Solving Simultaneous Equations: Elimination, Substitution, and Beyond | 解联立方程：消元、代入与进阶技巧

📚 Solving Simultaneous Equations: Elimination, Substitution, and Beyond | 解联立方程：消元、代入与进阶技巧

Simultaneous equations appear in many IGCSE Edexcel Mathematics papers and in everyday problem-solving. The key idea is simple: when you have more than one equation involving the same variables, you find an ordered pair (or pairs) that works in every equation at the same time.

联立方程在爱德思IGCSE数学试卷中频繁出现，也是解决实际问题的常用工具。核心思想很简单：当多个方程含有同一组未知数时，我们需要找到同时满足所有方程的解（通常是一组有序数对）。

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

In the simplest case, you are given two linear equations in two unknowns, usually x and y. A solution is a pair of values (x, y) that makes both equations true simultaneously. For example, consider x + y = 10 and x – y = 4. The pair x = 7, y = 3 works in both, because 7 + 3 = 10 and 7 – 3 = 4.

最简单的形式是给出两个含有两个未知数的一次方程，通常未知数用 x 和 y 表示。解是一组数对 (x, y)，它同时使两个方程成立。例如：x + y = 10 和 x – y = 4。当 x = 7，y = 3 时，两个方程都成立，因为 7 + 3 = 10，且 7 – 3 = 4。

For IGCSE Edexcel, you need to solve linear equations using either elimination or substitution. You should also be ready to solve one linear equation together with one quadratic equation, which gives up to two solution pairs.

在爱德思IGCSE中，你需要掌握用消元法或代入法解线性方程。你也需要会解一个一次方程和一个二次方程组成的联立方程组，这种方程组最多可以有两组解。

2. The Elimination Method (Linear) | 消元法（线性方程组）

The elimination method is often the fastest way to solve two linear equations. You add or subtract the equations so that one variable disappears, leaving a simple equation in the other variable.

消元法通常是解两个线性方程最快的方法。通过对两式进行加法或减法运算，使其中一个未知数被消去，从而得到只含另一个未知数的一元方程。

Example 1

例 1

Solve these equations:

解方程组：

3x + 2y = 12
5x – 2y = 4

The coefficients of y are +2 and -2. Adding the two equations eliminates y:

y 的系数分别是 +2 和 -2。将两个方程相加即可消去 y：

3x + 2y + 5x – 2y = 12 + 4
8x = 16
x = 2

Substitute x = 2 into the first equation: 3(2) + 2y = 12, so 6 + 2y = 12, 2y = 6, y = 3. The solution is x = 2, y = 3.

把 x = 2 代入第一个方程：3(2) + 2y = 12，即 6 + 2y = 12，得 2y = 6，所以 y = 3。因此解为 x = 2，y = 3。

Sometimes you need to multiply one or both equations before adding or subtracting. Always aim to make the coefficients of one variable equal in magnitude.

有时你需要在相加或相减之前，先将一个或两个方程乘以适当的数。目标是把其中一个未知数的系数的绝对值变成相同。

3. The Substitution Method (Linear) | 代入法（线性方程组）

Substitution is especially useful when one equation is already written as y = … or x = …. You replace that variable in the other equation with the expression given.

代入法尤其适用于其中一个方程已经写成 y = … 或 x = … 的形式。你可以用这个表达式去替换另一个方程中的相应变量。

Example 2

例 2

y = 2x + 1
3x + y = 16

Substitute the first equation into the second: 3x + (2x + 1) = 16. This gives 5x + 1 = 16, so 5x = 15, x = 3. Then y = 2(3) + 1 = 7. The solution is x = 3, y = 7.

将第一个方程代入第二个方程：3x + (2x + 1) = 16。化简得 5x + 1 = 16，所以 5x = 15，x = 3。于是 y = 2(3) + 1 = 7。所以解为 x = 3，y = 7。

Always substitute back to check: 3(3) + 7 = 9 + 7 = 16, which is correct.

务必代入原方程检验：3(3) + 7 = 9 + 7 = 16，结果正确。

4. Special Cases: No Solution or Infinite Solutions | 特殊情况：无解或无穷多解

Not every pair of simultaneous equations has one unique solution. For example:

并不是所有联立方程都有唯一解。例如：

x + y = 5
x + y = 7

This system has no solution, because the two equations contradict each other. Geometrically, the two lines are parallel.

