📚 Solving Simultaneous Equations | 联立方程的解法
Simultaneous equations are one of the most frequently tested topics in IGCSE Edexcel Mathematics. In this revision guide, we will break down the three main solution methods, look at word problems and quadratic forms, and highlight the exact pitfalls that cost marks in the examination room.
联立方程是 IGCSE Edexcel 数学中考频最高的知识点之一。在本复习指南中，我们将系统讲解三种主流解法，分析应用题与二次形式的联立方程，并指出考试中最容易失分的细节。
1. What Are Simultaneous Equations? | 什么是联立方程？
A simultaneous equation is a pair (or more) of equations that share two or more unknown variables. A valid solution must make every equation true at the same time, which is why the word “simultaneous” (at the same time) is used. For IGCSE Edexcel, you are expected to solve two linear equations, or one linear equation combined with one quadratic equation.
联立方程是共享两个或多个未知数的一组（两个或以上）方程。一个有效的解必须同时满足每一个方程，这就是”simultaneous”（同时）一词的含义。在 Edexcel IGCSE 考纲中，你需要掌握两个一次方程联立，以及一个一次方程配一个二次方程的联立。
Consider this very simple pair of equations:
请看下面这一组非常简单的方程：
x + y = 7
x − y = 3
If we add the two equations together, the y terms cancel: (x + y) + (x − y) = 7 + 3, so 2x = 10, giving x = 5. Substituting back, 5 + y = 7, so y = 2. The solution is x = 5, y = 2, or simply (5, 2).
将两个方程相加，y 项相互抵消：(x + y) + (x − y) = 7 + 3，得到 2x = 10，即 x = 5。代回去，5 + y = 7，所以 y = 2。最终解为 x = 5，y = 2，也可以写成坐标形式 (5, 2)。
Geometrically, each linear equation represents a straight line on the xy-plane. The solution of the system is exactly the point where the two lines cross. If the lines are parallel, there is no solution; if one equation is a multiple of the other, the lines coincide and there are infinitely many solutions. Understanding this picture helps you check whether your algebraic answer is sensible.
从几何角度看，每一个一次方程都代表 xy 平面上的一条直线。方程组的解正是两条直线的交点。若两直线平行，则无解；若一个方程是另一个方程的倍数，则两直线重合，有无数个解。理解这一几何图像有助于你判断代数答案是否合理。
2. Solving by Elimination | 消元法求解
The elimination method removes one variable by adding or subtracting the equations. This only works directly when the coefficients of one variable are the same size, or can be made the same by multiplying one or both equations first.
消元法通过对两个方程进行相加或相减，消去其中一个未知数。只有当某个未知数的系数大小相同时，才能直接相加减；如果系数不同，则需要先对方程两边乘以适当的数，使系数一致。
Follow these five steps every time:
每一步都按照以下五个步骤来操作：
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1. Line up the equations so that x terms, y terms and constants are in the same columns.
第一步：对齐方程。把 x 项、y 项和常数项对齐到各自的列中。
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2. Decide whether to add or subtract. If the coefficients have the same sign, subtract; if they have opposite signs, add.
第二步：决定相加还是相减。若系数符号相同则相减，若符号相反则相加。
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3. Solve the resulting one-variable equation.
第三步：解出剩下的一元方程。
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4. Substitute back into either original equation to find the second variable.
第四步：代回任意一个原方程，求出另一个未知数。
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5. Check your answer in the other original equation. This catches sign errors immediately.
第五步：代入另一个原方程验证。这样可以立刻发现符号错误。
Worked Example 1
例题 1
2x + 3y = 12
5x − 3y = 9
Here the y-coefficients are +3 and −3, so add the equations: 2x + 5x = 12 + 9, so 7x = 21, meaning x = 3. Substitute x = 3 into the first equation: 2(3) + 3y = 12, so 6 + 3y = 12, giving 3y = 6 and y = 2. Check in the second equation: 5(3) − 3(2) = 15 − 6 = 9 ✓. The solution is x = 3, y = 2.
此题中 y 的系数分别为 +3 与 −3，符号相反，所以将方程相加：2x + 5x = 12 + 9，得 7x = 21，即 x = 3。把 x = 3 代入第一个方程：2(3) + 3y = 12，即 6 + 3y = 12，解得 3y = 6，y = 2。代入第二个方程验证：5(3) − 3(2) = 15 − 6 = 9 ✓。答案 x = 3，y = 2。
Worked Example 2 (coefficients do not match)
例题 2（系数不一致）
3x + 2y = 13
2x + 5y = 16
Neither coefficient matches, so we multiply the first equation by 2 and the second by 3:
两方程没有相同的系数，因此将第一个方程乘以 2，第二个方程乘以 3：
6x + 4y = 26
6x + 15y = 48
Now subtract the first transformed equation from the second: (6x + 15y) − (6x + 4y) = 48 − 26, so 11y = 22, giving y = 2. Substitute into 3x + 2y = 13: 3x + 4 = 13, so 3x = 9 and x = 3. Check: 2(3) + 5(2) = 6 + 10 = 16 ✓.
现在用第二个变换后的方程减去第一个： (6x + 15y) − (6x + 4y) = 48 − 26，得 11y = 22，即 y = 2。代入 3x + 2y = 13：3x + 4 = 13，故 3x = 9，x = 3。验证：2(3) + 5(2) = 6 + 10 = 16 ✓。
3. Solving by Substitution | 代入法求解
The substitution method is the method of choice when one equation is already given with a single variable as the subject, such as y = … or x = …. It is also essential for linear-quadratic systems, which we will see in Section 6.
