Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations are the foundation of algebra and appear in nearly every IGCSE Mathematics exam paper. Mastering this topic not only secures easy marks but also builds the skills needed for simultaneous equations, inequalities, and word problems.

一元一次方程是代数的基石，几乎出现在每一份 IGCSE 数学试卷中。掌握这一主题不仅能稳拿基础分，还能为后续学习联立方程、不等式和文字应用题打下坚实技能基础。

1. What Is a Linear Equation? | 什么是一元一次方程

A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are constants and a ≠ 0. The variable appears only to the first power — no squares, cubes, or square roots of the variable are allowed.

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。变量只以一次幂出现——不允许有变量的平方、立方或平方根。

For example, 3x + 5 = 20 is a linear equation because x appears only to the first power. However, x² + 2x = 8 is not linear, because the x² term makes it quadratic.

例如，3x + 5 = 20 是一元一次方程，因为 x 只以一次幂出现。而 x² + 2x = 8 不是一次方程，因为 x² 项使其成为二次方程。

The goal of solving any linear equation is to find the value of the variable that makes the equation true. This value is called the solution or root of the equation.

解任何一元一次方程的目标，就是找到使方程成立的变量值。这个值称为方程的解或根。

2. The Golden Rule: Balance | 黄金法则：保持平衡

Think of an equation as a balance scale. The left-hand side (LHS) and the right-hand side (RHS) are perfectly balanced. Whatever you do to one side, you must do to the other side to keep the balance.

把方程想象成一台天平。等号左边（LHS）和右边（RHS）是完美平衡的。你对一边做的任何操作，必须对另一边做同样的操作，才能保持平衡。

This means you can add, subtract, multiply, or divide both sides by the same number (but never divide by zero). This principle is called preserving equality.

这意味着你可以对两边同时加、减、乘或除以同一个数（但绝不能除以零）。这一原则称为保持等式成立。

For example, if 2x = 10, dividing both sides by 2 gives x = 5. Because we divided the same way on both sides, the balance is maintained and the solution is correct.

例如，若 2x = 10，两边同时除以 2 得 x = 5。因为我们对两边做了相同的除法，平衡得以保持，解也是正确的。

3. Solving Simple Equations: ax = b | 解简单方程：ax = b

The simplest type of linear equation has the variable on one side with a coefficient. To solve it, divide both sides by the coefficient.

最简单的一元一次方程是变量在一边且有系数。解这类方程，只需两边同时除以系数即可。

If ax = b, then x = b ÷ a (provided a ≠ 0)

Example: Solve 7x = 42.
Divide both sides by 7: x = 42 ÷ 7 = 6.
Check: 7 × 6 = 42 ✓

示例：解 7x = 42。
两边除以 7：x = 42 ÷ 7 = 6。
检验：7 × 6 = 42 ✓

When the coefficient is a fraction, such as x⁄3 = 9, multiply both sides by the denominator: x = 9 × 3 = 27.

当系数是分数时，例如 x⁄3 = 9，两边乘以分母：x = 9 × 3 = 27。

4. Equations with Addition or Subtraction | 含加减法的一次方程

When the equation involves adding or subtracting a constant term, perform the inverse operation on both sides to isolate the variable.

当方程涉及加减常数项时，对两边执行逆运算来分离变量。

If x + a = b, then x = b – a
If x – a = b, then x = b + a

Example: Solve x + 8 = 15.
Subtract 8 from both sides: x = 15 – 8 = 7.
Check: 7 + 8 = 15 ✓

示例：解 x + 8 = 15。
两边减 8：x = 15 – 8 = 7。
检验：7 + 8 = 15 ✓

Example: Solve x – 5 = 12.
Add 5 to both sides: x = 12 + 5 = 17.
Check: 17 – 5 = 12 ✓

示例：解 x – 5 = 12。
两边加 5：x = 12 + 5 = 17。
检验：17 – 5 = 12 ✓

Remember: addition and subtraction are inverse operations, and they cancel each other out when used together.

记住：加法和减法互为逆运算，两者结合使用时会相互抵消。

5. Solving ax + b = c | 解 ax + b = c 型方程

This is the most common form you will encounter. The strategy is to undo the operations step by step, starting with the addition/subtraction first, then the multiplication/division.