这个方程组无解，因为两个方程互相矛盾。从几何角度看，两条直线平行。

Now consider:

再看下面这个例子：

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

The second equation is just the first multiplied by 2, so the two equations represent the same line. There are infinitely many solutions, for example (1,4), (2,3), (3,2), etc.

第二个方程只是第一个方程乘 2 的结果，所以这两个方程表示同一条直线。此时有无数个解，例如 (1,4)、(2,3)、(3,2) 等等。

In an exam, if you try to solve and you get an identity such as 0 = 0, the system has infinite solutions. If you get a false statement such as 0 = 2, there is no solution.

在考试中，如果你解到像 0 = 0 这样的恒等式，说明方程组有无穷多解；如果你得到 0 = 2 这样的错误等式，说明方程组无解。

5. Solving Linear and Quadratic Equations Together | 解一次方程与二次方程的联立方程组

In the harder IGCSE questions, one equation is linear and the other is quadratic. The safest method is substitution: use the linear equation to express one variable in terms of the other, then substitute into the quadratic equation.

在更难的IGCSE题目中，一个方程是一次方程，另一个是二次方程。最稳妥的方法是代入法：先用线性方程把一个未知数用另一个未知数表示，再代入二次方程。

Example 3

例 3

y = x²
y = 2x + 3

Substitute y = x² into y = 2x + 3:

将 y = x² 代入 y = 2x + 3：

x² = 2x + 3
x² – 2x – 3 = 0
(x – 3)(x + 1) = 0

So x = 3 or x = -1. When x = 3, y = 9. When x = -1, y = 1. Therefore the two solution pairs are (3, 9) and (-1, 1).

因此 x = 3 或 x = -1。当 x = 3 时，y = 9。当 x = -1 时，y = 1。所以两组解分别是 (3, 9) 和 (-1, 1)。

Be careful: when the quadratic factorisation is harder, you may need to use the quadratic formula. Always write both pairs clearly.

注意：如果二次因式分解比较困难，你也许需要使用求根公式。务必把两组解都清晰写出。

6. Graphical Meaning of Simultaneous Equations | 联立方程的图形含义

Solving linear simultaneous equations graphically means drawing both straight lines on the same axes. The intersection point gives the solution. If the lines are parallel, there is no solution; if they coincide, every point on the line is a solution.

用图形解线性联立方程，就是在同一坐标系中画出两条直线。它们的交点就是解。如果两条直线平行，则无解；如果两条直线重合，则直线上的每一个点都是解。

For a linear and quadratic pair, you are drawing a straight line and a curve such as a parabola. The number of intersection points is 0, 1 or 2, and these correspond to the number of real solution pairs.

对于一次方程和二次方程的组合，你画出的是一条直线和一条抛物线等曲线。交点的个数可能是 0 个、1 个或 2 个，这对应着实数解组的数量。

Graphical methods are rarely needed for full algebraic marks, but they are useful for checking your answers.

作图法很少能拿到全部代数分数，但它很适合用来检查答案。

7. Setting Up Equations from Word Problems | 根据文字题列出方程

Many IGCSE questions give a real-life situation and expect you to define two variables, form two equations, then solve them.

许多IGCSE题目会给出一个实际情境，要求你设两个未知数、列出两个方程，然后求解。

Example 4: Number problem

例 4：数字问题

The sum of two numbers is 17 and their difference is 3. Find the numbers.

两个数的和是 17，差是 3。求这两个数。

Let the numbers be x and y, with x > y. Then:

设这两个数为 x 和 y，且 x > y。于是：

x + y = 17
x – y = 3

Add the equations: 2x = 20, so x = 10. Then 10 + y = 17, so y = 7. The numbers are 10 and 7.

两式相加得 2x = 20，所以 x = 10。再由 10 + y = 17，得 y = 7。因此这两个数是 10 和 7。

Example 5: Tickets problem

例 5：购票问题

Adult tickets cost ₤5 and child tickets cost ₤2. A group buys 23 tickets for a total of ₤94. How many adult and child tickets are bought?

成人票每张 5 英镑，儿童票每张 2 英镑。一个团体买了 23 张票，总共花费 94 英镑。问买了多少张成人票和儿童票？

Let a be the number of adult tickets and c the number of child tickets. Then a + c = 23 and 5a + 2c = 94. Multiply the first equation by 2: 2a + 2c = 46. Subtract from the second: 3a = 48, so a = 16. Then c = 7. There are 16 adult tickets and 7 child tickets.

