Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations are one of the most important topics in KS3 Cambridge Mathematics. They appear throughout algebra, problem solving, and even in geometry and data handling. This article explains how to solve one-variable linear equations step by step, from simple one-step equations to equations with brackets and variables on both sides.

一元一次方程是 KS3 剑桥数学中最重要的主题之一。它们广泛出现在代数、问题解决,甚至几何和数据处理中。本文将逐步解释如何解一元一次方程,从简单的单步方程到含括号和变量在两侧的方程。

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 = c, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 1, so the graph of the equation is a straight line.

一元一次方程是能写成 ax + b = c 形式的方程,其中 a、b 和 c 是常数,且 a ≠ 0。变量 x 的最高次数是 1,因此方程的图像是一条直线。

For example, 3x + 5 = 20, 2x = 14, and x/4 – 1 = 2 are all linear equations. The goal is to find the value of x that makes the equation true.

例如,3x + 5 = 20、2x = 14 和 x/4 – 1 = 2 都是一元一次方程。我们的目标是求出使方程成立的 x 值。


2. The Balancing Method | 平衡法

An equation is like a balance scale. The left side and right side must always have the same value. Whatever you do to one side, you must also do to the other side.

方程就像一架天平。左边和右边必须始终保持相同的值。无论对方程的一边做什么,另一边也必须做同样的操作。

If you add 3 to the left side, you must add 3 to the right side. If you divide the left side by 2, you must divide the right side by 2. This keeps the equation balanced.

如果左边加 3,右边也必须加 3。如果左边除以 2,右边也必须除以 2。这样才能保持方程平衡。

The balancing method is the foundation for solving all linear equations and helps prevent mistakes.

平衡法是解所有一元一次方程的基础,有助于避免错误。


3. Inverse Operations | 逆运算

To isolate the variable x, we use inverse operations. Addition and subtraction are inverse operations, just as multiplication and division are inverse operations.

为了单独求出变量 x,我们使用逆运算。加法和减法互为逆运算,乘法和除法也互为逆运算。

For example, if the equation is x + 7 = 12, we subtract 7 from both sides because subtraction is the inverse of addition.

例如,如果方程是 x + 7 = 12,我们在两边同时减去 7,因为减法是加法的逆运算。

If the equation is 4x = 20, we divide both sides by 4 because division is the inverse of multiplication.

如果方程是 4x = 20,我们在两边同时除以 4,因为除法是乘法的逆运算。


4. Solving One-Step Equations | 解一步方程

One-step equations require only one inverse operation to solve. For x + 5 = 13, subtract 5 from both sides:

一步方程只需要一步逆运算就能解出。对于 x + 5 = 13,在两边同时减去 5:

x + 5 – 5 = 13 – 5 ⇒ x = 8

For 6x = 42, divide both sides by 6:

对于 6x = 42,两边同时除以 6:

6x ÷ 6 = 42 ÷ 6 ⇒ x = 7

Always write the solution as x = a number. This is called the solution set or the root of the equation.

总要把解写成 x = 某个数。这叫做方程的解或根。


5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations. For example, in 2x + 3 = 11, the variable x is first multiplied by 2, then 3 is added. To solve, reverse the order: first subtract 3, then divide by 2.

两步方程包含两个运算。例如在 2x + 3 = 11 中,变量 x 先乘以 2,然后加上 3。解方程时要逆序操作:先减去 3,再除以 2。

Step 1: Subtract 3 from both sides.

第一步:两边同时减去 3。

2x + 3 – 3 = 11 – 3 ⇒ 2x = 8

Step 2: Divide both sides by 2.

第二步:两边同时除以 2。

2x ÷ 2 = 8 ÷ 2 ⇒ x = 4

This reverse order is important: undo addition or subtraction before undoing multiplication or division.

这个逆序很重要:先撤销加减法,再撤销乘除法。


6. Equations with Brackets | 含括号的方程

When an equation contains brackets, expand them first using the distributive law. For example, solve 3(x + 2) = 21.

