📚 IGCSE Mathematics: Linear Equations, Simultaneous Equations and Inequalities | IGCSE 数学：线性方程、联立方程与不等式
Linear equations, simultaneous equations and inequalities form a central part of the IGCSE Mathematics algebra syllabus. These topics appear in both Core and Extended papers, often as structured questions in Paper 2 and Paper 4. Mastering the basic techniques of balancing equations, eliminating variables and reversing inequality signs under negative multiplication will give you a strong foundation for harder topics such as quadratic functions, linear programming and coordinate geometry.
线性方程、联立方程和不等式是 IGCSE 数学代数部分的核心内容。这些主题在核心和扩展试卷中都会出现，通常以结构化题目的形式出现在 Paper 2 和 Paper 4 中。掌握方程两边平衡、消元以及不等式乘除负数时改变不等号方向等基本技巧，将为你学习二次函数、线性规划和坐标几何等更难的专题打下坚实基础。
1. What Are Linear Equations? | 什么是线性方程？
A linear equation is an equation in which the highest power of the unknown variable is 1. For example, 2x + 5 = 17 is linear because x appears to the power 1. The general form is ax + b = c, where a, b and c are constants and a ≠ 0.
线性方程是指未知数的最高次数为 1 的方程。例如 2x + 5 = 17 就是线性方程，因为 x 的指数为 1。一般形式为 ax + b = c，其中 a、b、c 为常数，且 a ≠ 0。
The solution of a linear equation is the value of the variable that makes both sides equal. You can check your solution by substituting it back into the original equation.
线性方程的解就是使等号两边相等的变量值。你可以将解代回原方程来检验答案。
It is important to recognise the difference between linear and non-linear expressions. Any term containing x², x³, √x or 1 ÷ x makes the equation non-linear, which means quadratic methods or other techniques will be needed instead.
区分线性与非线性表达式很重要。任何含有 x²、x³、√x 或 1 ÷ x 的项都会使方程变成非线性方程，这意味着需要改用二次方程方法或其他技巧来求解。
| Equation | 方程 | Linear? | 是否线性 | Reason | 原因 |
| 3x − 7 = 8 | Yes | 是 | x has power 1 | x 的次数为 1 |
| 5(x + 2) = 30 | Yes | 是 | Expands to 5x + 10 = 30 | 展开后为 5x + 10 = 30 |
| x² + 2x = 8 | No | 否 | Contains x² | 含有 x² |
| 2 ÷ x = 6 | No | 否 | x in the denominator | x 在分母中 |
2. Solving One-Step and Two-Step Equations | 解一步与两步方程
To solve an equation, you must keep the equation balanced by performing the same operation on both sides. For a one-step equation such as x + 7 = 12, subtract 7 from both sides to obtain x = 5.
解方程时，你必须对等号两边同时进行相同的运算以保持平衡。对于一步方程，如 x + 7 = 12，两边同时减去 7，得到 x = 5。
For a two-step equation such as 2x + 5 = 17, first subtract 5 from both sides, then divide both sides by 2:
对于两步方程，如 2x + 5 = 17，先两边同时减 5，再两边同时除以 2：
2x + 5 − 5 = 17 − 5 → 2x = 12 → x = 6
Always use inverse operations: addition undoes subtraction, multiplication undoes division, and vice versa.
请始终使用逆运算：加法抵消减法，乘法抵消除法，反之亦然。
- Addition ↔ Subtraction | 加法 ↔ 减法
x + 9 = 15 → x = 15 − 9 = 6 - Multiplication ↔ Division | 乘法 ↔ 除法
4x = 28 → x = 28 ÷ 4 = 7 - Order of operations reversed | 运算顺序要倒过来
For 3x − 4 = 11, undo the − 4 first, then undo the × 3.
对于 3x − 4 = 11，先撤销 − 4，再撤销 × 3。
Many students lose marks by doing the operations in the wrong order. Remember the rule: undo the addition or subtraction first, then undo the multiplication or division.
很多学生因为运算顺序错误而失分。请记住规则：先撤销加或减，再撤销乘或除。
3. Equations with Brackets and Fractions | 含括号与分数的方程
When an equation contains brackets, the safest method is to expand the brackets first. For example, 3(x − 4) = 15 expands to 3x − 12 = 15, so 3x = 27 and x = 9.