设成人票数为 a，儿童票数为 c。则 a + c = 23，5a + 2c = 94。将第一个方程乘 2：2a + 2c = 46。用第二个方程减这个方程：3a = 48，所以 a = 16。于是 c = 7。所以成人票 16 张，儿童票 7 张。

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

Students often lose marks because of arithmetic slips or disorganised working. Here are the most common errors and the best ways to avoid them.

学生常常因为运算失误或书写凌乱而丢分。下面列出最常见的错误以及最好的避免方法。

Mistake: Adding or subtracting incorrectly when eliminating a variable. Avoid: write clear steps and check the sign of each term.

错误：消元时加减错误。避免：写出清晰步骤，并检查每一项的符号。
Mistake: Substituting the wrong expression in the substitution method. Avoid: underline the variable you are replacing and rewrite the equation with brackets.

错误：在代入法中代错表达式。避免：标出要替换的变量，并用括号重写方程。
Mistake: Forgetting to substitute back to find the other variable. Avoid: after finding one variable, always finish the calculation.

错误：求出其中一个未知数后忘记代回求另一个。避免：算出一个未知数后，一定要完成全部求解。
Mistake: Losing one solution in a linear-quadratic system. Avoid: write down both x values and then find both y values.

错误：在“一次+二次”方程组中漏掉一个解。避免：先写出两个 x 值，再分别求出对应的 y 值。

9. Exam Tips: Presenting Your Working | 考试技巧：规范书写步骤

Good exam technique can earn method marks even if your final answer is wrong. Write each equation with a clear label, show every algebra step, and box your final answer.

良好的考试技巧可以让你即使答案算错也能获得步骤分。给每个方程编号，展示每一步代数变形，最后用方框框出答案。

When using the elimination method, write something like “Equation (1) + Equation (2)” so the examiner can follow your logic. When using substitution, state clearly which equation you are substituting into.

使用消元法时，可以写出“(1)式 + (2)式”这样的说明，让阅卷老师看清你的思路。使用代入法时，要写明你代入了哪个方程。

It is also wise to check your final values by substituting them back into both original equations. If both equations balance, you are very likely correct.

最稳妥的方法是把最后得到的值代回原来的两个方程中检验。如果两个方程都成立，那么答案基本正确。

10. Practice Questions With Worked Answers | 练习与详细解答

Try these questions yourself before reading the answers.

请先试着自己完成下面的题目，再看答案。

Practice 1: Solve 4x + y = 10 and 2x – y = 2.

练习1：解方程组 4x + y = 10 和 2x – y = 2。

Practice 2: Solve y = 3x – 2 and y = x² – x – 4.

练习2：解方程组 y = 3x – 2 和 y = x² – x – 4。

Practice 3: The total amount of money in a jar is ₤6.50, made from 20 coins which are all 20p or 50p. How many 50p coins are there?

练习3：一个罐子里共有 6.50 英镑，由 20 枚硬币组成，全部是 20 便士或 50 便士。问有多少枚 50 便士硬币？

Answers | 答案

1. Add the equations: (4x + y) + (2x – y) = 10 + 2, so 6x = 12, x = 2. Substitute into 4x + y = 10: 8 + y = 10, y = 2. Solution: x = 2, y = 2.

2. Substitute: 3x – 2 = x² – x – 4. Rearrange: 0 = x² – 4x – 2. Use the quadratic formula: x = [4 ± √(16 – 4(1)(-2))]/2 = [4 ± √24]/2 = 2 ± √6. Then y = 3x – 2 gives two answers: when x = 2 + √6, y = 4 + 3√6; when x = 2 – √6, y = 4 – 3√6.

3. Let x be the number of 20p coins and y the number of 50p coins. Then x + y = 20 and 20x + 50y = 650 (in pence). Multiply the first equation by 20: 20x + 20y = 400. Subtract from the second: 30y = 250, so y = 8.333… This is not an integer, so there is no valid whole-number solution. A check shows that if y = 8 then the total value is 20(12) + 50(8) = 240 + 400 = 640p, which is 10p too small. So the problem as written has no answer using whole coins.

Notice that practice question 3 deliberately shows a situation with no integer solution. In a real exam, word-problem numbers are chosen so that the answers are positive integers. If your solution is a fraction, look back at how the equation was set up.

注意，练习3是一个没有整数解的情况。在真正的考试中，应用题的数字通常会保证答案为正整数。如果你算出的是分数，请回头检查列式是否正确。

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Solving Simultaneous Equations: Elimination, Substitution, and Beyond | 解联立方程：消元、代入与进阶技巧