Solving Simultaneous Equations | 联立方程的解法

📚 Solving Simultaneous Equations | 联立方程的解法

A simultaneous equation is a set of two or more equations that share the same unknown variables. When we solve them, we look for a pair (or more) of values that makes every equation true at the same time. This is one of the most examined topics in IGCSE Mathematics, appearing in both Paper 1 and Paper 2.

联立方程是一组共享相同未知数的两个或更多方程。解联立方程时，我们寻找一组能同时满足所有方程的值。这是 IGCSE 数学中最常考的考点之一，在 Paper 1 和 Paper 2 中都会出现。

---

1. What Are Simultaneous Equations? | 什么是联立方程

For example, consider the system below:

2x + y = 7
x − y = 2

The solution is x = 3, y = 1, because both equations hold at the same time: 2(3) + 1 = 7 and 3 − 1 = 2. A solution to a simultaneous equation system must satisfy all equations simultaneously.

方程组的解是 x = 3，y = 1，因为两个方程同时成立：2(3) + 1 = 7，且 3 − 1 = 2。联立方程组的解必须同时满足所有方程。

At IGCSE level you are expected to solve linear systems with two unknowns, and occasionally a system where one equation is linear and the other is quadratic. You should also be able to interpret solutions graphically.

在 IGCSE 阶段，你需要求解含两个未知数的线性方程组，偶尔还会遇到一个线性方程加一个二次方程的方程组。你还应能通过图像解释解的意义。

---

2. The Substitution Method | 代入消元法

The substitution method works best when one equation is already written with a single variable on one side, or when it can be rearranged easily.

代入消元法适用于其中一个方程已经写成一侧只有一个变量的形式，或可以轻松重排的情况。

• Rearrange one equation to make x or y the subject.
• Substitute this expression into the other equation.
• Solve the resulting one-variable equation.
• Substitute back to find the other variable.

• 重排其中一个方程，用另一个变量表示 x 或 y。
• 将该表达式代入另一个方程。
• 解所得的一元方程。
• 代回求出另一个变量。

Worked Example: Solve the system

y = x + 3
2x + y = 15

Since y is already the subject, substitute y = x + 3 into the second equation:

2x + (x + 3) = 15
3x + 3 = 15
3x = 12
x = 4

Then y = 4 + 3 = 7. The solution is x = 4, y = 7. Always check: 2(4) + 7 = 15 ✓

因为 y 已经是主项，将 y = x + 3 代入第二个方程：2x + (x + 3) = 15，即 3x + 3 = 15，解得 3x = 12，所以 x = 4。于是 y = 4 + 3 = 7。解为 x = 4，y = 7。务必检验：2(4) + 7 = 15 ✓

---

3. The Elimination Method | 加减消元法

The elimination method combines the two equations by adding or subtracting them so that one variable disappears. It is often the fastest method when both equations are in the form Ax + By = C.

加减消元法通过将两个方程相加或相减，使其中一个变量消去。当两个方程都写成 Ax + By = C 的形式时，这通常是最快的方法。

• Arrange both equations in the form Ax + By = C.
• Multiply one or both equations so that the coefficients of x or y match in size.
• Add or subtract to eliminate that variable.
• Solve for the remaining variable, then substitute back.

• 将两个方程整理成 Ax + By = C 的形式。
• 将其中一个或两个方程乘以适当的数，使 x 或 y 的系数绝对值相等。
• 相加或相减以消去该变量。
• 解出剩余变量，再代回求另一个变量。

Worked Example: Solve

3x + 2y = 12
2x + 2y = 8

Both equations have the same y coefficient, 2. Subtract the second equation from the first:

(3x + 2y) − (2x + 2y) = 12 − 8
x = 4

Substitute x = 4 into 2x + 2y = 8: 8 + 2y = 8, so 2y = 0, hence y = 0. The solution is x = 4, y = 0.

