Algebraic Expressions and Simple Equations | 代数表达式与简单方程

📚 Algebraic Expressions and Simple Equations | 代数表达式与简单方程

Algebra is a fundamental part of mathematics, and solving equations is a key skill at KS3 level. This article revises simple linear equations, including one-step, two-step, and equations with unknowns on both sides, aligned with the Cambridge Lower Secondary curriculum. Practice the methods and avoid common pitfalls to build confidence.

代数是数学的基础，解方程是KS3阶段的关键技能。本文复习简单线性方程，包括一步方程、两步方程以及未知数在两边的情况，与剑桥初中课程一致。通过练习各种方法和避免常见错误，你将建立起解题的信心。

1. Understanding Equations | 理解等式

An equation is a mathematical statement that shows two expressions are equal, using the equals sign ‘=’.

等式是使用等号 “=” 表示两个表达式相等的数学语句。

For example, 2x + 3 = 7 is an equation where the unknown x makes the left side equal to the right side.

例如，2x + 3 = 7 是一个方程，其中未知数 x 使左边等于右边。

The goal is to find the value of x that makes the equation true; this value is called the solution.

目标是求出使等式成立的 x 值，这个值叫作方程的解。

An equation behaves like a balanced scale: whatever you do to one side, you must do to the other to keep it balanced.

方程就像一个平衡的天平：你对一边做什么操作，就必须对另一边做同样的操作，以保持平衡。

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2. Solving One-step Equations | 解一步方程

For one-step equations, use the inverse operation to isolate the unknown.

对于一步方程，使用逆运算来分离未知数。

If x + 5 = 12, subtract 5 from both sides:

如果 x + 5 = 12，两边同时减去5：

x + 5 − 5 = 12 − 5 → x = 7

If x − 3 = 9, add 3 to both sides:

如果 x − 3 = 9，两边同时加上3：

x − 3 + 3 = 9 + 3 → x = 12

If 4x = 20, divide both sides by 4:

如果 4x = 20，两边除以4：

4x ÷ 4 = 20 ÷ 4 → x = 5

If x/3 = 6, multiply both sides by 3:

如果 x/3 = 6，两边乘以3：

(x/3) × 3 = 6 × 3 → x = 18

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3. Solving Two-step Equations | 解两步方程

For two-step equations like 2x + 3 = 11, first undo the addition or subtraction, then the multiplication or division.

对于两步方程如 2x + 3 = 11，先处理加减法，再处理乘除法。

Subtract 3 from both sides: 2x = 8, then divide by 2: x = 4.

两边减3：2x = 8，然后除以2：x = 4。

2x + 3 − 3 = 11 − 3 → 2x = 8 → x = 4

Always perform inverse operations in the reverse order to the order of operations (BIDMAS/BODMAS).

始终按照运算顺序（BIDMAS/BODMAS）的逆序执行逆运算。

Example: to solve 10 − 2y = 4, first subtract 10 from both sides: −2y = −6, then divide by −2: y = 3.

例如：解 10 − 2y = 4，先从两边减去10：−2y = −6，然后除以 −2：y = 3。

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4. Equations with Brackets | 带括号的方程

If an equation contains brackets, expand them first using the distributive law.

如果方程含有括号，先用分配律展开。

Example: 3(x + 2) = 15 → multiply 3 by both terms inside the bracket: 3x + 6 = 15.

例如：3(x + 2) = 15 → 用3乘以括号内的每一项：3x + 6 = 15。

Then solve: subtract 6: 3x = 9, divide by 3: x = 3.

然后求解：减去6：3x = 9，除以3：x = 3。

3x + 6 = 15 → 3x = 9 → x = 3

Remember to multiply every term inside the bracket by the factor outside, including when the factor is negative: −2(x − 4) = −2x + 8.

记住括号内的每一项都要乘以外面的因数，包括因数为负时：−2(x − 4) = −2x + 8。

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5. Unknown on Both Sides | 未知数在两边

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

当未知数出现在方程的两边时，将所有含未知数的项移到一边，常数项移到另一边。

Solve 5x − 3 = 2x + 6.