当方程含有括号时，最稳妥的方法是先展开括号。例如 3(x − 4) = 15 展开为 3x − 12 = 15，所以 3x = 27，x = 9。
then x = 9. Both methods are correct, and choosing the faster one is a useful exam skill.
有些学生喜欢两边先同时除以 3，得到 x − 4 = 5，再得到 x = 9。两种方法都正确，在考试中选择更快的方法是一项实用技巧。
For equations with fractions, multiply every term by the lowest common multiple of the denominators to clear the fractions. Consider (2x + 1) ÷ 3 = 5:
对于含分数的方程，可将每一项都乘以各分母的最小公倍数来消去分数。来看 (2x + 1) ÷ 3 = 5：
(2x + 1) ÷ 3 = 5 → 2x + 1 = 15 → 2x = 14 → x = 7
With two different denominators, multiply by their product or LCM. For x ÷ 2 + x ÷ 3 = 5, multiply every term by 6:
当有两个不同分母时，乘以它们的乘积或最小公倍数。对于 x ÷ 2 + x ÷ 3 = 5，每项乘以 6：
6 × (x ÷ 2) + 6 × (x ÷ 3) = 6 × 5 → 3x + 2x = 30 → 5x = 30 → x = 6
A common error is to multiply only one side of the equation by the LCM. Always multiply every single term on both sides, otherwise the balance is destroyed.
常见错误是只把方程的一边乘以最小公倍数。必须把两边每一个项都乘，否则等式平衡就被破坏。
4. Forming Equations from Word Problems | 从应用题列方程
Many IGCSE questions ask you to form an equation from a written situation and then solve it. The four-step routine below works for almost every case.
许多 IGCSE 题目要求你根据文字情境列出方程再求解。下面的四步法几乎适用于所有情况。
- Step 1: Define the unknown with a letter. | 第一步：用字母表示未知量。
Let x be the number of small boxes. | 设 x 为小盒子的数量。 - Step 2: Translate each sentence into algebra. | 第二步：把每一句话翻译成代数式。
"5 more than twice x" becomes 2x + 5. | "x 的两倍多 5" 写成 2x + 5。 - Step 3: Solve the equation using inverse operations. | 第三步：用逆运算解方程。
- Step 4: Check the answer against the original words. | 第四步：把答案代回原来的文字情境中检验。
For example, the perimeter of a rectangle is 46 cm and its length is 3 cm more than its width. Let the width be w, so the length is w + 3. Using the perimeter formula:
例如，一个长方形的周长是 46 cm，长比宽多 3 cm。设宽为 w，则长为 w + 3。利用周长公式：
2(w + w + 3) = 46 → 2(2w + 3) = 46 → 4w + 6 = 46 → 4w = 40 → w = 10
So the width is 10 cm and the length is 13 cm. Always write a concluding sentence with units, because IGCSE mark schemes often award a mark for a clearly stated answer.
所以宽是 10 cm，长是 13 cm。一定要写出带单位的结论句，因为 IGCSE 评分标准常常为表述清楚的答案单独给分。
5. What Are Simultaneous Equations? | 什么是联立方程？
Simultaneous equations are two (or more) equations that must be satisfied at the same time by the same values of the variables. A pair of linear simultaneous equations in two unknowns has the general form:
联立方程是指必须由同一组变量的值同时满足的两个（或更多）方程。含两个未知数的二元一次联立方程组的一般形式为：
a₁x + b₁y = c₁ and a₂x + b₂y = c₂
The solution is an ordered pair (x, y) that makes both equations true. For instance, x = 3, y = 2 satisfies both 2x + y = 8 and x − y = 1, so it is the solution of that pair.
方程组的解是一对有序数 (x, y)，它使两个方程同时成立。例如 x = 3, y = 2 同时满足 2x + y = 8 和 x − y = 1，因此它就是该方程组的解。
There are three standard methods: elimination, substitution and graphical solution. Core candidates must be confident with elimination and substitution; Extended candidates should also be able to interpret the graphical meaning of the solution.
