📚 Solving Linear Equations | 解一元一次方程
Linear equations are the foundation of IGCSE algebra. They appear in almost every topic, from graphs to word problems, and mastering them gives you a reliable toolkit for solving problems involving unknown values. This lesson walks through the key methods for solving one-variable linear equations, including equations with brackets, fractions, and variables on both sides.
一元一次方程是 IGCSE 代数的基础。它们几乎出现在每一个主题中，从图像到应用题，掌握它们能为你提供解决未知数问题的可靠工具。本课将逐步讲解一元一次方程的主要解法，包括带有括号、分数以及变量在等式两边的方程。
1. What is a Linear Equation? | 什么是线性方程？
nts and a ≠ 0. The graph of such an equation is a straight line, which is why it is called 'linear' CSE exams, you will usually need to find the value of the unknown, often represented by x.
一元一次方程是指可以写成 ax + b = 0 形式的方程，其中 a 和 b 是常数，且 a ≠ 0。这种方程的图像是一条直线，因此被称为“线性”。在 IGCSE 考试中，你通常需要求出未知数的值，通常用 x 表示。
ower of x is 1. The equation x² + 3 = 7 is not linear because the highest power of x is 2.
例如，2x + 6 = 14 是一元一次方程，因为 x 的最高次数是 1。方程 x² + 3 = 7 不是线性方程，因为 x 的最高次数是 2。
appear in the denominator or under a square root sign. Expressions such as 3/x + 2 = 5 or √x = 4 are not linear equations.
在线性方程中，变量不能出现在分母或平方根号下。像 3/x + 2 = 5 或 √x = 4 这样的表达式都不是线性方程。
2. The Balancing Method | 等式的平衡方法
To solve a linear equation, you must keep the equation balanced at all times. This means that whatever operation you perform on one side of the equation, you must also perform on the other side. Think of an equation as a set of scales: both sides must always remain equal.
要解一元一次方程，你必须始终保持等式两边平衡。这意味着你对等式一边进行的任何运算，也必须对另一边进行。把方程想象成一台天平：两边必须始终保持相等。
The goal is to isolate the variable on one side. You can do this by using inverse operations: addition undoes subtraction, subtraction undoes addition, multiplication undoes division, and division undoes multiplication.
目标是将变量单独放在一边。你可以通过使用逆运算来实现：加法抵消减法，减法抵消加法，乘法抵消除法，除法抵消乘法。
act 5 from both sides to undo the addition: x + 5 - 5 = 12 - 5, which simplifies to x = 7.
例如，如果有 x + 5 = 12，你在两边同时减去 5 来抵消加法：x + 5 - 5 = 12 - 5，化简后得到 x = 7。
15, you divide both sides by 3 to undo the multiplication: 3x ÷ 3 = 15 ÷ 3, giving x = 5.
同样，如果有 3x = 15，你在两边同时除以 3 来抵消乘法：3x ÷ 3 = 15 ÷ 3，得到 x = 5。
3. Solving Simple Equations | 解简单方程
raction, while 5x = 35 only requires division. The process is short, but accuracy matters.
简单的一元一次方程只涉及对变量的一种运算。例如，x + 8 = 13 只需要减法，而 5x = 35 只需要除法。过程很短，但准确性很重要。
3 - 8, so x = 5. Second, solve 5x = 35. Divide both sides by 5: 5x ÷ 5 = 35 ÷ 5, so x = 7.
让我们看两个逐步示例。首先，解 x + 8 = 13。两边同时减去 8：x + 8 - 8 = 13 - 8，所以 x = 5。其次，解 5x = 35。两边同时除以 5：5x ÷ 5 = 35 ÷ 5，所以 x = 7。
xample, solve -x = 9. Multiply both sides by -1: -x × (-1) = 9 × (-1), which gives x = -9.
有时变量的系数为负数。例如，解 -x = 9。两边同时乘以 -1：-x × (-1) = 9 × (-1)，得到 x = -9。
Always write the final answer as clearly as possible. In IGCSE exams, the answer alone is often not enough; you must show your working to earn method marks.
