Mastering Simultaneous Equations | 解方程组完全指南

📚 Mastering Simultaneous Equations | 解方程组完全指南

Simultaneous equations are a set of two or more equations that share two or more unknown variables. In IGCSE Mathematics, you are most often asked to solve two equations for two unknowns, typically x and y. The word ‘simultaneous’ means that the same values of x and y must satisfy both equations at the same time.

联立方程（方程组）是指包含两个或两个以上未知数的多个方程。在IGCSE数学中，最常见的问题是解两个方程求两个未知数，通常为 x 和 y。”联立”的含义是：同一组 x 和 y 的值必须同时满足所有方程。

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1. The Graphical Meaning | 图像意义

Each linear equation in two variables represents a straight line on the Cartesian plane. When you solve a pair of simultaneous linear equations, you are finding the coordinates of the point where the two lines intersect. This point is the only pair of values that lies on both lines.

每个含有两个变量的线性方程在直角坐标平面上都代表一条直线。当你解一对联立线性方程时，其实是在求两条直线的交点坐标。这个点是唯一同时位于两条直线上的数值对。

If the two lines are parallel, they never intersect, and the equations have no solution. If the two equations represent the same line, they intersect everywhere, meaning there are infinitely many solutions. Understanding this visual picture helps you interpret your algebraic results.

如果两条直线平行，它们永不相交，方程组无解。如果两个方程表示同一条直线，它们处处相交，意味着有无穷多组解。理解这一图像有助于你解释代数运算得出的结果。

Example: y = 2x + 1 and y = −x + 4 intersect at (1, 3).

例：y = 2x + 1 与 y = −x + 4 的交点为 (1, 3)。

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2. The Elimination Method | 消元法

The elimination method, also called the addition–subtraction method, works by making the coefficient of one variable the same in both equations, then adding or subtracting the equations to eliminate that variable.

消元法，也称加减消元法，其核心是使两个方程中某个未知数的系数相同，然后通过两式相加或相减来消去该未知数。

Step-by-step procedure:

分步步骤：

• Arrange both equations so that the x-terms, y-terms and constants line up vertically.
• Multiply one or both equations by constants so that the coefficients of x (or y) become equal in magnitude.
• Add the equations if the signs are opposite; subtract if the signs are the same.
• Solve the resulting one-variable equation.
• Substitute the value back into either original equation to find the other variable.

• 将两个方程整理为 x 项、y 项和常数项纵向对齐的形式。
• 将一个或两个方程乘以适当的常数，使 x（或 y）的系数绝对值相等。
• 若系数符号相反则两式相加；若符号相同则两式相减。
• 解得到的一元一次方程。
• 将求出的值代回任一原方程，求出另一个未知数。

Worked example:

例题：

3x + 2y = 13 … (1)
2x − 2y = 2 … (2)

Notice that the coefficients of y are +2 and −2, which are opposite signs. Adding the two equations eliminates y:

注意到 y 的系数分别为 +2 和 −2，符号相反。将两式相加即可消去 y：

(3x + 2x) + (2y − 2y) = 13 + 2
5x = 15
x = 3

Substitute x = 3 into equation (1):

将 x = 3 代入方程 (1)：

3(3) + 2y = 13
9 + 2y = 13
2y = 4
y = 2

Therefore the solution is x = 3, y = 2. Always verify by substituting into equation (2): 2(3) − 2(2) = 6 − 4 = 2 ✓

因此解为 x = 3，y = 2。一定要代入方程 (2) 进行验证：2(3) − 2(2) = 6 − 4 = 2 ✓

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3. The Substitution Method | 代入法

The substitution method involves rearranging one equation to make one variable the subject, then substituting that expression into the other equation. This method is especially useful when one equation has a coefficient of 1 for x or y.

代入法的步骤是：先将一个方程变形，使某个未知数用另一个未知数的表达式表示，然后将该表达式代入另一个方程。当一个方程中 x 或 y 的系数为 1 时，代入法尤其方便。

Worked example:

例题：

y = 2x − 1 … (1)
3x + 2y = 12 … (2)

Equation (1) already gives y in terms of x. Substitute y = 2x − 1 into equation (2):

方程 (1) 已经给出了 y 关于 x 的表达式。将 y = 2x − 1 代入方程 (2)：

3x + 2(2x − 1) = 12
3x + 4x − 2 = 12
7x = 14
x = 2

Now substitute x = 2 back into equation (1):

再将 x = 2 代回方程 (1)：

y = 2(2) − 1 = 3

The solution is x = 2, y = 3. Check in equation (2): 3(2) + 2(3) = 6 + 6 = 12 ✓

解为 x = 2，y = 3。代入方程 (2) 检验：3(2) + 2(3) = 6 + 6 = 12 ✓

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4. Solving Linear + Quadratic Systems | 线性与二次方程组

In this type of problem, one equation is linear and the other is quadratic. Because the quadratic equation contains an x² term, the substitution method is normally the simplest approach. The intersection of a straight line and a curve can yield two solutions, one solution or no solution.