解方程组有三种标准方法：消元法、代入法和图像法。核心卷考生必须熟练掌握消元法和代入法；扩展卷考生还应能解释解在图像上的意义。
6. Solving by Elimination | 用消元法求解
Elimination works by adding or subtracting the two equations so that one variable disappears. If the coefficients of one variable are already the same, subtract; if they are opposites, add.
消元法通过把两个方程相加或相减，使其中一个变量消失。如果某个变量的系数已经相同，就相减；如果互为相反数，就相加。
Worked example: solve 2x + 3y = 12 and 2x − y = 4. Both equations contain 2x, so subtract the second from the first:
例题：解 2x + 3y = 12 和 2x − y = 4。两个方程都含有 2x，因此用第一式减第二式：
(2x + 3y) − (2x − y) = 12 − 4 → 4y = 8 → y = 2
y = 2 into 2x − y = 4 gives 2x − 2 = 4, so 2x = 6 and x = 3. The solution is x = 3, y = 2.
把 y = 2 代入 2x − y = 4，得 2x − 2 = 4，所以 2x = 6，x = 3。方程组的解是 x = 3，y = 2。
If the coefficients do not match, multiply one or both equations first. For 3x + 2y = 16 and 2x + 3y = 14:
如果系数不匹配，先给一个或两个方程乘以适当的数。对于 3x + 2y = 16 和 2x + 3y = 14：
- Multiply equation 1 by 2: 6x + 4y = 32 | 第一式乘 2：6x + 4y = 32
- Multiply equation 2 by 3: 6x + 9y = 42 | 第二式乘 3：6x + 9y = 42
- Subtract: 5y = 10, so y = 2 | 相减：5y = 10，所以 y = 2
- Substitute back: 3x + 4 = 16, so x = 4 | 代回：3x + 4 = 16，所以 x = 4
The elimination method is usually the fastest approach when the coefficients are integers, and it is the method most often shown in IGCSE mark schemes.
当系数为整数时，消元法通常是最快的方法，也是 IGCSE 评分标准中最常展示的方法。
7. Solving by Substitution | 用代入法求解
. . Rearrange one equation to make a variable the subject, then replace that variable in the other equation.
当其中一个方程已经写成 x = ... 或 y = ... 的形式时，代入法最方便。先把一个方程变形，使某个变量成为主元，再把它代入另一个方程。
Worked example: solve y = 2x + 1 and 3x + y = 11. Substitute the first expression for y into the second equation:
例题：解 y = 2x + 1 和 3x + y = 11。把第一式中的 y 代入第二式：
3x + (2x + 1) = 11 → 5x + 1 = 11 → 5x = 10 → x = 2
ion is x = 2, y = 5. Substituting into the original equations confirms both are satisfied.
于是 y = 2(2) + 1 = 5。方程组的解是 x = 2，y = 5。代回原方程可验证两式均成立。
- Always substitute into the simpler equation to keep the arithmetic short. | 始终代入较简单的方程，让运算更简短。
- Remember to find the value of both variables before writing your answer. | 写答案前记得求出两个变量的值。
- Check by substituting into the equation you did not use. | 用你没用过的那个方程检验答案。
Substitution also prepares you for solving one linear and one quadratic equation simultaneously, a standard Extended topic.
代入法也为求解一个线性方程与一个二次方程的联立问题打下基础，这是扩展卷的标准考点。
8. Graphical Meaning and Number of Solutions | 图像意义与解的个数
Each linear equation represents a straight line on a graph. The solution of a pair of simultaneous equations is the coordinates of the point where the two lines intersect.
每个线性方程在图像上都表示一条直线。二元一次方程组的解就是两条直线交点处的坐标。
| Case | 情形 | Graph | 图像 | Number of solutions | 解的个数 |
| Different gradients | 斜率不同 | Lines cross once | 两直线相交于一点 | Exactly one | 恰有一个解 |
| Same gradient, different intercept | 斜率相同、截距不同 | Parallel lines | 平行直线 | No solution | 无解 |
| Same gradient, same intercept | 斜率相同、截距也相同 | The same line | 同一条直线 | Infinitely many | 无穷多解 |
You can also solve a pair of simultaneous equations graphically by drawing both lines accurately on a grid and reading off the intersection point. This is only reliable if your scale is sensible and your lines are drawn with a ruler.
