📚 Simultaneous Equations | 联立方程
Simultaneous equations are a pair (or more) of equations in two unknowns whose values must satisfy every equation at the same time. They appear in almost every IGCSE Mathematics paper, both as pure algebra questions and as problem-solving questions set in real-life contexts. Mastering them unlocks reliable marks across Paper 2 and Paper 4, and the same skills are needed in IGCSE Additional Mathematics and A-Level study.
联立方程是指两个（或多个）含有两个未知数的方程，其解必须同时满足所有方程。在 IGCSE 数学考试中，联立方程几乎每年必考，既出现在纯代数题中，也常以实际生活情境的应用题形式出现。掌握联立方程，意味着在 Paper 2 和 Paper 4 中都能稳稳拿到分数，所学的技巧在 IGCSE 附加数学和 A-Level 学习中同样不可或缺。
1. What Are Simultaneous Equations? | 什么是联立方程？
A single linear equation in two variables has infinitely many solutions. For example, the equation x + y = 10 includes the pairs (1, 9), (2, 8), (5, 5), (−3, 13) and so on. When a second equation is introduced, such as x − y = 4, the two equations together usually have exactly one common solution. That common pair is called the solution of the simultaneous equations.
只含两个未知数的一个线性方程有无数多组解。例如，方程 x + y = 10 包含 (1, 9)、(2, 8)、(5, 5)、(−3, 13) 等很多组解。当引入第二个方程，如 x − y = 4 时，两个方程通常只有一个共同的解。这组共同解就叫做联立方程组的解。
x + y = 10 and x − y = 4 → x = 7, y = 3
We can verify the solution quickly by substituting into both equations: 7 + 3 = 10 and 7 − 3 = 4. Because both statements are true, the pair (7, 3) is the correct solution. Always carry out this check in an exam; it is the simplest way to avoid losing easy marks.
我们可以很快验证这组解：把 x = 7, y = 3 分别代入两个方程，得到 7 + 3 = 10 和 7 − 3 = 4，两个等式都成立，所以 (7, 3) 是正确的解。考试中一定要养成代入检验的习惯，这是避免丢分最简单的方法。
2. Solving by Elimination | 消元法
The elimination method works by adding or subtracting the equations so that one variable disappears. Look at the coefficients of the same variable in both equations: if the signs are opposite, add the equations; if the signs are the same, subtract one equation from the other.
消元法的核心思想是通过将两个方程相加或相减，使其中一个未知数被消去。观察两个方程中同一个未知数的系数：如果符号相反，就把两个方程相加；如果符号相同，就把两个方程相减。
Consider the pair:
请看以下方程组：
2x + y = 7
x − y = 2
Here the y terms have coefficients +1 and −1, which are opposite in sign, so we add the equations: (2x + y) + (x − y) = 7 + 2, which simplifies to 3x = 9. Therefore x = 3. Substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. The solution is x = 3, y = 1.
这里 y 项的系数分别是 +1 和 −1，符号相反，因此我们把两个方程相加：(2x + y) + (x − y) = 7 + 2，化简得 3x = 9，所以 x = 3。把 x = 3 代入 x − y = 2，得 3 − y = 2，所以 y = 1。方程组的解为 x = 3，y = 1。
Add: 3x = 9 → x = 3; then y = 1
3. Elimination When Coefficients Differ | 系数不同时的消元法
In most exam questions, the two equations do not have matching coefficients. We must first multiply one or both equations by a suitable number so that one variable has the same magnitude of coefficient in both equations.
大多数考试题中，两个方程的系数并不相同。这时需要先给一个或两个方程乘以适当的数，使某个未知数在两个方程中的系数绝对值相等。
Solve the following pair:
解下列方程组：
2x + 3y = 8
3x + 2y = 7
Multiply the first equation by 3 to get 6x + 9y = 24. Multiply the second equation by 2 to get 6x + 4y = 14. Now the x coefficients are both 6, so subtract the second result from the first: (6x + 9y) − (6x + 4y) = 24 − 14, which gives 5y = 10, so y = 2. Substitute y = 2 into 2x + 3y = 8: 2x + 6 = 8, so x = 1. The solution is x = 1, y = 2.
将第一个方程乘以 3，得 6x + 9y = 24；将第二个方程乘以 2，得 6x + 4y = 14。此时 x 的系数都是 6，于是用第一个新方程减去第二个新方程：(6x + 9y) − (6x + 4y) = 24 − 14，得到 5y = 10，所以 y = 2。把 y = 2 代入 2x + 3y = 8：2x + 6 = 8，所以 x = 1。方程组的解为 x = 1，y = 2。
6x + 9y = 24 and 6x + 4y = 14 → 5y = 10 → y = 2, x = 1
4. Solving by Substitution | 代入法
The substitution method is especially useful when one variable is already the subject of an equation, or when a variable has coefficient 1. Rearrange one equation to express one variable in terms of the other, then substitute that expression into the second equation.
代入法特别适用于某个未知数已经是方程的主项，或者某个未知数系数为 1 的情况。先改写一个方程，把一个未知数用另一个未知数表示出来，再把这个表达式代入第二个方程。
Solve the pair:
解方程组：
y = 2x − 1
3x + 2y = 12
Because y is already given as 2x − 1, substitute this into the second equation: 3x + 2(2x − 1) = 12. Expanding gives 3x + 4x − 2 = 12, so 7x = 14 and x = 2. Then y = 2(2) − 1 = 3. The solution is x = 2, y = 3.