始终尽可能清晰地写出最终答案。在 IGCSE 考试中，仅写出答案往往不够；你必须展示解题过程才能获得方法分。
4. Equations with Brackets | 带括号的方程
When an equation contains brackets, the first step is to expand them using the distributive property. For example, 3(x + 4) becomes 3x + 12. Expanding brackets removes the grouping and allows you to collect like terms.
当方程中包含括号时，第一步是使用分配律将其展开。例如，3(x + 4) 展开后为 3x + 12。展开括号可以消除分组，使你能合并同类项。
ide: 3x + 12 = 27. Then subtract 12 from both sides: 3x = 15. Finally, divide by 3: x = 5.
考虑方程 3(x + 4) = 27。展开左边：3x + 12 = 27。然后两边同时减去 12：3x = 15。最后，两边同时除以 3：x = 5。
- 5) = 8. Expanding gives 6x - 10 = 8. Add 10 to both sides: 6x = 18. Divide by 6: x = 3.
注意负号。例如，解 2(3x - 5) = 8。展开得到 6x - 10 = 8。两边同时加 10：6x = 18。两边同时除以 6：x = 3。
- Always expand brackets before collecting like terms.
- When multiplying a bracket by a negative number, be sure to apply the sign to every term inside.
- Check your expanded form by substituting a value, but keep this as a quick mental check.
- 在合并同类项之前，一定要先展开括号。
- 当括号乘以负数时，确保将符号应用到括号内的每一项。
- 可以通过代入一个值来检查展开后的形式，但这只是一个快速心算检查。
5. Equations with Fractions | 带分数的方程
Equations with fractions often look harder than they really are. The most efficient method is to multiply every term on both sides by the lowest common multiple (LCM) of the denominators. This clears the fractions in one step.
带分数的方程通常看起来比实际更难。最有效的方法是先用所有分母的最小公倍数（LCM）乘以等式两边的每一项。这样一步就能消去分数。
iply every term by 3: 3 × (x/3) + 3 × 2 = 3 × 5. This gives x + 6 = 15. Subtract 6: x = 9.
例如，解 x/3 + 2 = 5。分母是 3，所以将每一项都乘以 3：3 × (x/3) + 3 × 2 = 3 × 5。得到 x + 6 = 15。两边同时减去 6：x = 9。
e the LCM of both. Solve (x - 1)/4 = 2. Multiply both sides by 4: x - 1 = 8. Add 1: x = 9.
对于分母不同的方程，使用所有分母的 LCM。解 (x - 1)/4 = 2。两边同时乘以 4：x - 1 = 8。两边同时加 1：x = 9。
+ 6×(x/3) = 6×5. This gives 3x + 2x = 30. Collect like terms: 5x = 30. Divide by 5: x = 6.
如果方程是 x/2 + x/3 = 5，2 和 3 的 LCM 是 6。将每一项都乘以 6：6×(x/2) + 6×(x/3) = 6×5。得到 3x + 2x = 30。合并同类项：5x = 30。两边同时除以 5：x = 6。
- Identify all denominators before multiplying.
- Multiply every term on both sides by the LCM, including whole numbers.
- Simplify each term carefully to avoid losing factors.
- 在相乘之前先找出所有分母。
- 用 LCM 乘以等式两边的每一项，包括整数项。
- 仔细化简每一项，避免丢失因子。
6. Variables on Both Sides | 变量在等式两边
When the variable appears on both sides of the equation, your first aim is to collect all variable terms on one side and all constant terms on the other. You can do this by adding or subtracting the same term on both sides.
当变量出现在等式的两边时，你的首要目标是将所有含变量的项移到一边，将所有常数项移到另一边。你可以通过在两边同时加上或减去相同的项来实现。
+ 6, which simplifies to 3x - 3 = 6. Then add 3 to both sides: 3x = 9. Divide by 3: x = 3.
例如，解 5x - 3 = 2x + 6。两边同时减去 2x：5x - 2x - 3 = 2x - 2x + 6，化简得到 3x - 3 = 6。然后两边同时加上 3：3x = 9。两边同时除以 3：x = 3。
- 3. Subtract 2x from both sides: 5 = -3, which is false. Therefore there is no solution.
