📚 Mastering Unit 2 Algebra: Expressions, Equations and Inequalities | 掌握第二单元代数：表达式、方程与不等式
Unit 2 in many Year 9 courses focuses on algebraic techniques: understanding terms, simplifying expressions, expanding and factorising, solving linear equations and inequalities, and using formulae. This article revises the key skills step by step, with examples written in simple notation. Use it alongside your homework pack to identify weak points and correct mistakes before the test.
许多九年级课程的第二单元都集中于代数技巧：理解项、化简表达式、展开与因式分解、解一元一次方程和不等式以及使用公式。本文逐步复习关键技能，并用简单符号给出例子。请配合你的作业包使用，找出薄弱点并在考试前纠正错误。
1. Algebraic Expressions and Terms | 代数表达式与项
In algebra, a term is a single number, a single letter, or a product of numbers and letters. For example, 4x, -3y², and 7 are all terms. A coefficient is the number part of a term, so in 4x the coefficient is 4.
在代数中，项是单个数字、单个字母或数字与字母的乘积。例如，4x、-3y² 和 7 都是项。系数是项中的数字部分，因此在 4x 中系数是 4。
An expression is made by combining terms using addition or subtraction, such as 4x - 3y² + 7. Expressions do not contain an equals sign. A variable is a letter that can take different values, while a constant is a term with a fixed value.
表达式是由项通过加法或减法组合而成的，例如 4x - 3y² + 7。表达式不包含等号。变量是可以取不同值的字母，而常数是数值固定的项。
Term: 4x, -3y², 7 | Expression: 4x - 3y² + 7
2. Simplifying Expressions by Collecting Like Terms | 合并同类项化简表达式
Like terms have exactly the same variable parts and powers. For example, 5x and -2x are like terms, but 5x and 3x² are not. To simplify, add or subtract the coefficients of like terms and keep the variable part unchanged.
同类项具有完全相同的变量部分和幂次。例如，5x 和 -2x 是同类项，但 5x 和 3x² 不是。化简时，将同类项的系数相加或相减，变量部分保持不变。
5x + 3y - 2x + 4y = (5 - 2)x + (3 + 4)y = 3x + 7y
When collecting like terms, rewrite the expression so that the variable terms are grouped together, then perform the arithmetic on the coefficients only. This reduces the chance of changing powers by mistake.
合并同类项时，重新排列表达式，使变量项分组在一起，然后只对系数进行算术运算。这样可以减少误改幂次的可能。
3. Expanding Brackets | 去括号展开
ab + ac. Always apply the sign in front of the bracket. For example, 3(2x - 5) = 6x - 15.
去括号展开是指用括号外的项乘以括号内的每一项。基本规则是 a(b + c) = ab + ac。始终应用括号前面的符号。例如，3(2x - 5) = 6x - 15。
iply every term in the first bracket by every term in the second, then collect like terms.
如果括号前是负号，它会改变括号内所有项的符号：-(x - 3) = -x + 3。对于双括号，将第一个括号中的每一项乘以第二个括号中的每一项，再合并同类项。
(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6
4. Factorising Simple Expressions | 简单因式分解
highest common factor (HCF) of all terms. For 6x + 9, the HCF is 3, so 6x + 9 = 3(2x + 3).
因式分解是展开的逆过程：你将表达式重写为因式的乘积。首先找出所有项的最高公因式（HCF）。对于 6x + 9，HCF 是 3，所以 6x + 9 = 3(2x + 3)。
For x² + 5x, the common factor is x, giving x(x + 5). Always check your answer by expanding the brackets to see whether you return to the original expression.
对于 x² + 5x，公因式是 x，得到 x(x + 5)。始终通过展开括号来检查答案，看是否回到原表达式。
6x + 9 = 3(2x + 3) | x² + 5x = x(x + 5)
5. Solving Linear Equations: The Basics | 解一元一次方程：基础
s to keep the equation balanced: whatever you do to one side, you must do to the other side.
一元一次方程是未知数只以一次幂出现的方程，例如 2x + 3 = 11。黄金法则是保持方程平衡：你对一边做的任何运算，也必须对另一边做。
des by 2 to get x = 4. Always check by substituting x = 4 back into the original equation.
求解时，按相反顺序逆运算。对于 2x + 3 = 11，先从两边减去 3，得到 2x = 8，然后两边除以 2，得到 x = 4。始终通过将 x = 4 代回原方程来检验。
2x + 3 = 11 → 2x = 8 → x = 4
6. Solving Equations with Unknowns on Both Sides | 解两边含未知数的方程
, subtract 3x from both sides to get 2x + 2 = 10, then subtract 2 to get 2x = 8, so x = 4.
当方程两边都有未知数时，将变量项移到一边，常数移到另一边。对于 5x + 2 = 3x + 10，从两边减去 3x，得到 2x + 2 = 10，再减去 2，得到 2x = 8，因此 x = 4。
A common choice is to move the smaller variable term so the coefficient of x is positive. You can also move constants first; the solution is the same as long as every operation is applied to both sides.
常见的选择是移动较小的变量项，使 x 的系数为正。你也可以先移动常数；只要每一步都对两边进行相同运算，解就是相同的。
5x + 2 = 3x + 10 → 2x + 2 = 10 → 2x = 8 → x = 4
7. Setting Up Equations from Word Problems | 根据文字题建立方程
Word problems require you to translate sentences into algebraic equations. First define the unknown with a letter, for example let n be the number. Then convert phrases: "three more than twice a number" becomes 2n + 3.
文字题要求你把句子翻译成代数方程。首先用字母定义未知数，例如设 n 为这个数。然后转换短语：“一个数的两倍再加 3”变成 2n + 3。
2 to find n = 6. Always answer in the context of the question, including units if needed.
如果题目说结果是 15，则方程为 2n + 3 = 15。通过减 3 再除以 2 求解，得到 n = 6。始终在题目情境下作答，如需要则包含单位。
