📚 Mastering Simultaneous Equations: Elimination, Substitution and Graphs | 解联立方程：消元法、代入法与图像法
Simultaneous equations are a cornerstone of IGCSE Mathematics. They appear in algebra, geometry, coordinate geometry, and even in word problems from everyday life. This article will guide you through every essential method, from the basic elimination and substitution techniques to the more challenging combination of linear and quadratic equations. You will also learn how to solve them graphically, how to interpret the number of solutions using the discriminant, and how to avoid common traps in the exam.
联立方程是 IGCSE 数学的核心内容。无论是在代数、几何、坐标几何，还是日常生活中的应用题，它们都会频繁出现。这篇文章将带你系统掌握每一种重要方法：从基础的消元法和代入法，到更具挑战性的线性方程与二次方程联立求解。你还会学到如何用图像法解题、如何利用判别式判断解的个数，以及在考试中如何避免常见易错点。
1. What Are Simultaneous Equations? | 什么是联立方程？
A single equation with two variables, such as y = 2x + 1, has infinitely many solutions. Each solution corresponds to a point on a straight line. When we have two equations that must be true at the same time, they are called simultaneous equations. The solution is the set of x and y values that satisfy both equations simultaneously, which geometrically represents the point where the two lines intersect.
单个含有两个变量的方程，例如 y = 2x + 1，通常有无穷多组解。每一组解都对应直线上的一个点。当我们拥有两个必须同时成立的方程时，它们称为联立方程。联立方程的解是同时满足这两个方程的 x 和 y 的值，从几何角度看，就是两条直线的交点坐标。
For example, consider the system:
x + y = 5, x − y = 1
A moment of inspection shows that x = 3 and y = 2, because 3 + 2 = 5 and 3 − 2 = 1. In the exam you will need a reliable method to find such solutions without guessing.
例如，考虑方程组：
x + y = 5, x − y = 1
稍加观察可得 x = 3，y = 2，因为 3 + 2 = 5 且 3 − 2 = 1。在考试中，你需要可靠的方法来求出解，而不是依赖猜测。
2. The Elimination Method | 消元法
Elimination is often the fastest method when both equations are linear. The idea is to add or subtract the equations so that one variable is eliminated, leaving a single equation in one unknown.
消元法通常适用于两个方程都为线性方程的情况，速度最快。其核心思想是通过将两个方程相加或相减，消去其中一个变量，从而得到只含一个未知数的一元方程。
Steps:
- Write both equations in the form ax + by = c.
- Make the coefficients of one variable the same (and opposite for addition) by multiplying one or both equations, if necessary.
- Add or subtract the equations to eliminate that variable.
- Solve the resulting linear equation for the remaining variable.
- Substitute the value back into one of the original equations to find the other variable.
解题步骤：
- 将两个方程都写成 ax + by = c 的形式。
- 通过适当乘以一个或两个方程，使得某一个变量的系数相同（若用加法则还需符号相反）。
- 将两个方程相加或相减，以消去该变量。
- 解出得到的一元一次方程，求出第一个变量。
- 将所求值代回任一原方程，求出另一个变量。
Let’s solve a worked example:
3x + 2y = 12, 2x − 3y = 8
To eliminate x, multiply the first equation by 2 and the second by 3:
6x + 4y = 24, 6x − 9y = 24
Subtract the second from the first: (6x + 4y) − (6x − 9y) = 24 − 24, which gives 13y = 0, so y = 0. Substitute y = 0 into 3x + 2(0) = 12, so 3x = 12, x = 4. The solution is x = 4, y = 0.
下面通过一个例子演示：
3x + 2y = 12, 2x − 3y = 8
为了消去 x，将第一个方程乘以 2，第二个方程乘以 3：
6x + 4y = 24, 6x − 9y = 24
两式相减：(6x + 4y) − (6x − 9y) = 24 − 24，得到 13y = 0，所以 y = 0。将 y = 0 代入 3x + 2(0) = 12，得 3x = 12，x = 4。因此解为 x = 4，y = 0。
3. The Substitution Method | 代入法
Substitution is particularly useful when one equation already has a variable on its own, such as y = mx + c, or when dealing with non-linear equations. The method involves replacing one variable with an expression from the other equation.