解 5x − 3 = 2x + 6。

Subtract 2x from both sides: 3x − 3 = 6.

两边减去2x：3x − 3 = 6。

Add 3: 3x = 9 → x = 3.

加3：3x = 9 → x = 3。

5x − 3 − 2x = 2x + 6 − 2x → 3x − 3 = 6 → 3x = 9 → x = 3

Always aim to have a positive coefficient of x at the end; if you end with −x = 4, multiply both sides by −1 to get x = −4.

最后尽量让 x 的系数为正；如果得到 −x = 4，两边乘以 −1 得 x = −4。

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6. Checking Your Solution | 检验解

Always substitute your answer back into the original equation to verify it is correct.

始终将答案代入原方程以验证其正确性。

For x = 4 in 2x + 3 = 11: left side = 2×4 + 3 = 8 + 3 = 11, which equals the right side.

对于 2x + 3 = 11 的解 x = 4：左边 = 2×4 + 3 = 8 + 3 = 11，与右边一致。

For the equation 5x − 3 = 2x + 6 with x = 3: left = 5×3 − 3 = 12, right = 2×3 + 6 = 12, so the solution is correct.

对于方程 5x − 3 = 2x + 6，x = 3：左边 = 5×3 − 3 = 12，右边 = 2×3 + 6 = 12，解正确。

This habit catches mistakes and builds confidence in your algebraic skills.

这个习惯可以发现错误并增强你对代数技能的信心。

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7. Forming Equations from Word Problems | 从应用题列方程

Read the problem carefully, identify the unknown quantity, and write an equation using a letter to represent it.

仔细读题，确定未知量，并用一个字母代表它来列出方程。

“The sum of a number and 5 is 12” → let the number be x: x + 5 = 12 → x = 7.

“一个数与5的和是12” → 设这个数为 x：x + 5 = 12 → x = 7。

“Three times a number minus 4 equals 11” → 3x − 4 = 11 → 3x = 15 → x = 5.

“一个数的三倍减4等于11” → 3x − 4 = 11 → 3x = 15 → x = 5。

“The perimeter of a rectangle is 30 cm. Its length is twice its width. Find the width.” Let width = w, then length = 2w, perimeter: 2(w + 2w) = 30 → 6w = 30 → w = 5 cm.

“一个长方形的周长是30厘米。它的长是宽的两倍。求宽。” 设宽为 w，则长为 2w，周长：2(w + 2w) = 30 → 6w = 30 → w = 5 厘米。

Translate words into algebraic expressions step by step, and always define your variable clearly.

逐步将文字转换为代数表达式，并清晰地定义你的变量。

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8. Common Mistakes to Avoid | 常见错误

Mistake 1: Forgetting to perform the same operation on both sides of the equation.

错误1：忘记在等号两边执行相同的运算。

Correct approach: x + 3 = 10 → x + 3 − 3 = 10 − 3. Never just subtract 3 from one side.

正确做法：x + 3 = 10 → x + 3 − 3 = 10 − 3。绝不要只从一边减去3。

Mistake 2: Mishandling negative signs. Example: solving −x = 5 incorrectly as x = 5; actually multiply by −1 to get x = −5.

错误2：错误处理负号。例如将 −x = 5 错解为 x = 5；实际上应乘以 −1 得 x = −5。

Mistake 3: Not expanding brackets correctly. Writing 2(x + 3) as 2x + 3 instead of 2x + 6.

错误3：括号展开不正确。将 2(x + 3) 写成 2x + 3 而非 2x + 6。

Mistake 4: Combining unlike terms, e.g. trying to add 3x and 4 as 7x, which is not valid.

错误4：合并不同类项，例如试图将 3x 和 4 相加为 7x，这是无效的。

Avoid rushing; write each step clearly and double-check your work.

避免着急；每一步都写清楚，并仔细检查。

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9. Practice Questions | 练习题

Try solving these equations and check your answers.

试解以下方程并检查答案。

(a) y + 7

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