有时变量项会完全抵消，从而得到一个真命题或假命题。例如，解 2x + 5 = 2x - 3。两边同时减去 2x：5 = -3，这是假的。因此该方程无解。
5 = 5, the equation is true for all values of x, and there are infinitely many solutions.
如果结果是一个真命题，例如 5 = 5，那么这个方程对所有 x 值都成立，因此有无限多个解。
- Always combine like terms on each side before moving terms across.
- Move the smaller variable term first to keep the coefficient positive where possible.
- Watch out for cases with no solution or infinitely many solutions.
- 在把项移到对面之前，一定要先合并各边的同类项。
- 尽可能先移较小的变量项，以保持系数为正。
- 注意无解或无限多解的情况。
7. Checking Your Answer | 检验你的答案
After solving a linear equation, you should always substitute your answer back into the original equation to verify that both sides are equal. This quick check can catch arithmetic mistakes and sign errors.
解完一元一次方程后，你应该始终将答案代入原方程，检验两边是否相等。这种快速检验可以发现计算错误和符号错误。
4(5 - 2) = 4 × 3 = 12. Since the left side equals the right side, the solution is correct.
例如，如果你解 4(x - 2) = 12 得到 x = 5，将 5 代入原方程：4(5 - 2) = 4 × 3 = 12。因为左边等于右边，所以解是正确的。
Checking is especially important in exam conditions. It only takes a few seconds, but it can prevent losing marks for a small slip. If the check fails, go back through your working line by line.
在考试环境中，检验尤其重要。它只需要几秒钟，但可以防止因小失误而失分。如果检验不通过，就逐行回顾你的解题过程。
le. For example, in the equation 7x = 49, the answer x = 7 makes sense because 7 × 7 = 49.
你也可以通过心算估算来检查答案是否合理。例如，在方程 7x = 49 中，答案 x = 7 是合理的，因为 7 × 7 = 49。
- Substitute the answer into the original equation, not into a rearranged step.
- Perform the check on both sides independently.
- If the check fails, re-check each step in reverse order.
- 将答案代入原方程，而不是代入某一步变形后的方程。
- 分别独立检查等式两边。
- 如果检验不通过，按相反顺序逐步复查。
8. Word Problems | 应用题
Many IGCSE exam questions ask you to form and solve a linear equation from a written problem. The key skill is translating words into algebraic expressions. Look for clues such as 'more than', 'less than', 'total', and 'times'.
许多 IGCSE 考试题目要求你根据文字描述列出并解一元一次方程。关键技能是将文字转化为代数表达式。注意“多于”、“少于”、“总共”和“倍”等提示词。
cm. Then the length is x + 3 cm. The perimeter is 2(width + length), so 2(x + x + 3) = 26.
例题：一个矩形的长比宽多 3 cm，周长为 26 cm。设宽为 x cm，那么长为 x + 3 cm。周长等于 2(宽 + 长)，因此 2(x + x + 3) = 26。
26. Subtract 6: 4x = 20. Divide by 4: x = 5. So the width is 5 cm and the length is 8 cm.
化简括号内：2(2x + 3) = 26。展开：4x + 6 = 26。两边同时减去 6：4x = 20。两边同时除以 4：x = 5。所以宽为 5 cm，长为 8 cm。
Sam is twice as old as his sister. In 5 years, their total age will be 37. Let the sister's current age be x. Then Sam' In 5 years, their ages will be x + 5 and 2x + 5. The equation is (x + 5) + (2x + 5) = 37.
另一种常见类型是年龄问题。例题：Sam 的年龄是他妹妹的两倍。5 年后，他们的年龄总和将是 37 岁。设妹妹现在的年龄为 x，那么 Sam 的年龄是 2x。5 年后，他们的年龄分别为 x + 5 和 2x + 5。方程是 (x + 5) + (2x + 5) = 37。
tract 10: 3x = 27. Divide by 3: x = 9. The sister is 9 years old, and Sam is 18 years old.
化简：3x + 10 = 37。两边同时减去 10：3x = 27。两边同时除以 3：x = 9。妹妹 9 岁，Sam 18 岁。
- Define your variable clearly at the start.