在这类问题中，一个方程是线性的，另一个是二次的。由于二次方程含有 x² 项，通常使用代入法最为简单。直线与曲线的交点可能有两个、一个或零个解。

Worked example:

例题：

y = x + 2 … (1)
y = x² − 3x + 4 … (2)

Substitute y = x + 2 into equation (2):

将 y = x + 2 代入方程 (2)：

x + 2 = x² − 3x + 4
0 = x² − 4x + 2

This quadratic does not factorise nicely, so use the quadratic formula:

该二次方程不易因式分解，因此使用求根公式：

x = [4 ± √(16 − 8)] ÷ 2 = [4 ± √8] ÷ 2 = 2 ± √2

The two x-values are 2 + √2 ≈ 3.41 and 2 − √2 ≈ 0.59. Substitute each into y = x + 2 to find the corresponding y-values. The solutions are approximately (3.41, 5.41) and (0.59, 2.59).

两个 x 值为 2 + √2 ≈ 3.41 和 2 − √2 ≈ 0.59。将每个值代入 y = x + 2 求出相应的 y 值。解约为 (3.41, 5.41) 和 (0.59, 2.59)。

Note that you must match each x-value with its corresponding y-value. A common mistake is to swap the pairs. Also, when substituting back, always use the linear equation because it is simpler and avoids confusion.

注意：必须将每个 x 值与对应的 y 值正确配对。常见错误是颠倒配对。另外，代回时务必使用一次方程，因为它更简单且不易出错。

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5. Solving Quadratic + Quadratic Systems | 二次与二次方程组

When both equations are quadratic, you need to be especially careful with algebraic manipulation. The most reliable approach is to set the two expressions for y equal to each other, since both equations can usually be rearranged into the form y = …

当两个方程都是二次方程时，代数运算需要格外小心。最可靠的方法是将两个 y 的表达式相等，因为两个方程通常都可以整理成 y = … 的形式。

Worked example:

例题：

y = x² − 2x + 1 … (1)
y = 2x² − x − 5 … (2)

Set the right-hand sides equal:

令两式右侧相等：

x² − 2x + 1 = 2x² − x − 5
0 = x² + x − 6
0 = (x + 3)(x − 2)

So x = −3 or x = 2. Substitute into equation (1):

因此 x = −3 或 x = 2。代入方程 (1)：

When x = −3, y = (−3)² − 2(−3) + 1 = 9 + 6 + 1 = 16.
When x = 2, y = 4 − 4 + 1 = 1.

当 x = −3 时，y = (−3)² − 2(−3) + 1 = 9 + 6 + 1 = 16。
当 x = 2 时，y = 4 − 4 + 1 = 1。

The two solutions are (−3, 16) and (2, 1). Always check both pairs in the second equation as well.

两组解为 (−3, 16) 和 (2, 1)。务必也把两组解代入第二个方程进行检验。

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6. Solving Equations with Fractions | 含分数方程组的解法

Some IGCSE questions present equations containing fractions, such as x/2 + y/3 = 4. Before applying elimination or substitution, multiply every term by the lowest common denominator (LCD) to clear the fractions.

有些IGCSE题目给出的方程含有分数，例如 x/2 + y/3 = 4。在应用消元法或代入法之前，需要先将每一项乘以最小公分母，以去掉分母。

Worked example:

例题：

x/2 + y/3 = 4 … (1)
x/4 − y/2 = 0.5 … (2)

Multiply equation (1) by 6 and equation (2) by 4:

将方程 (1) 乘以 6，方程 (2) 乘以 4：

3x + 2y = 24 … (1′)
x − 2y = 2 … (2′)

Now add (1′) and (2′) to eliminate y:

现在将 (1′) 与 (2′) 相加以消去 y：

4x = 26
x = 6.5

Substitute x = 6.5 into (2′):

将 x = 6.5 代入 (2′)：

6.5 − 2y = 2
2y = 4.5
y = 2.25

The solution is x = 6.5, y = 2.25. Checking in (1): 6.5/2 + 2.25/3 = 3.25 + 0.75 = 4 ✓

解为 x = 6.5，y = 2.25。代入 (1) 检验：6.5/2 + 2.25/3 = 3.25 + 0.75 = 4 ✓

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7. Solving Word Problems | 应用题建模

Examination problems often wrap simultaneous equations inside a real-world context. The key skill is translating words into algebra. Look for two unknown quantities and two independent pieces of information.

考试题目常常把方程组嵌入实际情境中。关键技能是将文字转化为代数表达式。寻找两个未知量和两条独立信息。

Example problem:

例题：

A shop sells apples and bananas. Two apples and three bananas cost $2.90. Five apples and two bananas cost $4.90. Find the cost of one apple and one banana.