代入法特别适用于其中一个方程已经隔离出某个变量（例如 y = mx + c）的情况，也常用于处理非线性方程。代入法的本质是用一个方程中的表达式替换另一个方程中的变量。
Steps:
- Rearrange one of the equations to make either x or y the subject.
- Substitute this expression into the other equation.
- Solve the resulting single-variable equation.
- Substitute the found value back into the rearranged equation to find the second variable.
步骤：
- 将其中一个方程变形，使其表示出 x 或 y 的表达式。
- 把这个表达式代入另一个方程。
- 解出这个一元方程。
- 将所求值代回变形后的方程，求出另一个变量。
Worked example:
y = 2x + 1, x + 2y = 11
The first equation already gives y = 2x + 1. Substitute into x + 2(2x + 1) = 11, so x + 4x + 2 = 11, giving 5x = 9, so x = 1.8. Then y = 2(1.8) + 1 = 4.6. The solution is x = 1.8, y = 4.6.
示例：
y = 2x + 1, x + 2y = 11
第一个方程已经给出 y = 2x + 1。将其代入 x + 2(2x + 1) = 11，得 x + 4x + 2 = 11，即 5x = 9，所以 x = 1.8。再求 y = 2(1.8) + 1 = 4.6。解为 x = 1.8，y = 4.6。
4. Solving Linear Simultaneous Equations by Graphs | 用图像法解线性联立方程
Graphically, every linear equation can be drawn as a straight line. The point where the two lines cross represents the unique solution of the system. If the lines are parallel, there is no solution; if they are the same line, there are infinitely many solutions.
从图像上看，每个线性方程都可以画成一条直线。两条直线的交点坐标就是方程组的唯一解。若两直线平行，则方程组无解；若两直线完全重合，则有无穷多组解。
Example: Draw the lines y = 2x + 1 and x + 2y = 11 on the same axes. From the previous section, the solution is (1.8, 4.6). The lines should intersect at that point.
例如：在同一坐标轴上画出 y = 2x + 1 和 x + 2y = 11 两条直线。由上节可知，解为 (1.8, 4.6)，两条直线的交点应当在这个点上。
To use this method accurately in the exam you must:
- Use graph paper and a sharp pencil.
- Choose a suitable scale so that the intersection is clearly visible.
- Plot at least three points for each line to ensure accuracy.
- Read the coordinates of the intersection carefully.
在考试中使用图像法时，你需要：
- 使用方格纸和削尖的铅笔。
- 选择合适的刻度，使交点清晰可见。
- 每条直线至少描出三个点，确保准确。
- 仔细读取交点的坐标。
5. Forming Equations from Word Problems | 从应用题建立方程
Many real-world problems require you to translate words into algebra. The key is to define variables clearly, then create two independent equations from the given information.
许多实际问题需要你把文字转化为代数式。关键是清晰定义变量，然后根据题目信息建立两个独立的方程。
For example: A shop sells apples and bananas. Two apples and three bananas cost $5, while one apple and four bananas cost $4. Let the cost of an apple be a and a banana be b. Then:
2a + 3b = 5, a + 4b = 4
Solve by elimination: multiply the second equation by 2 to get 2a + 8b = 8. Subtract the first: (2a + 8b) − (2a + 3b) = 8 − 5, so 5b = 3, b = 0.6. Substitute into a + 4(0.6) = 4, so a = 1.6. One apple costs $1.60 and one banana costs $0.60.
例如：一家商店出售苹果和香蕉。两个苹果和三个香蕉需要 5 美元，而一个苹果和四个香蕉需要 4 美元。设苹果单价为 a，香蕉单价为 b。则：
2a + 3b = 5, a + 4b = 4
用消元法：第二个方程乘以 2，得 2a + 8b = 8。两式相减：(2a + 8b) − (2a + 3b) = 8 − 5，即 5b = 3，b = 0.6。代入 a + 4(0.6) = 4，得 a = 1.6。因此一个苹果 1.60 美元，一个香蕉 0.60 美元。
When forming equations, always check that your equations describe the relationships correctly. Write down the meaning of each variable, and use units consistently.
在建立方程时，务必检查方程是否准确描述了题目中的关系。写清每个变量的含义，并保持单位一致。
6. Handling Fractions and Negative Numbers | 处理分数与负数
Simultaneous equations often involve fractions or negative coefficients. These can be simplified by multiplying each equation by the lowest common multiple of the denominators to obtain integer coefficients.
