📚 IGCSE Maths Series 300: Solving Linear Equations and Inequalities | IGCSE 数学系列 300：解一次方程与不等式
Linear equations and inequalities form the backbone of IGCSE algebra. In this series, you will learn how to solve equations systematically, handle brackets and fractions, move terms across the equals sign, and apply the same logic to inequalities. Mastery of these techniques is essential for higher-level topics such as simultaneous equations, graphs, and functions.
一次方程与不等式是 IGCSE 代数的基础。在本系列中，你将学习如何系统地解方程，处理括号和分数，将项移到等号另一边，并将同样的逻辑应用于不等式。掌握这些技巧对于联立方程、图像和函数等更高层次的主题至关重要。
1. Introduction to Linear Equations | 一次方程入门
A linear equation is an equation in which the highest power of the variable is 1. For example, 3x + 5 = 14 is linear, while x² + 2 = 6 is not because the variable x is raised to the power 2.
一次方程是指未知数的最高次数为 1 的方程。例如，3x + 5 = 14 是一次方程，而 x² + 2 = 6 不是，因为未知数 x 的次数为 2。
In IGCSE Mathematics, you are expected to solve linear equations fluently by using inverse operations and to check your answer by substituting it back into the original equation.
在 IGCSE 数学中，要求你能够熟练运用逆运算解一次方程，并通过将答案代入原方程进行检验。
An equation has two sides separated by an equals sign. The equals sign tells us that the two expressions have the same value, so any operation must preserve this equality.
方程由等号分成两边。等号表示两个表达式的值相同，因此任何运算都必须保持这种相等关系。
2. Balancing Method: The Golden Rule | 平衡法：黄金法则
The balancing method means that whatever you do to one side of an equation, you must do to the other side. This keeps the equation balanced, just like a pair of scales.
平衡法的含义是，对方程一边进行任何操作，必须对另一边进行相同操作，从而使方程保持平衡，就像一架天平一样。
x + 7 = 15 → x + 7 − 7 = 15 − 7 → x = 8
To undo the addition of 7, we subtract 7 from both sides. This isolates the variable x on the left-hand side.
为了消去加 7，我们在两边同时减去 7。这样就将未知数 x 单独留在左边。
Similarly, to solve 4x = 20, divide both sides by 4 because division is the inverse of multiplication.
类似地，解方程 4x = 20 时，两边同时除以 4，因为除法是乘法的逆运算。
4x = 20 → x = 20 ÷ 4 → x = 5
Always perform the same operation in the same step to avoid unbalanced work. The operation should be written on both sides clearly.
始终在同一步骤中对两边进行相同运算，以避免不平衡。运算应清晰地写在两边。
3. Equations with Brackets | 含括号的方程
When an equation contains brackets, the first step is usually to expand them using the distributive law. For example, 2(x + 3) means 2 multiplied by every term inside the bracket.
当方程含有括号时，第一步通常是用分配律展开括号。例如，2(x + 3) 表示 2 乘以括号内的每一项。
Expand first, then simplify like terms, and finally use the balancing method to solve for the unknown.
先展开，再合并同类项，最后使用平衡法解出未知数。
2(x + 3) = 14 → 2x + 6 = 14 → 2x = 8 → x = 4
Be careful with negative signs outside brackets. For instance, −(x − 5) is the same as −1(x − 5), which expands to −x + 5.
注意括号外是负号的情况。例如，−(x − 5) 相当于 −1(x − 5)，展开后得到 −x + 5。
After expanding brackets, always simplify both sides before moving terms. This reduces the chance of sign errors.
展开括号后，始终先化简两边，再移项。这样可以减少符号错误。
4. Equations with Fractions | 含分数的方程
Equations that contain fractions can be solved more easily by clearing the denominators. Multiply every term on both sides by the lowest common denominator of all the fractions.
含有分数的方程可以通过去分母来简化求解。将两边每一项都乘以所有分数的最小公分母。
For example, in the equation (x + 2)/3 = 5, multiply both sides by 3 to eliminate the denominator first.
例如，在方程 (x + 2)/3 = 5 中，先将两边乘以 3，以消去分母。
(x + 2)/3 = 5 → x + 2 = 15 → x = 13
If there are two fractions, such as x/4 + 1 = x/2, multiply every term by the lowest common denominator, which is 4 in this case.
如果有两个分数，例如 x/4 + 1 = x/2，将每一项乘以最小公分母，这里是 4。
x/4 + 1 = x/2 → x + 4 = 2x → 4 = x
Always put brackets around numerators if the numerator contains more than one term. This prevents mistakes when multiplying through.
当分子含有多项时，务必将分子用括号括起来。这样可以防止通分时出错。
5. Variables on Both Sides | 两边含未知数的方程
Sometimes the variable appears on both sides of the equation. The strategy is to collect all variable terms on one side and all constant terms on the other side.
