📚 Solving Equations and Inequalities | 解方程与不等式
Linear equations and inequalities are core topics in Year 9 mathematics. This unit develops the algebra skills needed for IGCSE and Checkpoint examinations, including balancing, expanding brackets, rearranging formulae and representing solution sets. Mastering these techniques now will make advanced topics such as simultaneous equations and quadratic inequalities much easier later.
一元一次方程和不等式是九年级数学的核心主题。本单元培养 IGCSE 和 Checkpoint 考试所需的代数技能，包括等式平衡、展开括号、移项变形以及表示解集。现在掌握这些技巧，将使以后学习联立方程和二次不等式等内容变得容易得多。
1. What Are Linear Equations? | 什么是一元一次方程？
is a linear equation because x appears only to the first power. The goal is to find the value of x that makes the statement true. This value is called the solution or root of the equation.
一元一次方程是指未知数的最高次数为 1 的代数等式。例如，2x + 5 = 17 就是一元一次方程，因为 x 只出现一次方。我们的目标是求出使等式成立的 x 值，这个值叫做方程的解或根。
of BODMAS until x is isolated. This gives x = 5, which should be checked by substitution.
考虑方程 3x − 2 = 13。等号左边和右边相等。要解这个方程，我们需要按照 BODMAS 的逆序进行逆运算，直到把 x 单独留在一边。最后得到 x = 5，并通过代入原式检验。
3x − 2 = 13 → 3x = 15 → x = 5
2. The Balancing Method | 等式平衡法
he equation true. For example, if x + 6 = 20, subtract 6 from both sides to obtain x = 14.
平衡法把方程看作一架天平。你对等号一边做的任何运算，必须同时施加到另一边。这一规则能保持等式成立。例如，如果 x + 6 = 20，两边同时减去 6，就得到 x = 14。
Always write each new line underneath the previous one and keep the equals signs aligned.
当涉及乘法或除法时，使用逆运算。对于 5x = 35，两边同时除以 5。如果 x/4 = 9，两边同时乘以 4。书写时，每一行放在上一行下方，并保持等号对齐。
5x = 35 → x = 35 ÷ 5 = 7
3. Equations with Brackets | 含有括号的方程
mes 8x − 12 = 28. Then add 12 to both sides to get 8x = 40, and divide by 8 to find x = 5.
要解含有括号的方程，首先用乘法分配律展开括号。例如，4(2x − 3) = 28 可化为 8x − 12 = 28。然后两边加 12，得到 8x = 40，再除以 8，求得 x = 5。
x + 10 = 3x − 3. Rearranging gives x = 13. Checking by substitution verifies the solution.
有时等号两边都有括号。先分别展开两边，再合并同类项。例如，2(x + 5) = 3(x − 1) 展开为 2x + 10 = 3x − 3。移项后得到 x = 13。代入原式可检验解的正确性。
2(x + 5) = 3(x − 1) → 2x + 10 = 3x − 3 → x = 13
4. Variables on Both Sides | 两边都含有未知数的方程
act 3x from both sides: 4x + 4 = 20. Then subtract 4: 4x = 16. Finally divide by 4: x = 4.
当未知数出现在等号两边时，用平衡法把所有含 x 的项移到一边，所有常数项移到另一边。例如，7x + 4 = 3x + 20，先两边减去 3x，得到 4x + 4 = 20。再两边减 4，得到 4x = 16。最后除以 4，x = 4。
A common error is to move a term without changing its sign. Remember: when a term crosses the equals sign, its operation reverses. Addition becomes subtraction, and multiplication becomes division. Practise writing each step clearly to avoid sign errors.
一个常见错误是移项时忘记变号。记住：当一项越过等号时，它的运算要变为逆运算。加法变减法，乘法变除法。请练习把每一步写清楚，以避免符号错误。
5. Introduction to Inequalities | 不等式导论
or equal to). Solving an inequality means finding all values of x that make the statement true. The solution is usually a range of values rather than a single number.
不等式使用以下符号之一比较两个表达式：<（小于）、≤（小于或等于）、>（大于）和 ≥（大于或等于）。解不等式就是找出所有能使不等式成立的 x 值。解通常是一个范围，而不是单一数值。
ms. On a number line, < and > use an open circle, while ≤ and ≥ use a closed circle.
例如，x > 3 表示任何大于 3 的数都是解，如 3.1、4、10 和 100。然而，x ≥ 3 包括 3 本身。这个区别在考试中很重要。在数轴上，< 和 > 用空心圆圈，而 ≤ 和 ≥ 用实心圆圈。
6. Solving Linear Inequalities | 解一元一次不等式
Solving a linear inequality follows the same steps as solving an equation, with one crucial exception: if you multiply or divide both sides by a negative number, you must reverse the direction of the inequality symbol. For example, −2x < 8 becomes x > −4 after dividing by −2 and reversing the sign.
解一元一次不等式的步骤与解方程相同，但有一个关键例外：如果两边同时乘或除以一个负数，必须反转不等号的方向。例如，−2x < 8 除以 −2 并反转符号后，得到 x > −4。
a value within the range, such as x = 2, and substitute: 3(2) − 5 = 1 ≤ 10, which is true.
考虑不等式 3x − 5 ≤ 10。两边加 5，得到 3x ≤ 15。两边除以 3，得到 x ≤ 5。解集包含所有小于或等于 5 的实数。要检验，可以在范围内选一个值，例如 x = 2，代入：3(2) − 5 = 1 ≤ 10，成立。
- Use inverse operations in the correct order.
- Reverse the sign only when multiplying or dividing by a negative value.
- Always express the final answer with x on the left if required.
- 按正确顺序使用逆运算。
- 仅在乘或除以负数时反转符号。
- 如需要，将最终答案写成 x 在左边。
7. Representing Inequalities on
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