某商店出售苹果和香蕉。2个苹果和3根香蕉共2.90美元，5个苹果和2根香蕉共4.90美元。求1个苹果和1根香蕉的价格。

Let a = cost of one apple and b = cost of one banana. Then:

设 a = 1个苹果的价格，b = 1根香蕉的价格。则有：

2a + 3b = 2.90 … (1)
5a + 2b = 4.90 … (2)

Multiply (1) by 2 and (2) by 3 to equalise b-coefficients:

将 (1) 乘以 2，(2) 乘以 3，使 b 的系数相等：

4a + 6b = 5.80 … (1′)
15a + 6b = 14.70 … (2′)

Subtract (1′) from (2′):

用 (2′) 减去 (1′)：

11a = 8.90
a = 0.8091… ≈ 0.81

Substitute a = 0.81 into (1):

将 a = 0.81 代入 (1)：

2(0.81) + 3b = 2.90
1.62 + 3b = 2.90
3b = 1.28
b ≈ 0.427 ≈ 0.43

So one apple costs approximately $0.81 and one banana costs approximately $0.43. Always state your final answer in the context of the question.

因此1个苹果约0.81美元，1根香蕉约0.43美元。最后一定要结合题意写出答案。

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8. Common Mistakes and How to Avoid Them | 常见错误与避坑指南

Many marks are lost in simultaneous equation questions due to small algebraic slips. Here are the most frequent errors and how to avoid them.

许多分数因为小的代数失误而丢失。以下是最高频的错误及规避方法。

Mistake | 错误	Correction | 更正
Forgetting to multiply every term when clearing fractions	Multiply every single term by the LCD, including the constant.
去分母时漏乘某些项	将每一项（包括常数项）都乘以公分母。
Sign errors when subtracting equations	Write each equation in brackets, then distribute the minus sign.
两式相减时符号出错	用括号写出每个方程，再逐项分配负号。
Mixing up x and y solution pairs	Always write solutions as (x, y) and test both equations.
弄混 x 与 y 的解对	始终按 (x, y) 形式书写并代入两式检验。
Substituting into the wrong equation for quadratic systems	Use the linear equation whenever one exists.
解二次方程组时代回错误的方程	只要存在一次方程就代回一次方程。

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9. Exam Techniques and Timing | 考试技巧与时间分配

Simultaneous equation questions are usually worth 3 to 6 marks in the IGCSE paper. A well-organised written solution helps you earn method marks even if you make a numerical slip.

方程组题目在IGCSE试卷中通常占3至6分。条理清晰的书写过程能帮助你在数值出错时仍获得方法分。

• Show every step of the elimination or substitution process, not just the final answer.
• Label your equations (1) and (2), and clearly state which equations you add or subtract.
• Always check your final answer in both original equations; this takes only 10 seconds.
• If time is short, attempt the first few steps — they often earn method marks.

• 展示消元或代入过程中的每一步，而不仅仅写最终答案。
• 将方程标号为 (1) 和 (2)，并清楚写出相加或相减的是哪两个方程。
• 始终将最终答案代入两个原方程验算；这只需10秒钟。
• 如果时间紧张，尝试写出前几步——这些通常能获得方法分。

For the extended paper (IGCSE), be prepared for simultaneous equations involving one linear and one quadratic equation. Practise until the substitution steps feel automatic.

对于扩展卷（IGCSE），要准备好一个线性方程与一个二次方程联立的题目。反复练习，直到代入步骤成为本能反应。

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10. Practice Problems | 实战练习

Work through the following problems on your own before checking the answers below.

先独立完成以下题目，再对照下方答案。

Problem 1 | 题1: Solve: 5x + 3y = 19 and 2x − y = 1.

Problem 1 | 题1：解方程组：5x + 3y = 19，2x − y = 1。

Problem 2 | 题2: Solve: y = 3x − 5 and y = x² − 2x − 1.

Problem 2 | 题2：解方程组：y = 3x − 5，y = x² − 2x − 1。

Problem 3 | 题3: The sum of two numbers is 28 and their difference is 12. Find the numbers.

Problem 3 | 题3：两个数之和为28，差为12，求这两个数。

Answers | 答案：

Problem 1: x = 2, y = 3.
Problem 2: Substitute 3x − 5 = x² − 2x − 1 → x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0, so x = 1 or x = 4. Corresponding y-values: y = −2 and y = 7. Solutions: (1, −2) and (4, 7).
Problem 3: Let a + b = 28 and a − b = 12. Adding gives 2a = 40, so a = 20 and b = 8.

题1：x = 2，y = 3。
题2：代入得 3x − 5 = x² − 2x − 1，即 x² − 5x + 4 = 0，因式分解得 (x − 1)(x − 4) = 0，故 x = 1 或 x = 4。对应 y 值为 y = −2 和 y = 7。解为 (1, −2) 和 (4, 7)。
题3：设 a + b = 28，a − b = 12。两式相加得 2a = 40，所以 a = 20，b = 8。

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By mastering elimination, substitution and the graphical interpretation, you can confidently tackle any simultaneous equation question the IGCSE exam presents. Practice a wide variety of questions, including word problems and linear–quadratic systems, and always check your answers.

通过掌握消元法、代入法以及图像含义，你就能自信地应对IGCSE考试中的任何方程组问题。广泛练习各类题型，包括应用题和线性—二次混合系统，并养成检验答案的习惯。

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