Linear Simultaneous Equations | 线性联立方程

📚 Linear Simultaneous Equations | 线性联立方程

Linear simultaneous equations are pairs or groups of linear equations that share the same variables. In A-Level Mathematics for Edexcel, you are expected to solve two linear equations in two unknowns with confidence, interpret the solution graphically, and recognise when a system has no unique solution.

线性联立方程是指含有相同变量的两个或多个线性方程。在 Edexcel A-Level 数学中，你需要熟练掌握两个未知数、两个线性方程的求解，能够从图像上解释解，并识别方程组何时没有唯一解。

1. What Are Linear Simultaneous Equations? | 什么是线性联立方程？

A linear equation in two variables has the general form ax + by = c, where a, b and c are constants and a, b are not both zero. When two such equations are considered together, they form a linear simultaneous system.

二元线性方程的一般形式为 ax + by = c，其中 a、b、c 为常数，且 a、b 不同时为零。当两个这样的方程放在一起时，就构成线性联立方程组：

a₁x + b₁y = c₁
a₂x + b₂y = c₂

The solution is an ordered pair (x, y) that satisfies both equations at the same time.

方程组的解是一个有序数对 (x, y)，它能同时满足两个方程。

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2. The Elimination Method | 消元法

Elimination is usually the fastest method when the coefficients of one variable are already equal or opposite. For example:

当某一个变量的系数已经相等或互为相反数时，消元法通常是最快的方法。例如：

2x + 3y = 12
4x − 3y = 6

Adding the two equations eliminates y because 3y + (−3y) = 0:

将两个方程相加可以消去 y，因为 3y + (−3y) = 0：

6x = 18 ⇒ x = 3

Substitute x = 3 into the first equation to find y:

将 x = 3 代入第一个方程求 y：

2(3) + 3y = 12 ⇒ 6 + 3y = 12 ⇒ 3y = 6 ⇒ y = 2

The solution is (3, 2).

解为 (3, 2)。

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3. Choosing a Multiplier | 选择乘数

If coefficients do not match, multiply one or both equations by suitable constants. Aim to make the coefficients of one variable equal in magnitude and opposite in sign.

如果系数不匹配，需要给一个或两个方程乘上合适的常数。目标是将某一个变量的系数变为大小相等、符号相反。

Solve:

求解：

3x + 2y = 13
2x + 5y = 16

Choose to eliminate x. Multiply the first equation by 2 and the second by 3:

选择消去 x。将第一个方程乘 2，第二个方程乘 3：

6x + 4y = 26
6x + 15y = 48

Subtracting the first from the second gives 11y = 22, so y = 2. Then x = 3.

用第二个方程减去第一个方程，得到 11y = 22，所以 y = 2，再得 x = 3。

Always choose the variable whose multipliers lead to the smallest manageable numbers.

始终选择使乘数最小、最易计算的变量。

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4. The Substitution Method | 代入法

Substitution is useful when one equation is already solved for one variable, or when a coefficient is 1. Rearrange one equation to express one variable in terms of the other, then substitute.

当其中一个方程已经表示出某一个变量，或者某个变量的系数为 1 时，代入法很有用。先将一个方程变形，把一个变量用另一个变量表示，再代入。

Solve:

求解：

y = 2x − 3
3x + 4y = 10

Substitute y = 2x − 3 into the second equation:

将 y = 2x − 3 代入第二个方程：

3x + 4(2x − 3) = 10 ⇒ 3x + 8x − 12 = 10 ⇒ 11x = 22 ⇒ x = 2

Then y = 2(2) − 3 = 1, so the solution is (2, 1).

然后 y = 2(2) − 3 = 1，所以解为 (2, 1)。

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5. Graphical Interpretation | 图像解释

Each linear equation represents a straight line. Solving simultaneous equations means finding the point where the two lines intersect.

每个线性方程都表示一条直线。解联立方程就是求两条直线的交点。

For the system 2x + y = 5 and x − y = 1, the solution is (2, 1). This is exactly the intersection point of the two lines.

对于方程组 2x + y = 5 与 x − y = 1，解为 (2, 1)，这正是两条直线的交点。

If the lines cross, there is one unique solution. If they are parallel, there is no solution. If they coincide, there are infinitely many solutions.