因为 y 已经写成 2x − 1，直接把该式代入第二个方程：3x + 2(2x − 1) = 12。展开得 3x + 4x − 2 = 12，所以 7x = 14，x = 2。再代回得 y = 2(2) − 1 = 3。方程组的解为 x = 2，y = 3。
3x + 2(2x − 1) = 12 → 7x = 14 → x = 2, y = 3
5. Solving by Graphical Method | 图像法
Every linear equation can be drawn as a straight line on a coordinate grid. If we plot both lines on the same axes, the coordinates of the point where the two lines intersect give the solution of the simultaneous equations.
每个线性方程都可以在坐标平面内画成一条直线。如果把两条线画在同一坐标系中，两条直线交点的坐标就是联立方程组的解。
Take the examples above: y = 2x − 1 and 3x + 2y = 12. Rewrite the second equation as y = 6 − 1.5x. Plot both lines; they intersect at the point (2, 3), so the solution is x = 2, y = 3.
取上面的例子：y = 2x − 1 和 3x + 2y = 12。把第二个方程改写成 y = 6 − 1.5x。画出这两条直线，它们的交点是 (2, 3)，所以解为 x = 2，y = 3。
In an exam, graphical solutions drawn by hand are often approximate, so you must check them by substitution. Remember also that if the lines are parallel, there is no solution, and if the two equations represent the same line, there are infinitely many solutions.
在考试中，手绘图像得到的解往往是近似值，必须通过代入来检验。还要特别注意：若两条直线平行，则方程组无解；若两个方程表示同一条直线，则方程组有无数多组解。
6. Solving Word Problems | 应用题解法
Word problems require us to translate ordinary language into algebra. The key steps are: read carefully, define variables clearly, write two equations from the given information, solve them, and then answer in context with correct units.
应用题的关键是把自然语言翻译成代数语言。解题步骤为：仔细审题、明确设定未知数、根据已知条件列出两个方程、解方程组，最后结合题目情境写出带单位的答案。
Example: Two pens and one notebook cost $7. One pen and three notebooks cost $11. Find the price of one pen and one notebook.
例题：两支笔和一个笔记本共 7 美元；一支笔和三个笔记本共 11 美元。求一支笔和一个笔记本的单价。
Let x be the price of one pen and y be the price of one notebook. Then 2x + y = 7 and x + 3y = 11. From the first equation, y = 7 − 2x. Substitute into the second: x + 3(7 − 2x) = 11, which gives x + 21 − 6x = 11, so −5x = −10 and x = 2. Then y = 7 − 4 = 3. One pen costs $2 and one notebook costs $3.
设一支笔的单价为 x，一个笔记本的单价为 y。则 2x + y = 7，x + 3y = 11。由第一个方程得 y = 7 − 2x，代入第二个方程：x + 3(7 − 2x) = 11，即 x + 21 − 6x = 11，所以 −5x = −10，x = 2。于是 y = 7 − 4 = 3。一支笔 2 美元，一个笔记本 3 美元。
Pen = $2, Notebook = $3
7. Special Cases: No Solution and Infinite Solutions | 特殊情况：无解与无数解
Not every pair of simultaneous equations has exactly one solution. If the elimination step produces a statement like 0 = 4, which is impossible, then the two lines are parallel and the system has no solution.
并不是每组联立方程都恰好有一个解。如果在消元过程中得到 0 = 4 这类不可能成立的等式，说明两条直线平行，方程组无解。
Consider 2x + y = 5 and 2x + y = 9. Subtracting gives 0 = 4, which is false, so there is no common point and no solution.
看方程组 2x + y = 5 和 2x + y = 9。相减得 0 = 4，这是矛盾的，所以两条直线没有交点，方程组无解。
On the other hand, if elimination gives 0 = 0, then the two equations are actually the same line. For example, 2x + y = 5 and 4x + 2y = 10 are equivalent because the second is the first multiplied by 2. In this case there are infinitely many solutions.
反之，如果消元得到 0 = 0，说明两个方程实际上表示同一条直线。例如 2x + y = 5 与 4x + 2y = 10 等价，因为第二个方程是第一个方程乘以 2 得到的。此时方程组有无数多组解。
8. Non-Linear Simultaneous Equations | 非线性联立方程
For Extended IGCSE candidates, a common question pairs one linear equation with one quadratic equation. Use substitution: replace the linear expression into the quadratic, solve the resulting quadratic, then find the corresponding y values.
对于 IGCSE 扩展课程的学生，常考题型是一个线性方程配一个二次方程。解题方法仍然是代入：把线性表达式代入二次方程，解出所得的二次方程，再分别求出对应的 y 值。
Solve y = x + 1 and y = x² − 5.
解方程组 y = x + 1 和 y = x² − 5。
x + 1 = x² − 5 → x² − x − 6 = 0 → (x − 3)(x + 2) = 0
So x = 3 or x = −2. Substituting back into y = x + 1 gives y = 4 when x = 3, and y = −1 when x = −2. The two solutions are (3, 4) and (−2, −1). Notice that a linear–quadratic system can have zero, one, or two solutions.
因此 x = 3 或 x = −2。代回 y = x + 1：当 x = 3 时 y = 4；当 x = −2 时 y = −1。两个解分别为 (3, 4) 和 (−2, −1)。注意，线性—二次方程组可能有零个、一个或两个解。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
Candidates often lose marks not because they cannot do the algebra, but because of avoidable errors. The following list covers the most frequent pitfalls and the habits that will protect your marks.
考生丢分往往不是因为不会做代数，而是
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