📚 Mastering Simultaneous Equations | 掌握联立方程
Simultaneous equations are a pair (or more) of equations that share the same unknown variables. In IGCSE Mathematics, solving these equations is a core skill tested across both Paper 2 and Paper 4, and it appears frequently in both calculator and non-calculator papers. Understanding this topic well can secure easy marks, as the methods are systematic and reliable.
联立方程是指两个（或更多）共享相同未知数的方程。在 IGCSE 数学中，解联立方程是核心技能之一，在 Paper 2 和 Paper 4 中都有考查，并且经常出现在允许和不允许使用计算器的试卷中。透彻理解这一主题可以轻松得分，因为其方法系统且可靠。
1. What Are Simultaneous Equations? | 什么是联立方程？
A linear equation in two variables, such as 2x + y = 10, has infinitely many solutions. For example, (1, 8), (2, 6) and (3, 4) all satisfy it. But when we pair it with another linear equation, such as x − y = 2, the two equations together usually have exactly one common solution.
二元一次方程，例如 2x + y = 10，有无数个解。例如，(1, 8)、(2, 6) 和 (3, 4) 都满足该方程。但当它与另一个线性方程（如 x − y = 2）配对时，这两个方程通常恰好有一个公共解。
2x + y = 10
x − y = 2
The solution is x = 4 and y = 2. Substitute these values into both equations: 2(4) + 2 = 10 ✓ and 4 − 2 = 2 ✓. The ordered pair (4, 2) satisfies both equations simultaneously, which is why we call them simultaneous equations.
解为 x = 4，y = 2。将这两个值代入两个方程：2(4) + 2 = 10 ✓，4 − 2 = 2 ✓。有序数对 (4, 2) 同时满足两个方程，因此我们称之为联立方程。
There are three main methods you need to master: elimination, substitution and graphical. In the IGCSE exams, the elimination method is often the most efficient for linear pairs, while substitution is essential when one equation is quadratic.
你需要掌握三种主要方法：消元法、代入法和图像法。在 IGCSE 考试中，对于线性方程对，消元法通常最高效；而当其中一个方程是二次方程时，代入法则必不可少。
2. The Elimination Method | 消元法
The elimination method works by adding or subtracting the equations to remove one variable. Here is a step-by-step example using the following pair of equations.
消元法的核心是通过相加或相减两个方程来消去一个未知数。下面以一个方程组为例，逐步说明。
3x + 2y = 13
2x − y = 4
Step 1: Make the coefficients of one variable equal in magnitude. Multiply the second equation by 2 so that the y coefficients become +2 and −2.
步骤1：使某个未知数的系数绝对值相等。将第二个方程乘以 2，使 y 的系数变为 +2 和 −2。
4x − 2y = 8
Step 2: Add the two equations to eliminate y. Since −2y + 2y = 0, we get 7x = 21, so x = 3.
步骤2：将两个方程相加，消去 y。因为 −2y + 2y = 0，得到 7x = 21，所以 x = 3。
Step 3: Substitute x = 3 back into one of the original equations, say 2x − y = 4. Then 6 − y = 4, so y = 2.
步骤3：将 x = 3 代回原方程组中的任意一个方程，例如 2x − y = 4。于是 6 − y = 4，解得 y = 2。
Step 4: Check the answer in the other equation: 3(3) + 2(2) = 9 + 4 = 13 ✓. The solution is x = 3, y = 2.
步骤4：将答案代入另一个方程验证：3(3) + 2(2) = 9 + 4 = 13 ✓。解为 x = 3，y = 2。
When the signs of the equal coefficients are the same, subtract the equations; when they are opposite, add them. Always write down which operation you are using, because examiners award marks for clear working.
当相等系数符号相同时，用减法消元；当符号相反时，用加法消元。务必写清楚你使用的是哪种运算，因为阅卷官会根据清晰的步骤给分。
3. The Substitution Method | 代入法
The substitution method is particularly useful when one equation is already written with a variable as the subject, or when dealing with a linear equation plus a quadratic equation. Let us solve the following pair.