两个方程的 y 系数相同，都是 2。用第一个方程减去第二个方程：x = 4。将 x = 4 代入 2x + 2y = 8：8 + 2y = 8，故 2y = 0，即 y = 0。解为 x = 4，y = 0。

When the coefficients are not the same, multiply first. For example, to eliminate x from 2x + y = 5 and x − 2y = 0, multiply the second equation by 2 to obtain 2x − 4y = 0, then subtract.

当系数不同时，先乘以适当的数。例如，要从 2x + y = 5 和 x − 2y = 0 中消去 x，可将第二个方程乘以 2，得到 2x − 4y = 0，然后相减。

---

4. Graphical Interpretation | 图像解释

Each linear equation represents a straight line on the x-y plane. The simultaneous solution is the point where the two lines intersect, i.e. the point whose coordinates satisfy both equations.

每个线性方程在 x-y 平面上代表一条直线。联立方程组的解就是两条直线的交点，即其坐标同时满足这两个方程的点。

To draw a line quickly, rearrange into the form y = mx + c, or find the x-intercept and y-intercept. For example, the line 2x + y = 7 passes through (0, 7) and (3.5, 0).

要快速画出直线，可将方程重排为 y = mx + c 的形式，或者求出 x 截距和 y 截距。例如，直线 2x + y = 7 经过 (0, 7) 和 (3.5, 0) 两点。

When solving by graph, draw both lines accurately using the same scale, then read the intersection point. The coordinates of the intersection give x and y.

用图像法求解时，需用同一比例尺准确画出两条直线，然后读出交点坐标。交点的横纵坐标就是 x 和 y 的值。

---

5. Special Cases: No Solution and Infinite Solutions | 特殊情况：无解与无穷多解

No solution: If the two lines are parallel, they never meet, so the system has no solution. This happens when the equations have the same gradient but different y-intercepts.

无解：如果两条直线平行，它们永不相交，所以方程组无解。当两个方程的斜率相同但 y 截距不同时，就会发生这种情况。

x + y = 5
x + y = 8

Subtracting gives 0 = 3, which is impossible. In the elimination method, arriving at a false statement such as 0 = 3 means the lines are parallel and there is no solution.

相减得到 0 = 3，这是不可能的。在加减消元法中，如果得出 0 = 3 这样的假命题，就说明两直线平行，方程组无解。

Infinite solutions: If the two equations represent the same line, every point on that line is a solution.

无穷多解：如果两个方程代表同一条直线，那么该直线上的每一个点都是解。

x + y = 5
2x + 2y = 10

Here the second equation is exactly double the first. Subtracting gives 0 = 0, a true statement, meaning infinitely many solutions.

这里第二个方程正好是第一个方程的两倍。相减得到 0 = 0，这是真命题，意味着有无穷多组解。

---

6. Solving Word Problems | 解应用题

Exam questions often present a real-world scenario. The key skill is translating words into algebraic equations.

考试题常给出实际情境。关键技能是将文字转化为代数方程。

Worked Example: The sum of two numbers is 25 and their difference is 3. Find the two numbers.

例题：两个数之和为 25，之差为 3。求这两个数。

Let the two numbers be x and y. Then:

x + y = 25
x − y = 3

Add the two equations: 2x = 28, so x = 14. Then y = 25 − 14 = 11. The numbers are 14 and 11.

设两个数为 x 和 y，则 x + y = 25，x − y = 3。两式相加：2x = 28，所以 x = 14。则 y = 25 − 14 = 11。这两个数是 14 和 11。

Always define your variables clearly in the exam. This gains method marks even if you make a small arithmetic slip later.

考试中务必清楚地定义变量。即使后面出现小的计算失误，也能获得方法分。

---

7. Non-Linear Simultaneous Equations | 非线性联立方程组

In higher-tier IGCSE papers, you may meet one linear equation and one quadratic equation. Use the substitution method: substitute the linear expression into the quadratic equation.