联立方程中经常出现分数或负系数。此时可以将每个方程乘以分母的最小公倍数，从而把分数化为整数系数，简化计算。
Example:
x/2 + y/3 = 4, x/3 − y/4 = 1/6
Multiply the first equation by 6: 3x + 2y = 24. Multiply the second by 12: 4x − 3y = 2. Now solve the integer system. Multiply the first by 3: 9x + 6y = 72. Multiply the second by 2: 8x − 6y = 4. Add: 17x = 76, so x = 76/17. Substitute to find y = (24 − 3x)/2 = (24 − 228/17)/2 = (408/17 − 228/17)/2 = (180/17)/2 = 90/17. Hence x = 76/17, y = 90/17.
示例：
x/2 + y/3 = 4, x/3 − y/4 = 1/6
第一个方程乘以 6：3x + 2y = 24。第二个方程乘以 12：4x − 3y = 2。现在解整数方程组。第一个方程乘以 3：9x + 6y = 72。第二个方程乘以 2：8x − 6y = 4。两式相加：17x = 76，所以 x = 76/17。代入得 y = (24 − 3x)/2 = (24 − 228/17)/2 = (408/17 − 228/17)/2 = (180/17)/2 = 90/17。因此解为 x = 76/17，y = 90/17。
When dealing with negative numbers, remember to apply the sign rules carefully, especially when subtracting an expression containing a negative term.
当处理负数时，要特别注意符号规则，尤其是在减去含有负项的表达式时，不要漏变号。
7. Solving One Linear and One Quadratic Equation | 解一个线性方程和一个二次方程
At IGCSE, you are expected to solve a system where one equation is linear and the other is quadratic. Such a system can have 0, 1, or 2 solutions. The substitution method is normally the most straightforward.
在 IGCSE 中，你需要解决一个线性方程与一个二次方程组成的方程组。这类方程组可能有 0 个、1 个或 2 个解。通常使用代入法最为直接。
Example:
y = x² − 3x + 1, y = 2x − 1
Set the two expressions for y equal: x² − 3x + 1 = 2x − 1. Rearrange: x² − 5x + 2 = 0. Use the quadratic formula:
x = [5 ± √(25 − 8)] / 2 = (5 ± √17) / 2
So x ≈ 4.56 or x ≈ 0.44. Substitute each back into y = 2x − 1: when x ≈ 4.56, y ≈ 8.12; when x ≈ 0.44, y ≈ −0.12. Thus the two intersection points are approximately (4.56, 8.12) and (0.44, −0.12).
示例：
y = x² − 3x + 1, y = 2x − 1
令两个 y 表达式相等：x² − 3x + 1 = 2x − 1。整理得 x² − 5x + 2 = 0。使用求根公式：
x = [5 ± √(25 − 8)] / 2 = (5 ± √17) / 2
因此 x ≈ 4.56 或 x ≈ 0.44。将其代回 y = 2x − 1：当 x ≈ 4.56 时，y ≈ 8.12；当 x ≈ 0.44 时，y ≈ −0.12。所以两个交点约为 (4.56, 8.12) 和 (0.44, −0.12)。
8. The Discriminant and Types of Solutions | 判别式与解的类型
When we substitute a linear equation into a quadratic, we obtain a quadratic equation in x. The discriminant, Δ = b² − 4ac, tells us how many real solutions exist:
当我们将线性方程代入二次方程后，会得到一个关于 x 的二次方程。判别式 Δ = b² − 4ac 可以告诉我们实数解的个数：
| Δ = b² − 4ac | Number of Real Solutions | 几何意义 |
| Δ > 0 | 2 distinct solutions | 直线与抛物线相交于两点 |
| Δ = 0 | 1 repeated solution (tangent) | 直线与抛物线相切于一点 |
| Δ < 0 | 0 real solutions | 直线与抛物线没有交点 |
In the previous example, a = 1, b = −5, c = 2, so Δ = 25 − 8 = 17 > 0, confirming two real solutions. This is a quick way to check your answers before calculating them explicitly.
在上一个例子中，a = 1，b = −5，c = 2，因此 Δ = 25 − 8 = 17 > 0，说明确实有两个实数解。这是一个在完整计算前快速验证结果是否合理的方法。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Students frequently lose marks on simultaneous equations for a few preventable reasons. Recognising these traps can dramatically improve your score.