代入法特别适用于某个方程已经以某变量为未知量形式给出时，或者在处理一个线性方程与一个二次方程的组合时。我们来解下面这个方程组。
2x + 3y = 12
x − y = 1
Step 1: Rearrange the second equation to make x the subject: x = y + 1.
步骤1：将第二个方程变形，用 y 表示 x：x = y + 1。
Step 2: Substitute x = y + 1 into the first equation. This gives 2(y + 1) + 3y = 12.
步骤2：将 x = y + 1 代入第一个方程，得到 2(y + 1) + 3y = 12。
Step 3: Expand and solve. 2y + 2 + 3y = 12, so 5y = 10 and y = 2.
步骤3：展开并求解。2y + 2 + 3y = 12，因此 5y = 10，y = 2。
Step 4: Substitute y = 2 back into x = y + 1 to get x = 3. The solution is x = 3, y = 2, which matches the elimination result.
步骤4：将 y = 2 代回 x = y + 1，得到 x = 3。解为 x = 3，y = 2，与消元法的结果一致。
The key advantage of substitution is that it generalises to non-linear systems. Once you substitute the linear expression into the quadratic, you obtain a single equation in one variable that can be solved by factorising or using the quadratic formula.
代入法的关键优势在于它可以推广到非线性方程组。一旦将线性表达式代入二次方程，你就能得到一个只含一个未知数的一元方程，可以通过因式分解或求根公式来求解。
4. The Graphical Method | 图像法
Graphically, each linear equation represents a straight line. The solution to a pair of simultaneous equations is the point where the two lines intersect. To use this method, first rearrange each equation into the form y = mx + c.
在图像上，每个线性方程都表示一条直线。联立方程的解就是两条直线的交点。要使用此方法，首先将每个方程变形为 y = mx + c 的形式。
Consider the equations y = 2x − 1 and y = −x + 5. Plotting both lines on the same axes, they intersect at the point (2, 3). Therefore the solution is x = 2, y = 3.
考虑方程 y = 2x − 1 和 y = −x + 5。在同一直角坐标系中画出这两条直线，它们相交于点 (2, 3)。因此解为 x = 2，y = 3。
The graphical method also reveals important special cases. If the lines are parallel, there is no solution. If they are the same line, there are infinitely many solutions.
图像法还能揭示重要的特殊情况。如果两条直线平行，则无解。如果是同一条直线，则有无穷多组解。
| Situation | Graphical Meaning | Number of Solutions |
| Lines intersect at a point | Different gradients | Exactly one |
| Lines are parallel | Same gradient, different y-intercept | None |
| Lines coincide | Same gradient and same y-intercept | Infinitely many |
In the IGCSE non-calculator paper, the graphical method is rarely the fastest, but you may be asked to draw lines and read off the intersection point. Accuracy of drawing and reading is essential.
在 IGCSE 非计算器试卷中，图像法通常不是最快的方法，但你可能会被要求画线并读出交点。作图和读图必须准确。
5. Word Problems in Context | 情境应用题
Many exam questions test simultaneous equations through real-life contexts. The first step is always to define variables clearly, then translate the information into two equations.
许多考试题目通过实际情境来考查联立方程。第一步始终是清晰地定义未知数，然后将信息转化为两个方程。
Example: The sum of two numbers is 15 and their difference is 3. Find the numbers.
例题：两个数之和为 15，之差为 3。求这两个数。
Let the larger number be x and the smaller number be y. Then:
设较大的数为 x，较小的数为 y。则有：
x + y = 15
x − y = 3
Adding the equations gives 2x = 18, so x = 9. Substituting back gives y = 6. Therefore the two numbers are 9 and 6.
两式相加得 2x = 18，所以 x = 9。代回得 y = 6。因此这两个数为 9 和 6。
Example 2: An adult ticket costs $a and a child ticket costs $c. Two adults and three children pay $34, while one adult and two children pay $19. Find the prices.