在 IGCSE 高等卷（Higher Tier）中，你可能会遇到一个线性方程加一个二次方程的情况。使用代入法：将线性表达式代入二次方程。

Worked Example: Solve

y = x + 2
y = x²

Substitute y = x + 2 into y = x²:

x + 2 = x²
x² − x − 2 = 0
(x − 2)(x + 1) = 0

So x = 2 or x = −1. When x = 2, y = 4; when x = −1, y = 1. The solutions are (2, 4) and (−1, 1).

将 y = x + 2 代入 y = x²：x + 2 = x²，即 x² − x − 2 = 0，因式分解得 (x − 2)(x + 1) = 0。所以 x = 2 或 x = −1。当 x = 2 时，y = 4；当 x = −1 时，y = 1。解为 (2, 4) 和 (−1, 1)。

Graphically, the line cuts the curve at two points, giving two solutions. This is why a quadratic system can have up to two pairs of solutions.

从图像上看，直线与曲线在两个点相交，因此得到两组解。这就是为什么二次方程组最多可以有两组解。

---

8. Common Exam Mistakes | 常见考试错误

• Sign errors when subtracting: When subtracting one equation from another, every term must change sign, not just the first term.
• Forgot to substitute back: Finding one variable is only half the solution; you must substitute back to find the second variable.
• No final check: Always check your answer in both original equations.
• Incorrect multiplication: When multiplying an equation, multiply every term on both sides by the same number.

• 相减时符号错误：一个方程减去另一个方程时，每一项都要变号，而不只是第一项。
• 忘记代回：求出一个变量只是完成一半；必须代回求出第二个变量。
• 未进行最终检验：始终将答案代回两个原方程中检验。
• 乘法错误：方程两边乘以一个数时，必须每一项都乘以该数。

A good habit is to write your two values clearly and substitute them into both equations before moving to the next question.

一个好习惯是清晰写出两个值，并在做下一题之前将它们代入两个方程进行检验。

---

9. Practice Questions | 练习题

Try these questions yourself, then check the answers in the table below.

请自行尝试以下题目，然后对照下面表格中的答案检查。

1. Solve by elimination: 5x + y = 17 and 3x − y = 7.
2. Solve by substitution: y = 2x − 1 and 3x + 2y = 12.
3. Solve the system: 2x + 3y = 13 and 5x + 3y = 28.
4. State whether this system has no solution or infinite solutions: y = 3x + 1 and 2y = 6x + 2.

1. 用加减消元法解：5x + y = 17 和 3x − y = 7。
2. 用代入消元法解：y = 2x − 1 和 3x + 2y = 12。
3. 解方程组：2x + 3y = 13 和 5x + 3y = 28。
4. 判断下列方程组无解还是有无穷多解：y = 3x + 1 和 2y = 6x + 2。

Question | 题目	Answer | 答案
1	x = 3, y = 2
2	x = 2, y = 3
3	x = 5, y = 1
4	Infinite solutions | 无穷多解

---

10. Summary | 总结

Choose the method that suits the question. Use elimination when coefficients already match or can match easily; use substitution when one variable is already isolated or when one equation is quadratic.

选择适合题目的方法。当系数已经相同或容易配成相同时用加减消元法；当某个变量已经被分离，或有一个方程是二次方程时，用代入消元法。

Method | 方法	Best Used When | 适用时机	Key Skill | 关键技能
Elimination | 加减消元	Both equations are linear, coefficients match easily	Careful addition and subtraction of equations
Substitution | 代入消元	One variable is already isolated, or a quadratic is involved	Substituting and solving quadratics
Graphical | 图像法	Approximate answer is acceptable	Drawing straight lines accurately

Remember the golden rule: always check your answer by substituting it back into the original equations. This simple step saves many marks.

请记住黄金法则：始终将答案代回原方程检验。这个简单的步骤能帮你避免失分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）