当方程含有括号时,先用分配律展开括号。例如解 3(x + 2) = 21。

Expand the left side: 3 × x + 3 × 2 = 3x + 6.

展开左边:3 × x + 3 × 2 = 3x + 6。

3x + 6 = 21

Then subtract 6 from both sides: 3x = 15.

然后两边同时减去 6:3x = 15。

Finally divide by 3: x = 5.

最后两边同时除以 3:x = 5。

You can also divide first if the number outside the bracket divides the right side evenly, but expanding is usually safer.

如果括号外的数能整除右边,也可以先除以该数,但展开通常更稳妥。


7. Equations with Variables on Both Sides | 变量在两侧的方程

Some equations have the variable x on both sides, such as 5x + 2 = 3x + 10. The first step is to collect the x terms on one side.

有些方程的变量 x 同时出现在两边,例如 5x + 2 = 3x + 10。第一步是把含 x 的项移到一边。

Subtract 3x from both sides: 5x – 3x + 2 = 3x – 3x + 10, which simplifies to 2x + 2 = 10.

两边同时减去 3x:5x – 3x + 2 = 3x – 3x + 10,化简为 2x + 2 = 10。

Then subtract 2 from both sides: 2x = 8.

然后两边同时减去 2:2x = 8。

Finally divide by 2: x = 4.

最后两边同时除以 2:x = 4。

Always check which side has the smaller coefficient if you want to avoid negative coefficients; here 3x is smaller than 5x, so subtract 3x.

如果不想出现负系数,总是检查哪一边的系数较小;这里 3x 小于 5x,所以减去 3x。


8. Equations with Fractions | 含分数的方程

Equations with fractions can be solved by multiplying every term by the lowest common multiple (LCM) of the denominators. For example, solve x/3 + 2 = 5.

含有分数的方程可以通过将每一项乘以所有分母的最小公倍数 (LCM) 来求解。例如,解 x/3 + 2 = 5。

Multiply every term by 3: (x/3) × 3 + 2 × 3 = 5 × 3.

每一项乘以 3:(x/3) × 3 + 2 × 3 = 5 × 3。

x + 6 = 15

Then subtract 6 from both sides: x = 9.

然后两边同时减去 6:x = 9。

If there are two fractions, such as (x + 1)/2 = (x – 3)/4, multiply by the LCM of 2 and 4, which is 4.

如果有两个分数,例如 (x + 1)/2 = (x – 3)/4,则乘以 2 和 4 的最小公倍数 4。

This gives 2(x + 1) = x – 3, then expand and solve.

得到 2(x + 1) = x – 3,然后展开并求解。


9. Checking Your Solution | 检验你的解

After finding a solution, substitute it back into the original equation to check that both sides are equal. For x = 4 in 2x + 3 = 11:

找到解之后,把它代回原方程,检验两边是否相等。例如 x = 4 代入 2x + 3 = 11:

2 × 4 + 3 = 8 + 3 = 11 ✓

The left side equals the right side, so the solution is correct.

左边等于右边,所以解是正确的。

Checking is especially useful in exams because it can reveal arithmetic errors.

检验在考试中特别有用,因为它能发现计算错误。

If the check fails, go back and re-solve the equation carefully.

如果检验不通过,请回过头来仔细重新解方程。


10. Word Problems Leading to Linear Equations | 应用题转化为一元一次方程

Many KS3 problems ask you to form an equation from a real-life situation. Identify the unknown and call it x, then translate words into mathematical symbols.

许多 KS3 题目要求根据实际情境列出方程。先确定未知数并用 x 表示,然后把文字转换为数学符号。

For example, “Three more than twice a number is 15.” Let the number be x. Then 2x + 3 = 15.

例如,“一个数的两倍再加 3 等于 15。”设这个数为 x,则列出 2x + 3 = 15。

Solve: 2x + 3 = 15, subtract 3: 2x = 12, divide by 2: x = 6.

解方程:2x + 3 = 15,减去 3:2x = 12,除以 2:x = 6。

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