Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations appear in almost every IGCSE Mathematics examination, and they are the building blocks for solving problems in algebra, geometry, and statistics. This revision guide follows the logic of a clear teacher’s book: explain, demonstrate, practise, and check.

一次方程几乎出现在每一场 IGCSE 数学考试中，也是处理代数、几何与统计问题的基石。本复习指南按照教师用书的思路展开：讲解、示范、练习、检验。

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

A linear equation in one variable is an equation where the unknown has exponent 1. For example, 3x – 4 = 11 is linear, but x² + 3x = 0 is not linear because it contains x².

含一个未知数的一次方程，是指未知数的最高次数为 1 的方程。例如，3x – 4 = 11 是一次方程，而 x² + 3x = 0 不是一次方程，因为它含有 x²。

The general form of a simple linear equation is:

简单一次方程的一般形式是：

ax + b = c, where a ≠ 0

ax + b = c，其中 a ≠ 0

Here x is the unknown, while a, b and c are constants. The condition a ≠ 0 is important because if a = 0, there is no x term left.

其中 x 是未知数，a、b、c 是常数。条件 a ≠ 0 非常重要，因为如果 a = 0，方程中就没有 x 项了。

2. The Balancing Method | 等式平衡法

An equation is like a balance scale. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side so that the scale stays balanced.

方程就像一架天平。如果你对一边做加法、减法、乘法或除法，必须对另一边做完全相同的运算，才能保持天平平衡。

Example 1: Solve x + 5 = 12.

例 1：解 x + 5 = 12。

x + 5 – 5 = 12 – 5 → x = 7

Subtract 5 from both sides to isolate the variable. The answer is x = 7.

两边同时减去 5，以分离未知数，所以 x = 7。

Example 2: Solve x – 4 = 9.

例 2：解 x – 4 = 9。

x – 4 + 4 = 9 + 4 → x = 13

Add 4 to both sides because addition is the inverse operation of subtraction.

两边同时加上 4，因为加法是减法的逆运算。

3. Inverse Operations and Exact Solutions | 逆运算与精确解

To solve an equation, we reverse the operations that are applied to the unknown. Addition and subtraction are inverse operations; multiplication and division are also inverse operations.

解方程时，我们要反推施加在未知数上的运算。加法与减法互为逆运算，乘法与除法也互为逆运算。

Example: Solve x/4 = 9.

例：解 x/4 = 9。

x/4 × 4 = 9 × 4 → x = 36

Since x was divided by 4, multiply both sides by 4 to get x = 36.

因为 x 被除以 4，所以两边同时乘以 4，得到 x = 36。

This idea also works when the unknown is multiplied by a number. To solve 7x = 42, divide both sides by 7.

这个思路同样适用于未知数乘以一个数的情况。解 7x = 42 时，两边同时除以 7。

7x ÷ 7 = 42 ÷ 7 → x = 6

4. Collecting Like Terms | 合并同类项

Before solving an equation, simplify each side by collecting like terms. Like terms have the same variable and the same power. Constants are also like terms.

在解方程之前，要先合并各边同类项来化简。同类项是指具有相同未知数且次数相同的项；常数之间也是同类项。

Example: Solve 2x + 3x + 4 = 19.

例：解 2x + 3x + 4 = 19。

5x + 4 = 19

5x + 4 – 4 = 19 – 4 → 5x = 15

5x ÷ 5 = 15 ÷ 5 → x = 3

First add the x terms: 2x + 3x = 5x. Then use the balancing method to isolate x.

先合并含 x 的项：2x + 3x = 5x。再用等式平衡法分离 x。

5. Variable Terms on Both Sides | 未知数在等号两边

When the unknown appears on both sides of the equation, move all variable terms to one side and all constant terms to the other side.

当未知数同时出现在等号两边时，把含未知数的项移到一边，把所有常数项移到另一边。

Example: Solve 5x – 3 = 2x + 9.