例题2：成人票价为 $a，儿童票价为 $c。两名成人和三名儿童共付费 34 美元，而一名成人和两名儿童共付费 19 美元。求票价。
2a + 3c = 34
a + 2c = 19
Multiply the second equation by 2: 2a + 4c = 38. Subtract the first equation: (2a + 4c) − (2a + 3c) = 38 − 34, giving c = 4. Substitute into a + 2(4) = 19, so a = 11. The adult ticket is $11 and the child ticket is $4.
将第二个方程乘以 2：2a + 4c = 38。减去第一个方程：(2a + 4c) − (2a + 3c) = 38 − 34，得 c = 4。代入 a + 2(4) = 19，所以 a = 11。成人票价为 11 美元，儿童票价为 4 美元。
When solving word problems, always end with a clear sentence stating the answer in the original context. This shows the examiner that you understand the meaning of your solution.
解决应用题时，务必用一句话明确写出最终答案并回到原情境中。这向阅卷官表明你理解解的实际意义。
6. Simultaneous Equations with Quadratics | 含二次项的联立方程
In the Extended IGCSE syllabus, you must also solve a system where one equation is linear and the other is quadratic. The recommended method is substitution, because it produces a single quadratic equation in one variable.
在 IGCSE 扩展课程中，你还需要解一个线性方程与一个二次方程组成的方程组。推荐使用代入法，因为代入后能得到一个一元二次方程。
Example: Solve the following pair of equations.
例题：解下面这个方程组。
y = x² − 3x + 2
y = 2x − 4
Since both expressions equal y, set them equal to each other:
因为两个表达式都等于 y，将它们设为相等：
x² − 3x + 2 = 2x − 4
Rearrange to collect all terms on one side:
移项将所有项集中到一边：
x² − 5x + 6 = 0
Factorise: (x − 2)(x − 3) = 0, so x = 2 or x = 3. Substitute each value into the linear equation y = 2x − 4. When x = 2, y = 0. When x = 3, y = 2. The solutions are (2, 0) and (3, 2).
因式分解：(x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。将每个值代入线性方程 y = 2x − 4。当 x = 2 时，y = 0。当 x = 3 时，y = 2。解为 (2, 0) 和 (3, 2)。
It is important to check both solutions in the original quadratic equation. For (2, 0): 2² − 3(2) + 2 = 0 ✓. For (3, 2): 3² − 3(3) + 2 = 2 ✓. Both are correct.
务必在原二次方程中检验两个解。对于 (2, 0)：2² − 3(2) + 2 = 0 ✓。对于 (3, 2)：3² − 3(3) + 2 = 2 ✓。两个解都正确。
Use the discriminant to predict the number of intersections. For the resulting quadratic ax² + bx + c = 0, if b² − 4ac > 0, there are two real solutions; if b² − 4ac = 0, there is exactly one solution and the line is a tangent; if b² − 4ac < 0, there are no real intersection points.
利用判别式可以预判交点的个数。对于得到的一元二次方程 ax² + bx + c = 0，若 b² − 4ac > 0，则有两个实数解；若 b² − 4ac = 0，则恰有一个解，此时直线为切线；若 b² − 4ac < 0，则没有实数交点。
7. Common Mistakes and Exam Tips | 常见错误与考试技巧
Experience shows that students often lose marks on simultaneous equations due to a small number of repeated errors. Here is a checklist of pitfalls to avoid and habits to adopt.
经验表明，学生在联立方程上丢分往往是由于少数重复出现的错误。以下是一份需要避免的陷阱和应养成的好习惯清单。
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Mistake: Forgetting to multiply every term of an equation when scaling it. Always multiply each term on both sides.
错误：在对方程进行倍数缩放时，忘记乘每一个项。务必同时乘以等式两边的每一项。
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Mistake: Sign errors when subtracting equations. Write the subtraction carefully, and consider adding instead if signs are opposite.
错误：两式相减时出现符号错误。仔细书写减法过程，如果符号相反，可以考虑改用加法。
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Mistake: Confusing the x and y values after solving. Label your final answer clearly as x = … and y = … .
错误：Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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