有时未知数出现在方程的两边。解题策略是将所有含未知数的项移到一边，将所有常数项移到另一边。
Use inverse operations to move terms across the equals sign. Remember that when a term changes side, its sign changes.
利用逆运算将项移过等号。记住，项移到另一边时，符号要改变。
5x + 2 = 2x + 14 → 5x − 2x = 14 − 2 → 3x = 12 → x = 4
It is often easier to collect variables on the left and constants on the right, but you may choose the opposite if it gives fewer negative coefficients.
通常将未知数项移到左边、常数项移到右边较简便，但如果另一方向可以避免负系数，也可以选择相反方向。
After moving terms, simplify both sides before carrying out the final division.
移项后，先化简两边，再进行最后的除法运算。
6. Forming Equations from Word Problems | 由文字题建立方程
Many IGCSE questions ask you to form and solve an equation from a written situation. Start by defining a variable with a clear statement, such as “Let x be the unknown number.”
许多 IGCSE 题目要求你根据文字描述建立并解方程。首先清晰地设定变量，例如“设 x 为未知数”。
Translate key phrases into algebraic expressions: “three more than a number” becomes x + 3, “twice a number” becomes 2x, and “the result is” becomes an equals sign.
将关键词组翻译成代数表达式：“比一个数大 3”为 x + 3，“一个数的两倍”为 2x，“结果是”为等号。
Example: When 4 is added to three times a number, the result is 19. The equation is 3x + 4 = 19.
例子：一个数的三倍加上 4 等于 19。方程是 3x + 4 = 19。
Solving this gives x = 5, and you should check that 3(5) + 4 = 19 is true before writing the final answer.
解得 x = 5，在写出最终答案前应检验 3(5) + 4 = 19 是否正确。
For age, perimeter, or money problems, define the smallest unknown carefully and express all other quantities in terms of that variable.
对于年龄、周长或金钱问题，要谨慎定义最小的未知量，并用该变量表示所有其他量。
7. Introduction to Linear Inequalities | 一次不等式入门
A linear inequality is similar to a linear equation, but the equals sign is replaced by one of four inequality signs: less than, greater than, less than or equal to, and greater than or equal to.
一次不等式与一次方程相似，但等号被四种不等号之一代替：小于、大于、小于等于和大于等于。
< , > , ≤ , ≥
In IGCSE, the symbols ≤ and ≥ mean the boundary value is included, whereas < and > mean the boundary value is not included.
在 IGCSE 中，符号 ≤ 和 ≥ 表示边界值包含在内，而 < 和 > 表示边界值不包含在内。
The solution to an inequality is usually a range of values, not a single number. For example, x > 3 means any number larger than 3 works.
不等式的解通常是一个取值范围，而不是一个数。例如，x > 3 表示任何大于 3 的数都成立。
Treat an inequality like an equation when manipulating, but remember one critical exception involving multiplication or division by a negative number.
操作不等式时可以像方程一样处理，但有一个关键例外：乘以或除以负数时需要注意。
8. Solving Linear Inequalities | 解一次不等式
To solve a linear inequality, use the same inverse operations as for equations. However, if you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign.
解一次不等式时，使用与方程相同的逆运算。但是，如果将两边乘以或除以一个负数，必须反转不等号的方向。
−2x < 8 → x > 8 ÷ (−2) → x > −4
In the example above, dividing both sides by −2 changes the < sign into a > sign. Forgetting this is one of the most common errors in IGCSE algebra.
在上面的例子中，两边除以 −2 后，< 号变成 > 号。忘记这一点是 IGCSE 代数中最常见的错误之一。
When the unknown appears on both sides of an inequality, collect variable terms on one side first, then isolate the variable as usual.
当未知数出现在不等式两边时，先将含未知数的项移到一边，然后照常解出未知数。
3x + 1 ≤ 7x − 11 → 1 + 11 ≤ 7x − 3x → 12 ≤ 4x → x ≥ 3
Always check your final inequality by choosing a value from the solution set and testing it in the original inequality.
始终从解集中选取一个值代入原不等式进行检验，以确认最终不等式正确。
9. Representing Inequalities on a Number Line | 在数轴上表示不等式
Solutions to inequalities are often shown on a number line. Use an open circle for strict inequalities such as x > 2 or x < 5, meaning the endpoint is not included.
不等式的解常用数轴表示。对于严格不等式，如 x > 2 或 x < 5，使用空心圆圈，表示端点不包含在内。
Use a closed or filled circle for inclusive inequalities such as x ≥ 2 or x ≤ 5, meaning the endpoint is included.
对于包含边界的不等式，如 x ≥ 2 或 x ≤ 5，使用实心圆点，表示端点包含在内。
The arrow on the number line should point in the direction of all possible values. For x > 2, draw an open circle at 2 and shade to the right.