代入法适用于其中一个方程已经将某个变量单独放在等号一侧的情况，例如 y = … 或 x = …。在处理一次与二次方程的联立时，代入法也是必用的方法，详见第 6 节。
Worked Example 3
例题 3
2x + y = 9
3x + 4y = 21
Step 1: Make y the subject of the first equation: y = 9 − 2x.
第一步：将第一个方程改写为 y = 9 − 2x。
Step 2: Substitute y = 9 − 2x into the second equation: 3x + 4(9 − 2x) = 21.
第二步：把 y = 9 − 2x 代入第二个方程：3x + 4(9 − 2x) = 21。
Step 3: Expand and solve: 3x + 36 − 8x = 21, so −5x = −15, giving x = 3.
第三步：展开并求解：3x + 36 − 8x = 21，即 −5x = −15，得 x = 3。
Step 4: Substitute back: y = 9 − 2(3) = 3. The solution is x = 3, y = 3.
第四步：代回求 y：y = 9 − 2(3) = 3。解为 x = 3，y = 3。
When substituting, always use brackets around the expression you are replacing. For example, write 4(9 − 2x) instead of 4 · 9 − 2x, because the whole expression is multiplied by 4. This single habit removes most expansion errors.
代入时，一定要把被替换的表达式用括号括起来。例如写 4(9 − 2x)，而不是 4 · 9 − 2x，因为整个表达式都要乘以 4。养成这个习惯可以消除大多数展开错误。
4. Choosing the Best Method | 如何选择最合适的方法
Examiners rarely tell you which method to use. Choosing the fastest method saves time for Section B questions, so consider the structure of the equations before you start.
考试通常不会指定你必须用哪种方法。选择最快的方法可以为后面的题目节省时间，因此在下笔前先观察方程的结构。
| Situation | Recommended method | Why |
| One variable already isolated, e.g. y = 2x + 1 | Substitution | No rearrangement needed |
| Coefficients of one variable are equal or opposite | Elimination | One line of addition/subtraction removes a variable |
| Small coefficients, e.g. 3x + 2y and 4x + 5y | Elimination after multiplying | Avoids fractions until the final step |
| One linear and one quadratic equation | Substitution | Produces a single quadratic equation |
| You only need an approximate answer | Graphical | Intersection of graphs gives the estimate |
There is no penalty for using substitution instead of elimination, as long as your working is clear. In fact, substitution is the safer method for many students because it has a fixed rhythm: isolate, replace, expand, solve, substitute back.
用代入法代替消元法不会扣分，只要过程清晰即可。实际上，代入法对很多学生来说更安全，因为它有固定的节奏：隔离变量、代入、展开、解方程、回代。
5. Word Problems | 应用题
Word problems in IGCSE require you to translate everyday language into algebra. The key is to define your variables clearly at the top of your answer: state what x and y represent before writing any equations.
IGCSE 应用题要求你把日常语言转化为代数式。关键是先在答案开头明确设未知数：写下 x 和 y 分别代表什么，然后再列方程。
Worked Example 4
例题 4
A cinema sells adult tickets at £a each and child tickets at £c each. Two adult tickets and three child tickets cost £42. One adult ticket and one child ticket cost £16. Find the price of each type of ticket.
某电影院成人票每张 £a，儿童票每张 £c。两张成人票与三张儿童票共 £42；一张成人票与一张儿童票共 £16。求两种票各自的价格。
Step 1: Define variables. Let a = price of an adult ticket and c = price of a child ticket.
第一步：设未知数。设 a 为成人票价，c 为儿童票价。
Step 2: Write the equations.
第二步：列方程。
2a + 3c = 42
a + c = 16
Step 3: Solve. From the second equation, a = 16 − c. Substitute: 2(16 − c) + 3c = 42, which gives 32 − 2c + 3c = 42, so c = 10. Then a = 16 − 10 = 6.
第三步：求解。由第二个方程得 a = 16 − c。代入：2(16 − c) + 3c = 42，展开得 32 − 2c + 3c = 42，故 c = 10。于是 a = 16 − 10 = 6。
Step 4: Answer the question in words. An adult ticket costs £6 and a child ticket costs £10. Read the question again to make sure you state the units and use the correct notation.
第四步：用文字作答。成人票每张 £6，儿童票每张 £10。答完后再读一遍题目，确保写上了单位并使用正确的符号。
Common translation mistakes include reversing the coefficients (writing 3a + 2c instead of 2a + 3c) and forgetting to define variables. Drawing a quick table of the given information can prevent both of these errors.
常见的翻译错误包括把系数写反（写成 3a + 2c 而不是 2a + 3c），以及忘记设未知数。养成把信息整理成小表格的习惯，能同时避免这两种错误。
6. One Linear and One Quadratic Equation | 一次与二次方程联立
When one equation is quadratic (containing x²) and the other is linear, always use substitution: make one variable the subject of the linear equation and substitute it into the quadratic. This produces a quadratic equation in one variable, which may have two, one, or zero real solutions.
当其中一个方程是二次方程（含有 x²），另一个是一次方程时，一定要用代入法：先从一次方程中解出一个变量，再代入二次方程。这样会得到一个一元二次方程，它可能有 2 个、1 个或 0 个实数解。
Worked Example 5
例题 5Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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