Solving Simultaneous Equations | 解联立方程组

📚 Solving Simultaneous Equations | 解联立方程组

Simultaneous equations are a cornerstone of IGCSE Mathematics. They appear in nearly every exam paper, either as a direct algebraic question or as part of a word problem. Mastering them will earn you reliable marks across multiple topics.

联立方程是 IGCSE 数学的基石,几乎每张考卷都会出现,既可能直接考查代数解法,也可能出现在应用题中。掌握联立方程,能帮助你在多个专题中稳定拿分。


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

A simultaneous equation system involves two variables, usually x and y, and two separate equations. A solution is an ordered pair (x, y) that makes both equations true at the same time.

联立方程组含有两个未知数(通常为 x 和 y)以及两个独立方程。解就是一组有序数对 (x, y),它能同时使两个方程成立。

For example, consider the system:

例如,考虑以下方程组:

2x + y = 7

x – y = 2

Substituting x = 3 and y = 1 into both equations confirms the solution, because 2(3) + 1 = 7 and 3 – 1 = 2.

将 x = 3、y = 1 代入两个方程即可验证:2(3) + 1 = 7,且 3 – 1 = 2。


2. The Substitution Method | 代入消元法

Substitution is ideal when one variable has a coefficient of 1. Follow these steps:

当某个未知数的系数为 1 时,代入消元法尤为方便。步骤如下:

  • Rearrange one equation to make x or y the subject.
  • 将其中一个方程变形,用另一个未知数表示 x 或 y。
  • Substitute this expression into the other equation.
  • 把该表达式代入另一个方程。
  • Solve the resulting single-variable linear equation.
  • 解出这个一元一次方程。
  • Back-substitute to find the second variable.
  • 回代求出另一个未知数。

Worked example: Solve y = 2x – 1 and 3x + 2y = 12.

例:解方程组 y = 2x – 1 与 3x + 2y = 12。

Since y is already the subject, substitute y = 2x – 1 into 3x + 2y = 12:

由于 y 已单独表示,将 y = 2x – 1 代入 3x + 2y = 12:

3x + 2(2x – 1) = 12

3x + 4x – 2 = 12

7x = 14, so x = 2

Then y = 2(2) – 1 = 3. The solution is x = 2, y = 3.

于是 y = 2(2) – 1 = 3,解为 x = 2,y = 3。


3. The Elimination Method | 加减消元法

Elimination is powerful when both equations are in the form ax + by = c. The goal is to add or subtract the equations so that one variable cancels out.

当两个方程都形如 ax + by = c 时,加减消元法非常有效。核心目标是通过相加或相减,使某个未知数消去。

  • Make the coefficients of one variable equal, using multiplication if necessary.
  • 必要时通过乘法,使某个未知数的系数相等。
  • Add or subtract the equations to eliminate that variable.
  • 将两式相加或相减,消去该未知数。
  • Solve for the remaining variable, then back-substitute.
  • 解出剩下的未知数,再回代求解。

Worked example: Solve 2x + 3y = 13 and 5x + 3y = 22.

例:解方程组 2x + 3y = 13 与 5x + 3y = 22。

Both equations have the same 3y coefficient, so subtract the first equation from the second:

两式中 3y 的系数相同,因此用第二式减去第一式:

(5x – 2x) + (3y – 3y) = 22 – 13

3x = 9, so x = 3

Substitute x = 3 into 2x + 3y = 13: 6 + 3y = 13, hence 3y = 7 and y = 7/3.

将 x = 3 代入 2x + 3y = 13:6 + 3y = 13,故 3y = 7,y = 7/3。


4. Choosing the Right Method | 如何选择合适的方法

Both methods always work, but certain question types favour one over the other.

两种方法都普遍适用,但不同题型各有偏好。

Scenario 题型 Recommended Method 推荐方法 Reason 理由
One equation has y = or x = 某个方程已写成 y = 或 x = Substitution 代入法 Avoids extra rearranging 省去多余变形
Coefficients already match 同类项系数已相同 Elimination 加减消元法 One step gives the answer 一步即可得解
Fractions or decimals appear 含分数或小数 Clear denominators first 先化为整数系数 Simplifies all arithmetic 简化全部计算

Whichever method you choose, show full working. Method marks are often awarded even for a small arithmetic slip.

无论选择哪种方法,都要写出完整过程。即使最终计算有误,步骤分也常常可以拿到。


5. Word Problems | 联

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