学生在联立方程题中常因几个可避免的原因失分。识别这些常见陷阱可以显著提高你的分数。
- Forgetting to reverse signs when subtracting. When you subtract one equation from another, every term of the second equation must be subtracted. Always write the expression in parentheses.
- Sign errors with negative coefficients. Double-check each step using a small verification: put the solution back into both original equations.
- Substituting back into the transformed (changed) equation. Use the original equation, or check the final pair in both originals.
- Not rearranging equations correctly. For substitution, make sure the expression is correctly isolated and all terms are handled.
- Forgetting to include both solutions. When solving non-linear systems, a quadratic may have two roots; both need to be substituted back.
- 相减时忘记变号。用一个方程减去另一个方程时，第二个方程的每一项都要变号。写式子时应加上括号。
- 负系数导致的符号错误。每一步都要仔细检验，最后把解代回两个原方程进行验证。
- 代回变形后的方程而不是原方程。应该使用原方程，或在最后将解代入两个原方程加以确认。
- 变形时出错。使用代入法时，要确保变量的表达式隔离正确，且所有项都被正确处理。
- 遗漏解。解非线性方程组时，二次方程通常有两个根，两个根都要代回求解。
10. Practice Questions with Worked Solutions | 练习题与详解
Here are three typical IGCSE exam-style questions. Try them yourself before reading the solutions.
以下三道典型的 IGCSE 考试风格练习。请先尝试自己解答，再对照解析。
Question 1: Solve by elimination: 5x + 3y = 19, 2x − y = 1.
Solution: Multiply the second equation by 3: 6x − 3y = 3. Add the first: (5x + 3y) + (6x − 3y) = 19 + 3, so 11x = 22, x = 2. Substitute into 2(2) − y = 1, so y = 3. Answer: (2, 3).
第1题：用消元法解：5x + 3y = 19, 2x − y = 1。
解析：第二个方程乘以 3：6x − 3y = 3。与第一个方程相加：(5x + 3y) + (6x − 3y) = 19 + 3，得 11x = 22，x = 2。代入 2(2) − y = 1，得 y = 3。答案：(2, 3)。
Question 2: Solve by substitution: y = x − 4, x² + y² = 10.
Solution: Substitute y = x − 4 into the quadratic: x² + (x − 4)² = 10. Expand: x² + x² − 8x + 16 = 10, so 2x² − 8x + 6 = 0, x² − 4x + 3 = 0. Factor: (x − 1)(x − 3) = 0, so x = 1 or 3. Then y = −3 or −1. Solutions: (1, −3) and (3, −1).
第2题：用代入法解：y = x − 4, x² + y² = 10。
解析：将 y = x − 4 代入二次方程：x² + (x − 4)² = 10。展开得 x² + x² − 8x + 16 = 10，整理为 2x² − 8x + 6 = 0，即 x² − 4x + 3 = 0。因式分解得 (x − 1)(x − 3) = 0，因此 x = 1 或 3。于是 y = −3 或 −1。解为 (1, −3) 和 (3, −1)。
Question 3: The sum of two numbers is 14 and their difference is 2. Find the numbers.
Solution: Let the numbers be a and b. a + b = 14, a − b = 2. Add: 2a = 16, a = 8. Then b = 6. The numbers are 8 and 6.
第3题：两个数的和为 14，差为 2，求这两个数。
解析：设这两个数为 a 和 b。联立 a + b = 14，a − b = 2。两式相加得 2a = 16，a = 8，则 b = 6。因此两个数为 8 和 6。
11. Summary and Tips for the Exam | 总结与考试技巧
Simultaneous equations are manageable if you follow a consistent method and check your work. In the exam, always decide which method is fastest: elimination for clean integer coefficients, substitution for equations already solved for a variable or for non-linear cases, and graphical methods only when asked or as a check.
联立方程只要用统一的方法并验算，就不难掌握。在考试中，先判断哪种方法最合适：系数整齐时用消元法；方程已经表达出某个变量或含二次方程时用代入法；只有在题目要求作图或将图像作为检验时，才用图像法。
- Read the question carefully and identify whether it is linear/linear or linear/quadratic.
- Write down all equations clearly, and state what your variables represent in word problems.
- After finding the solution, substitute both values back into both original equations.
- Show all working steps; many marks are awarded for method, not just the final
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