- Write all given relationships in terms of that variable.
- Check that your final answer matches all conditions in the problem.
- 在一开始就明确定义变量。
- 用该变量表示所有给定的关系。
- 检查最终答案是否满足题目中的所有条件。
9. Common Mistakes | 常见错误
e other side. For example, in x + 4 = 9, the correct step is x = 9 - 4 = 5, not x = 9 + 4.
即使是成绩好的学生在解一元一次方程时也可能犯粗心错误。一个常见错误是将一项移到另一边时忘记变号。例如，在 x + 4 = 9 中，正确的步骤是 x = 9 - 4 = 5，而不是 x = 9 + 4。
/2 by 2, leaving 3 unchanged. The correct step is to multiply every term by 2: x + 6 = 14.
另一个错误是没有将运算应用到每一项。例如，在解 x/2 + 3 = 7 时，有些学生只将 x/2 乘以 2，而 3 保持不变。正确的步骤是将每一项都乘以 2：x + 6 = 14。
Sign errors with brackets are also frequent. For example, -2(x - 3) should expand to -2x + 6, not -2x - 6. The negative sign applies to both terms inside the bracket.
带括号时的符号错误也很常见。例如，-2(x - 3) 应该展开为 -2x + 6，而不是 -2x - 6。负号要应用于括号内的两项。
Finally, some students forget to check their answers. A quick substitution can reveal many of these errors before the examiner does.
最后，有些学生忘记检验答案。一次快速代入就能在阅卷老师之前发现许多此类错误。
| Common Mistake | 常见错误 | Correct Approach | 正确做法 |
|---|---|
| Moving a term without changing sign | Use inverse operations to keep the equation balanced |
| Multiplying only part of the equation | Multiply every term on both sides |
| Expanding negative brackets incorrectly | Apply the negative sign to every term inside |
| Skipping the checking step | Substitute your answer into the original equation |
By being aware of these common errors, you can avoid them in your own working and gain extra marks in exams.
通过了解这些常见错误，你可以在自己的作答中避免它们，并在考试中获得额外分数。
10. Exam-style Practice | 考试风格练习
Practising exam-style questions is the best way to build confidence. Try these problems without looking at the worked solutions first. For each one, show all your steps clearly.
练习考试风格的题目是建立信心的最佳方式。先尝试独立完成这些题目，不要看详细解答。每道题都要清楚地展示所有步骤。
uestion 1: Solve 4x - 7 = 13. First add 7 to both sides: 4x = 20. Then divide by 4: x = 5.
题目 1：解 4x - 7 = 13。首先两边同时加上 7：4x = 20。然后两边同时除以 4：x = 5。
ion 2: Solve 3(2x + 1) = 27. Expand: 6x + 3 = 27. Subtract 3: 6x = 24. Divide by 6: x = 4.
题目 2：解 3(2x + 1) = 27。展开：6x + 3 = 27。两边同时减去 3：6x = 24。两边同时除以 6：x = 4。
Question 3: Solve (x + 4)/2 = 9. Multiply both sides by 2: x + 4 = 18. Subtract 4: x = 14.
题目 3：解 (x + 4)/2 = 9。两边同时乘以 2：x + 4 = 18。两边同时减去 4：x = 14。
5 = 3x + 11. Subtract 3x from both sides: 4x - 5 = 11. Add 5: 4x = 16. Divide by 4: x = 4.
题目 4：解 7x - 5 = 3x + 11。两边同时减去 3x：4x - 5 = 11。两边同时加上 5：4x = 16。两边同时除以 4：x = 4。
29. Let the number be n. The equation is 5n - 6 = 29. Add 6: 5n = 35. Divide by 5: n = 7.
题目 5：一个数乘以 5 后再减去 6，结果是 29。设这个数为 n。方程是 5n - 6 = 29。两边同时加上 6：5n = 35。两边同时除以 5：n = 7。
- Attempt all questions before checking the solutions.
- Write down each operation you perform on both sides.
- Use the final answer to verify against the original equation.
- 在查看答案之前先尝试完成所有题目。
- 写出你在等式两边进行的每一步运算。
- 用最终答案代入原方程进行验证。
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