如果两条直线相交，则有唯一解；如果平行，则无解；如果重合，则有无穷多解。

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

Inconsistent equations have no solution. This occurs when the lines are parallel but have different intercepts. Algebraically, elimination produces a false statement such as 0 = 5.

不相容方程组无解。这发生在两条直线平行但截距不同的情形。代数上，消元后会出现 0 = 5 这样的矛盾等式。

Example:

例如：

2x + y = 5
4x + 2y = 9

Doubling the first equation gives 4x + 2y = 10, but the second says 4x + 2y = 9, which is impossible.

将第一个方程乘 2 得 4x + 2y = 10，但第二个方程为 4x + 2y = 9，这是不可能的。

Dependent equations have infinitely many solutions because both equations describe the same line. Elimination leads to a true statement such as 0 = 0.

相关方程组有无穷多解，因为两个方程描述的是同一条直线。消元后会得到 0 = 0 这类恒等式。

Example:

例如：

3x + 6y = 12
x + 2y = 4

The first is 3 times the second, so every point on x + 2y = 4 is a solution.

第一个方程是第二个方程的 3 倍，所以 x + 2y = 4 上的每个点都是解。

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7. Using Simultaneous Equations in Context | 实际情境中的应用

Many Edexcel exam questions are word problems that require you to define variables and construct equations. Typical contexts include buying items, mixing ingredients, and age problems.

许多 Edexcel 考试题是应用题，需要你定义变量并建立方程。常见情境包括购买物品、混合原料以及年龄问题。

Example: Two coffees and three sandwiches cost £12.40. Four coffees and one sandwich cost £10.80. Find the individual prices.

例题：两杯咖啡和三个三明治共 12.40 英镑，四杯咖啡和一个三明治共 10.80 英镑。求各自的单价。

Let a coffee be £x and a sandwich be £y:

设一杯咖啡为 £x，一个三明治为 £y：

2x + 3y = 12.40
4x + y = 10.80

Solving gives x = 1.80 and y = 2.80, so a coffee costs £1.80 and a sandwich costs £2.80.

解得 x = 1.80，y = 2.80，即咖啡 1.80 英镑，三明治 2.80 英镑。

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8. Checking Your Solutions | 检验解

Always substitute your x and y values back into both original equations, not just the one you used for substitution.

一定要将 x 和 y 的值代回两个原始方程，而不只是代回你用于代入的那个方程。

For the solution (3, 2) in 2x + 3y = 12 and 4x − 3y = 6:

例如对解 (3, 2)，在方程 2x + 3y = 12 与 4x − 3y = 6 中检验：

• 2(3) + 3(2) = 6 + 6 = 12, which is correct.
• 4(3) − 3(2) = 12 − 6 = 6, which is correct.

Checking catches arithmetic slips and ensures your solution is valid for both equations.

检验可以发现计算错误，并确保解对两个方程都成立。

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9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One common mistake is subtracting only the variable terms and forgetting to subtract the constant. Always write the full subtraction carefully.

一个常见错误是只减变量项而忘记减常数项。一定要完整地书写减法过程。

Another mistake is using the transformed equation to check the solution. Always return to the original equations.

另一个常见错误是用变形后的方程检验解。一定要回到原始方程进行检验。

Also, when multiplying an equation, multiply every term on both sides by the same constant.

此外，给方程乘常数时，两边每一项都要乘以同一个常数。

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10. Exam-Style Example | 考试题型示例

Solve the simultaneous equations:

求解下列联立方程组：

5x − 2y = 11
3x + 4y = 9

Choose to eliminate y by multiplying the first equation by 2:

选择消去 y，将第一个方程乘 2：

10x − 4y = 22
3x + 4y = 9

Adding gives 13x = 31, so x = 31/13. This is not an integer, but exam solutions can be fractions. Substitute to find y:

两式相加得 13x = 31，所以 x = 31/13。这虽然不是整数，但考试中的解可以是分数。代入求 y：

5(31/13) − 2y = 11 ⇒ 155/13 − 2y = 143/13 ⇒ 2y = 12/13 ⇒ y = 6/13

So the exact solution is (31/13, 6/13).

因此精确解为 (31/13, 6/13)。

In an exam, write the exact fraction unless the question asks for decimals.

考试中除非题目要求小数，否则应写出精确分数。

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