数轴上的箭头应指向所有可能取值的方向。对于 x > 2，在 2 处画空心圆，并向右画阴影。
When representing a range such as −1 ≤ x < 4, draw a filled circle at −1, an open circle at 4, and shade the line between them.
表示范围如 −1 ≤ x < 4 时，在 −1 处画实心圆，在 4 处画空心圆，并给两者之间的线段涂上阴影。
Number line diagrams are frequently tested in IGCSE because they show whether you understand the difference between inclusive and strict inequalities.
数轴图在 IGCSE 中经常考查，因为它能显示你是否理解包含与不包含边界的区别。
10. Compound Inequalities | 复合不等式
A compound inequality combines two inequalities into one statement. For example, 1 < x ≤ 5 means x is greater than 1 and less than or equal to 5.
复合不等式将两个不等式合并为一个表述。例如，1 < x ≤ 5 表示 x 大于 1 且小于等于 5。
You can solve a compound inequality by performing the same operation on all three parts simultaneously.
解复合不等式时，可以对三个部分同时进行相同运算。
1 < 2x + 1 ≤ 5 → 0 < 2x ≤ 4 → 0 < x ≤ 2
Remember to reverse both inequality signs if you multiply or divide all three parts by a negative number.
如果将所有三个部分同时乘以或除以负数，请记住反转两个不等号的方向。
Compound inequalities are useful for setting bounds on quantities such as length, temperature, or cost, and they appear regularly in IGCSE word problems.
复合不等式常用于为长度、温度或成本等量设定边界，并且经常出现在 IGCSE 文字题中。
11. Common Misconceptions and Exam Tips | 常见误区与应试技巧
One major misconception is treating the inequality sign as an equals sign without considering the direction change when dividing by a negative number. Always write the sign reversal step explicitly.
一个主要误区是将不等号当作等号来用，却没有考虑除以负数时方向的改变。务必明确写出反转符号的步骤。
Another common error is applying an operation to only some terms on a side. When clearing fractions or expanding brackets, every term must be considered.
另一个常见错误是只对一边的部分项进行运算。在去分母或展开括号时，每一项都必须考虑。
In word problems, many students forget to define the variable before forming an equation. Examiners award method marks for a clear variable definition and a correct initial equation.
在文字题中，许多学生忘记先定义变量再建立方程。考官会为清晰的变量定义和正确的初始方程给予方法分。
When checking an answer, substitute it back into the original equation or inequality. This is the fastest way to detect arithmetic or sign errors before moving to the next question.
检查答案时，将其代入原方程或不等式。这是进入下一题之前发现算术或符号错误的最快方法。
Always present final answers in the form requested by the question, such as x = 4 or x > −3, and include units if the question involves real-life measurements.
始终按照题目要求的形式写出最终答案，例如 x = 4 或 x > −3，如果题目涉及实际测量，还要注明单位。
12. Practice Questions and Summary | 练习题与总结
Try solving the following equation step by step: 5(x − 2) = 3x + 4. First expand the bracket, then collect like terms, and finally divide to find x.
尝试逐步解下列方程：5(x − 2) = 3x + 4。首先展开括号，然后合并同类项，最后除以系数求出 x。
5(x − 2) = 3x + 4 → 5x − 10 = 3x + 4 → 2x = 14 → x = 7
Now solve the inequality −3x + 5 < 20 and represent the solution on a number line. Remember to reverse the sign when dividing by −3.
现在解不等式 −3x + 5 < 20，并在数轴上表示解集。记住在除以 −3 时要反转不等号。
−3x + 5 < 20 → −3x < 15 → x > 15 ÷ (−3) → x > −5
For a compound inequality, solve 2 ≤ 4x − 6 < 14 and give the final range in its simplest form.
对于复合不等式，解 2 ≤ 4x − 6 < 14，并给出化简后的最终范围。
2 ≤ 4x − 6 < 14 → 8 ≤ 4x < 20 → 2 ≤ x < 5
To master this topic, practise forming equations from written situations and checking every solution. Linear equations and inequalities underpin much of IGCSE algebra, so confidence here will help you across the syllabus.
要掌握本主题，请多练习从文字描述中建立方程并检验每一个解。一次方程与不等式是 IGCSE 代数许多内容的基础，因此熟练掌握这部分将有助于你应对整个课程。
In this series, we covered solving equations using the balancing method, expanding brackets, clearing fractions, handling variables on both sides, and applying the same rules to inequalities. Review these steps before attempting past paper questions.
在本系列中，我们学习了用平衡法解方程、展开括号、去分母、处理两边含未知数的情况，以及将同样的规则应用于不等式。在做真题前，请复习这些步骤。
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