Solving Linear Equations at KS3 | KS3 解一元一次方程

📚 Solving Linear Equations at KS3 | KS3 解一元一次方程

Linear equations are the foundation of algebra at Cambridge KS3. They appear in every Checkpoint test and in many real-life problems, from working out unknown lengths to sharing amounts. This article reviews the key methods, from one-step equations to equations with brackets and fractions, so you can solve them accurately and confidently.

一元一次方程是剑桥 KS3 代数的基础。它们出现在每一次 Checkpoint 测试和许多实际问题中，例如求未知长度或分配数量。本文复习关键方法，从一步方程到含括号和分数的方程，帮助你准确、自信地求解。

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

uch as 2x + 3 = 11, where the unknown x is not raised to any power higher than 1. The word 'linear' t line. At KS3, we focus on finding the value of the unknown that makes the equation true.

一元一次方程是形如 2x + 3 = 11 的方程，其中未知数 x 的最高次数为 1。"linear" 一词表示方程的图像是一条直线。在 KS3 阶段，我们重点是求出使方程成立的未知数的值。

Equations often contain numbers, variables, and operations. The expression on the left of the equals sign has the same value as the expression on the right. Solving the equation means finding every value of the variable that keeps this balance true.

方程通常包含数字、变量和运算。等号左边的表达式与右边的表达式具有相同的值。解方程就是找到所有使这种平衡成立的变量值。

• 3a + 2 = 14
• 5(y - 4) = 3y + 2
• x/3 + 1 = 7

These are all linear equations because the variable only appears to the power 1 and is never multiplied by itself.

这些都是一元一次方程，因为变量只出现一次方，并且不会与自己相乘。

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2. Keeping Equations Balanced | 保持方程平衡

The golden rule of equation solving is to do the same operation to both sides. An equation is like a balance scale: whatever you add, subtract, multiply or divide on one side must also be done on the other side. This keeps the equation true while you isolate the unknown.

解方程的黄金法则是对方程两边同时进行相同的运算。方程就像一架天平：无论在一侧加、减、乘、除什么，都必须对另一侧做同样操作。这样在分离未知数的同时方程仍然成立。

If two expressions are equal, then adding the same number to both expressions, subtracting the same number, multiplying by the same non-zero number, or dividing by the same non-zero number will keep them equal.

如果两个表达式相等，那么两边同时加上相同的数、减去相同的数、乘以相同的非零数或除以相同的非零数，它们仍然相等。

If a = b, then a + c = b + c, a − c = b − c, a × c = b × c, and a ÷ c = b ÷ c (c ≠ 0)

如果 a = b，那么 a + c = b + c、a − c = b − c、a × c = b × c、a ÷ c = b ÷ c（c ≠ 0）

This rule is the reason we can move terms from one side to another. For example, to undo an addition of 5, we subtract 5 from both sides. To undo a multiplication by 3, we divide both sides by 3.

这条法则就是我们能把项从一边移到另一边的原因。例如，要消去加 5，我们两边同时减 5；要消去乘 3，我们两边同时除以 3。

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3. One-Step Equations | 一步方程

des. To solve 4x = 24, divide both sides by 4. To solve x/5 = 3, multiply both sides by 5.

在一步方程中，只需要一次逆运算。例如，解 x + 7 = 15 时，两边同时减 7；解 4x = 24 时，两边同时除以 4；解 x/5 = 3 时，两边同时乘以 5。

• x + 9 = 20 → x = 11
• 6x = 42 → x = 7
• x/4 = 8 → x = 32

Always use the inverse operation of the operation that is already in the equation. Addition and subtraction are inverse operations; multiplication and division are inverse operations.

始终使用方程中已有运算的逆运算。加法和减法互为逆运算；乘法和除法互为逆运算。

x + a = b → x = b − a

ax = b → x = b ÷ a (a ≠ 0)

x/a = b → x = a × b (a ≠ 0)

In Checkpoint questions, one-step equations often appear inside a larger problem, so it is important to recognise them quickly and solve them without hesitation.

在 Checkpoint 题目中，一步方程常出现在较大的问题里，因此快速识别并毫不犹豫地求解是很重要的。

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4. Two-Step Equations | 两步方程

ision. Subtract 3 from both sides to get 2x = 8, then divide both sides by 2 to get x = 4.

两步方程包含两次运算，例如 2x + 3 = 11。首先消去加法或减法，然后消去乘法或除法。两边减 3 得 2x = 8，然后两边除以 2 得 x = 4。

Order matters: always remove the constant term that is added or subtracted before removing the coefficient that is multiplied or divided. This follows the reverse of the order of operations.

顺序很重要：先消去加或减的常数项，再消去乘或除的系数。这遵循运算顺序的逆过程。

Solve 5y − 7 = 23

Add 7: 5y = 30

Divide by 5: y = 6

解 5y − 7 = 23：两边加 7 得 5y = 30，两边除以 5 得 y = 6

= c, subtract b from both sides first, then divide by a. The solution is x = (c − b) ÷ a.

对于形如 ax + b = c 的方程，先两边减去 b，再除以 a。解为 x = (c − b) ÷ a。

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5. Equations with Brackets | 含括号的方程

+ 12 = 27. Then subtract 12 from both sides to get 3x = 15, and divide by 3 to get x = 5.

当方程含有括号时，先用分配律展开。例如，3(x + 4) = 27 展开为 3x + 12 = 27。然后两边减 12 得 3x = 15，再除以 3 得 x = 5。

• 2(a − 5) = 12 → 2a − 10 = 12 → 2a = 22 → a = 11
• −(b + 3) = −7 → −b − 3 = −7 → −b = −4 → b = 4

Be careful with negative signs outside a bracket. A negative sign means multiply every term inside the bracket by −1, so −(b + 3) becomes −b − 3, not −b + 3.

注意括号外的负号。负号表示将括号内每一项乘以 −1，所以 −(b + 3) 变成 −b − 3，而不是 −b + 3。

Expanding brackets is an essential skill because many Cambridge KS3 questions combine brackets with other operations. After expanding, you can use the same two-step method to finish solving.

展开括号是一项基本技能，因为许多剑桥 KS3 题目会把括号与其他运算结合起来。展开后，你可以继续用同样的两步法求解。

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6. Equations with Variables on Both Sides | 两边都含未知数的方程

x + 10. Subtract 3x from both sides to get 2x + 2 = 10. Then subtract 2: 2x = 8, so x = 4.

如果未知数出现在两边，先合并同类项，使未知数只出现在一边。例如，5x + 2 = 3x + 10。两边减 3x 得 2x + 2 = 10；再减 2 得 2x = 8，因此 x = 4。

You can choose to eliminate the smaller variable term or the larger one. Usually it is easier to keep the variable term positive, so subtract the smaller variable coefficient from both sides.

你可以选择消去较小或较大的变量项。通常保持变量项为正更容易，因此从两边减去较小的变量系数。

Solve 7 − 2m = m + 13

Add 2m: 7 = 3m + 13

Subtract 13: −6 = 3m

Divide by 3: m = −2

解 7 − 2m = m + 13：两边加 2m 得 7 = 3m + 13；两边减 13 得 −6 = 3m；两边除以 3 得 m = −2

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