例：解 5x – 3 = 2x + 9。

5x – 3 – 2x = 2x + 9 – 2x → 3x – 3 = 9

3x – 3 + 3 = 9 + 3 → 3x = 12

3x ÷ 3 = 12 ÷ 3 → x = 4

Subtract 2x from both sides so that the variable terms are on the left. Then add 3 and divide by 3.

两边同时减去 2x，使含未知数的项集中在左边；然后两边加 3，再除以 3。

Sometimes the variable terms disappear completely. For example, 2x + 1 = 2x + 3 has no solution because subtracting 2x gives 1 = 3, which is false. In contrast, 3x + 6 = 3(x + 2) is true for every value of x, so it is called an identity.

有时未知数项会完全消失。例如，2x + 1 = 2x + 3 无解，因为两边同时减去 2x 后得到 1 = 3，这是矛盾。而 3x + 6 = 3(x + 2) 对所有 x 都成立，所以它被称为恒等式。

6. Equations with Brackets | 含括号的方程

Use the distributive law to remove brackets before collecting like terms. The distributive law states that a(b + c) = ab + ac.

要先用乘法分配律去掉括号，再合并同类项。分配律是指 a(b + c) = ab + ac。

Example: Solve 3(2x + 1) = 15.

例：解 3(2x + 1) = 15。

6x + 3 = 15

6x + 3 – 3 = 15 – 3 → 6x = 12

6x ÷ 6 = 12 ÷ 6 → x = 2

Multiply each term inside the bracket by 3, then continue using inverse operations.

用 3 去乘括号内的每一项，然后继续使用逆运算解方程。

Be careful with a negative sign outside the bracket. For example, -2(x – 5) = -2x + 10, not -2x – 10.

要特别小心括号外是负号的情况。例如，-2(x – 5) = -2x + 10，而不是 -2x – 10。

7. Equations with Fractions | 含分数的方程

For equations with fractions, you can multiply every term by the lowest common multiple (LCM) of the denominators. This removes the fractions and makes the equation easier to solve.

对于含分数的方程，可以将每一项都乘以分母的最小公倍数（LCM），这样可以去掉分母，使方程更易求解。

Example: Solve x/3 + x/2 = 5.

例：解 x/3 + x/2 = 5。

The LCM of 3 and 2 is 6. Multiply every term by 6.

3 和 2 的最小公倍数是 6。将每一项都乘以 6。

6 × (x/3) + 6 × (x/2) = 6 × 5 → 2x + 3x = 30

5x = 30 → x = 6

Check: 6/3 + 6/2 = 2 + 3 = 5, so the solution is correct.

检验：6/3 + 6/2 = 2 + 3 = 5，所以解正确。

In general, multiply every term, including constants, by the same LCM. Do not multiply only the fractions.

一般情况下，包括常数项在内的每一项都要乘以相同的 LCM，不能只乘分数项。

8. Constructing Equations from Word Problems | 从实际问题列方程

One of the most useful skills in mathematics is translating a word problem into an equation. Start by choosing a letter for the unknown, then write down the relationships described in the words.

数学中最有用的技能之一，就是把实际问题转化成方程。先选择一个字母表示未知数，然后把题目中描述的关系写成方程。

Example: A rectangle has a length that is 5 cm more than its width, and its perimeter is 34 cm. Find the width and length.

例：一个长方形的长比宽多 5 cm，周长为 34 cm。求宽和长。

Let the width be w cm. Then the length is w + 5 cm.

设宽为 w cm，则长为 w + 5 cm。

2[(w + 5) + w] = 34

2[2w + 5] = 34 → 4w + 10 = 34

4w = 24 → w = 6

So the width is 6 cm and the length is 6 + 5 = 11 cm.

因此宽为 6 cm，长为 6 + 5 = 11 cm。

Always state the units in your final answer and check that the value makes sense in the original problem.

在最终答案中一定要写单位，并检查所得数值是否符合原题

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Solving Linear Equations | 解一元一次方程