这是最常见的形式。解题策略是逐步撤销运算，先处理加减法，再处理乘除法。

Step 1: Subtract b from both sides → ax = c – b
Step 2: Divide both sides by a → x = (c – b) / a

Example: Solve 3x + 4 = 19.
Step 1: Subtract 4 from both sides: 3x = 19 – 4 = 15.
Step 2: Divide both sides by 3: x = 15 ÷ 3 = 5.
Check: 3(5) + 4 = 15 + 4 = 19 ✓

示例：解 3x + 4 = 19。
第一步：两边减 4：3x = 19 – 4 = 15。
第二步：两边除以 3：x = 15 ÷ 3 = 5。
检验：3(5) + 4 = 15 + 4 = 19 ✓

Why do we handle subtraction before division? Because the addition/subtraction is “outside” the multiplication. Think of it as: 3 groups of x, plus 4 more. To find one group, we must first remove the extra 4.

为什么要先处理加减再处理乘除？因为加减法在乘法“外面”。可以这样想：3 组 x，再加 4 个。要求出一组 x 的值，必须先移除多出来的 4 个。

6. Variables on Both Sides | 变量在等号两边

When the variable appears on both sides of the equation, the strategy is to collect all variable terms on one side and all constant terms on the other side.

当变量出现在等号两边时，策略是将所有含变量的项集中到一边，所有常数项集中到另一边。

General strategy: Move all x-terms to LHS and constants to RHS

Example: Solve 5x + 3 = 2x + 15.
Step 1: Subtract 2x from both sides: 3x + 3 = 15.
Step 2: Subtract 3 from both sides: 3x = 12.
Step 3: Divide by 3: x = 4.
Check: 5(4) + 3 = 23 and 2(4) + 15 = 23 ✓

示例：解 5x + 3 = 2x + 15。
第一步：两边减 2x：3x + 3 = 15。
第二步：两边减 3：3x = 12。
第三步：除以 3：x = 4。
检验：5(4) + 3 = 23，2(4) + 15 = 23 ✓

Tip: To avoid negative coefficients, move the smaller x-term to the other side. This keeps the coefficient of x positive, making the algebra simpler.

小技巧：为了避免负系数，把较小的 x 项移到另一边。这样 x 的系数保持为正，运算更简便。

7. Equations with Brackets | 含括号的一次方程

When brackets appear, the first step is to expand them using the distributive property: a(b + c) = ab + ac.

当出现括号时，第一步是使用分配律展开括号：a(b + c) = ab + ac。

Step 1: Expand all brackets
Step 2: Collect like terms
Step 3: Solve the resulting equation

Example: Solve 2(3x – 1) = 4x + 10.
Step 1: Expand: 6x – 2 = 4x + 10.
Step 2: Subtract 4x from both sides: 2x – 2 = 10.
Step 3: Add 2 to both sides: 2x = 12.
Step 4: Divide by 2: x = 6.
Check: 2(18 – 1) = 34 and 4(6) + 10 = 34 ✓

示例：解 2(3x – 1) = 4x + 10。
第一步：展开：6x – 2 = 4x + 10。
第二步：两边减 4x：2x – 2 = 10。
第三步：两边加 2：2x = 12。
第四步：除以 2：x = 6。
检验：2(18 – 1) = 34，4(6) + 10 = 34 ✓

Be especially careful with negative signs outside brackets: -(2x – 3) means -2x + 3, not -2x – 3. Every term inside the brackets changes sign.

尤其小心括号外的负号：-(2x – 3) 表示 -2x + 3，而不是 -2x – 3。括号内每一项都要变号。

8. Equations with Fractions | 含分数的一次方程

When fractions appear in the equation, the most efficient method is to multiply both sides by the lowest common denominator (LCD) to eliminate all fractions in one step.

当方程中出现分数时，最有效的方法是两边同时乘以最小公分母（LCD），一步消除所有分数。

Multiply every term on both sides by the LCD, then solve normally

Example: Solve x/2 + 3 = x/3 + 5.
Step 1: LCD of 2 and 3 is 6. Multiply every term by 6:
6(x/2) + 6(3) = 6(x/3) + 6(5)
3x + 18 = 2x + 30.
Step 2: Subtract 2x: x + 18 = 30.
Step 3: Subtract 18: x = 12.
Check: 12/2 + 3 = 9 and 12/3 + 5 = 9 ✓

示例：解 x/2 + 3 = x/3 + 5。
第一步：2 和 3 的最小公分母是 6。每一项乘以 6：
6(x/2) + 6(3) = 6(x/3) + 6(5)
得 3x + 18 = 2x + 30。
第二步：两边减 2x：x + 18 = 30。
第三步：两边减 18：x = 12。
检验：12/2 + 3 = 9，12/3 + 5 = 9 ✓

When the equation contains a fraction like (2x + 1)/3 = 5, simply multiply both sides by 3 to get 2x + 1 = 15, then solve: 2x = 14, so x = 7.

