n + 3. Equation: 2n + 3 = 19 → n = 8.
示例：“一个数的两倍加 3 等于 19。” 设这个数为 n。两倍为 2n，再加 3 得到 2n + 3。方程：2n + 3 = 19 → n = 8。
Problems involving age, perimeter, or money are common. Always define your variable clearly and check that the solution makes sense in context.
涉及年龄、周长或金钱的问题很常见。始终清晰地定义变量，并检查解在上下文中是否合理。
8. Checking Your Solutions | 检验答案
Substituting your answer back into the original equation is more than a routine—it is a powerful way to catch mistakes. Both sides of the equation must yield the same numerical value.
将答案代回原方程不仅是一项例行操作，更是发现错误的强大方法。方程的两边必须得出相同的数值。
hen left side = 4(5 - 1) = 16, right side = 2(5) + 6 = 16. The check confirms correctness.
如果原方程是 4(x - 1) = 2x + 6，你求出 x = 5，那么左边 = 4(5 - 1) = 16，右边 = 2(5) + 6 = 16。检验确认了正确性。
If the sides do not match, retrace your steps. Common errors include arithmetic slips or forgetting to apply an operation to every term.
如果两边不相等，重新检查你的步骤。常见错误包括算术差错或忘记对每一项进行运算。
9. Common Mistakes to Avoid | 常见错误避免
Students often make mistakes when they forget to perform an operation on both sides, or when they misapply the distributive property with negative signs. Another pitfall is mishandling fractions.
学生常犯的错误是忘记在两边进行相同的运算，或者在负号前错误使用分配律。另一个陷阱是处理分数不当。
For –2(x + 3) = 10, expanding correctly gives –2x – 6 = 10, not –2x + 6. Watch the sign. Also, when solving x/3 = 4, remember to multiply by 3, not divide.
对于 –2(x + 3) = 10，正确展开得到 –2x – 6 = 10，而不是 –2x + 6。注意符号。另外，解 x/3 = 4 时，记住要乘以 3，而不是除以 3。
Using inverse operations in the wrong order with two-step equations is another frequent error. Always undo addition or subtraction before multiplication or division unless brackets indicate otherwise.
在两步方程中颠倒逆运算的顺序是另一个常见错误。除非括号有不同指示，始终先消去加法或减法，再消去乘法或除法。
10. Practice Problems and Solutions | 练习题与解答
Here are some targeted exercises to solidify your understanding, similar to those on p204:
以下是一些针对性练习，以巩固你的理解，类似于第 204 页的题目：
Solve 6p – 2 = 16. Answer: p = 3. (Add 2, then divide by 6.)
解 6p – 2 = 16。答案：p = 3。（加 2，然后除以 6。）
Solve 3(y + 4) = 27. Answer: y = 5. (Expand, subtract 12, divide by 3.)
解 3(y + 4) = 27。答案：y = 5。（展开，减去 12，除以 3。）
Solve 2x – 7 = x + 3. Answer: x = 10. (Subtract x, add 7.)
解 2x – 7 = x + 3。答案：x = 10。（减去 x，加上 7。）
Solve (x – 1)/4 = 3/2. Answer: x = 7. (Cross-multiply: 2(x – 1) = 12 → x = 7.)
解 (x – 1)/4 = 3/2。答案：x = 7。（交叉相乘：2(x – 1) = 12 → x = 7。）
re than the width. Find the width. Equation: 2(w + 4 + w) = 30 → 4w + 8 = 30 → w = 5.5 cm.
文字题：一个矩形的周长是 30 厘米。长比宽多 4 厘米。求宽。方程：2(w + 4 + w) = 30 → 4w + 8 = 30 → w = 5.5 厘米。
Attempt these without looking at the answers first. Then verify each solution by substitution.
先不要看答案尝试这些题目。然后通过代入检验每一个解。
11. Summary and Key Takeaways | 总结与要点
Solving linear equations is a skill built on keeping the equation balanced. Remember to use inverse operations, expand brackets carefully, and clear fractions by multiplying by the LCD. Always check your answer to be sure.
解线性方程是一项建立在保持等式平衡基础上的技能。记住使用逆运算，小心展开括号，并通过乘以 LCD 来去分母。始终检查你的答案以确保正确。
Mastering these techniques now will make topics like simultaneous equations, inequalities, and graphs much easier in later years. With consistent practice, you'll find that even the trickiest equations become manageable.
现在掌握这些技巧，将会让你在后续学习联立方程、不等式和图像时轻松许多。通过持续练习，你会发现即使最棘手的方程也会变得容易处理。
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