你也可以通过在网格上准确画出两条直线并读出交点坐标来图解二元一次方程组。只有比例尺选择合理、用直尺画线时，这种方法才可靠。
These ideas lead directly to linear programming, where inequalities define a region and you must find the maximum or minimum value of an expression at one of its vertices.
这些思想直接通向线性规划：用不等式定义区域，再在区域的某个顶点处求出表达式的最值。
9. Word Problems with Simultaneous Equations | 联立方程应用题
Examination questions often hide a pair of simultaneous equations inside a real-life context, such as shopping, ages, coins or mixtures. The technique is to name both unknowns and write one equation for each piece of information.
考试题常把方程组隐藏在现实情境中，例如购物、年龄、硬币或混合问题。技巧是给两个未知量命名，并针对每条信息各写一个方程。
Example: 3 pens and 2 pencils cost $4.30, while 2 pens and 3 pencils cost $3.70. Let a pen cost p dollars and a pencil cost q dollars:
例题：3 支钢笔和 2 支铅笔共 4.30 美元，2 支钢笔和 3 支铅笔共 3.70 美元。设钢笔每支 p 美元，铅笔每支 q 美元：
3p + 2q = 4.30 and 2p + 3q = 3.70
.10. Subtracting gives 5q = 2.50, so q = 0.50, and then 3p + 1.00 = 4.30, giving p = 1.10.
把第一式乘 2、第二式乘 3，使 p 的系数相同：6p + 4q = 8.60 和 6p + 9q = 11.10。相减得 5q = 2.50，所以 q = 0.50；再由 3p + 1.00 = 4.30 得 p = 1.10。
- Write a clear "Let ..." statement for both unknowns. | 对两个未知量都写出清楚的"设……"。
- Number your equations so you can refer to them in your working. | 给方程编号，方便在解题过程中引用。
- Finish with a sentence that answers the question in context. | 最后用一句话在情境中回答问题。
in 12 years he will be twice as old, the two equations are F = 3S and F + 12 = 2(S + 12).
年龄问题同样常见。如果父亲现在的年龄是儿子的 3 倍，12 年后是儿子的 2 倍，则两个方程是 F = 3S 和 F + 12 = 2(S + 12)。
10. What Are Inequalities? | 什么是不等式？
n or equal to). Unlike an equation, an inequality usually has a whole range of solutions rather than a single value.
不等式用四个符号之一比较两个表达式：<（小于）、>（大于）、≤（小于或等于）和 ≥（大于或等于）。与方程不同，不等式通常有一整段范围的解，而不是单一的值。
The method for solving a linear inequality is almost identical to solving a linear equation: perform the same operation on both sides to isolate the unknown.
解线性不等式的方法与解线性方程几乎完全相同：对两边进行相同的运算，从而把未知数单独留下。
> 4 2x ≤ 10 → x ≤ 5 5x − 4 ≥ 16 → 5x ≥ 20 → x ≥ 4
not included, while a filled circle is used for ≤ and ≥, showing the endpoint is included.
在数轴上，< 和 > 用空心圆圈表示端点不包含在内，而 ≤ 和 ≥ 用实心圆点表示端点包含在内。
- x > 4: open circle at 4, arrow pointing right. | x > 4：在 4 处画空心圆，箭头向右。
- x ≤ 5: filled circle at 5, arrow pointing left. | x ≤ 5：在 5 处画实心圆，箭头向左。
- −2 ≤ x < 3: a line segment between the two points, filled at −2 and open at 3. | −2 ≤ x < 3：两点之间的线段，−2 处实心、3 处空心。
Inequalities are used to describe constraints in real life, such as "the total mass must not exceed 20 kg" (m ≤ 20) or "at least 15 students are needed" (n ≥ 15).
不等式常用于描述现实中的限制条件，例如"总质量不得超过 20 kg"（m ≤ 20）或"至少需要 15 名学生"（n ≥ 15）。
11. The Negative Rule and Two-Sided Inequalities | 负数变号法则与双边不等式
Adding or subtracting the same number on both sides never changes the direction of an inequality sign. However, multiplying or dividing both sides by a negative number reverses the sign.
在两边同时加或减同一个数，不会改变不等号方向。但是，两边同时乘或除以一个负数时，不等号方向必须反转。
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