当方程中含有 (2x + 1)/3 = 5 这样的分数时，只需两边乘以 3 得 2x + 1 = 15，然后求解：2x = 14，所以 x = 7。

9. Word Problems | 应用题

Many students find word problems challenging, but the key is to translate the English sentence into a mathematical equation step by step.

许多学生觉得应用题有挑战性，但关键是把文字句子逐步翻译成数学方程。

Step 1: Define the variable
Step 2: Write an equation from the given information
Step 3: Solve the equation
Step 4: Answer the question in words

Example: John is 3 years older than his sister Mary. The sum of their ages is 25. How old is Mary?
Step 1: Let Mary’s age = x. Then John’s age = x + 3.
Step 2: x + (x + 3) = 25.
Step 3: 2x + 3 = 25 → 2x = 22 → x = 11.
Step 4: Mary is 11 years old. John is 14. Check: 11 + 14 = 25 ✓

示例：约翰比他妹妹玛丽大 3 岁。两人年龄之和为 25。玛丽多少岁？
第一步：设玛丽年龄为 x，则约翰年龄为 x + 3。
第二步：x + (x + 3) = 25。
第三步：2x + 3 = 25 → 2x = 22 → x = 11。
第四步：玛丽 11 岁。约翰 14 岁。检验：11 + 14 = 25 ✓

Look for keywords: “sum” means addition, “difference” means subtraction, “product” means multiplication, and “is” usually means equals.

注意关键词：“和”表示加法，“差”表示减法，“积”表示乘法，“是”通常表示等号。

10. Special Cases and Identities | 特殊情况与恒等式

Not every equation has exactly one solution. There are three possible cases:

并非每个方程都只有一个解。共有三种可能情况：

Case	Example	Solution Type
Exactly one solution	2x + 1 = 7	x = 3 (unique solution)
No solution	2x + 3 = 2x – 5	Contradiction (3 = -5, impossible)
Infinite solutions	3(x + 2) = 3x + 6	True for all x (identity)

For the no-solution case, when you simplify, the variable disappears and you are left with a false statement. For the infinite case, the variable disappears and you are left with a true statement.

在无解的情况下，化简后变量消失，剩下一个不成立的等式。在无穷多解的情况下，变量消失后剩下一个恒真的等式。

In the IGCSE exam, the identity case often appears as “show that this is an identity” — you need to expand and simplify both sides to demonstrate they are equivalent.

在 IGCSE 考试中，恒等式的情况常以“证明这是恒等式”的形式出现——你需要展开并化简两边，证明它们是等价的。

11. Common Mistakes to Avoid | 常见错误提醒

Being aware of frequent errors can save you valuable marks in the exam.

了解常见错误可以帮助你在考试中保住宝贵的分数。

Forgetting to apply operations to both sides: Always perform the same operation on the entire LHS and entire RHS, not just one term.

忘记对两边同时操作：始终对完整的左边和右边执行相同运算，而不是只对某一项。
Sign errors when removing brackets: -(x – 4) is -x + 4, not -x – 4.

去括号时符号错误：-(x – 4) 等于 -x + 4，而不是 -x – 4。
Incorrectly multiplying fractions: When multiplying both sides by the LCD, multiply every term, including constants, not just the fractional terms.

分数乘法错误：两边乘以最小公分母时，要乘以每一个项，包括常数项，而不仅仅是分数项。
Not checking the answer: Always substitute your solution back into the original equation. This takes 10 seconds and confirms you are correct.

不检验答案：始终把解代回原方程。这只需要 10 秒钟，却能确认你的答案是否正确。
Division by zero: If you try to divide both sides by a term that could be zero (such as x itself), you may lose a solution. Never divide by a variable without stating the condition x ≠ 0.

除以零：如果你试图两边同时除以一个可能为零的项（如 x 本身），可能会丢失解。切勿除以一个变量而不说明 x ≠ 0 的条件。

12. Exam-Style Practice Problems | 考试风格练习

Here are five IGCSE-style problems to test your understanding. Try to solve each one before looking at the answer.

以下是五道 IGCSE 风格的练习题，用来测试你的理解。先尝试独立解答，再对照答案。

Problem	Answer
1. Solve 4x – 7 = 29	x = 9
2. Solve 6x + 5 = 2x + 17	x = 3
3. Solve 3(2x – 4) = 24	x = 6
4. Solve x/4 + 5 = 9	x = 16
5. A number multiplied by 5, then increased by 3, equals 43. Find the number.	x = 8

If you got all five correct, you have mastered the core skills. If not, revisit the relevant section above and try again.

如果你五题全对，说明你已掌握了核心技能。如果有错误，请回顾上文相应章节并再次尝试。

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Solving Linear Equations | 解一元一次方程