Solving Linear and Nonlinear Simultaneous Equations | 线性与非线性联立方程组的解法

📚 Solving Linear and Nonlinear Simultaneous Equations | 线性与非线性联立方程组的解法

Simultaneous equations are a set of equations with multiple variables that are solved together because they share the same variables. In IGCSE Mathematics, you will encounter both linear simultaneous equations and nonlinear simultaneous equations, where at least one equation is quadratic. This guide walks through the key methods, common pitfalls, and exam-style techniques.

联立方程组是一组共享相同变量的方程，需要同时求解。在 IGCSE 数学中，你将会遇到线性联立方程组和非线性联立方程组，其中至少有一个方程是二次方程。本指南将讲解核心解法、常见错误以及考试技巧。

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

A linear simultaneous equation system consists of two or more linear equations, for example:

线性联立方程组由两个或两个以上的线性方程组成，例如：

2x + 3y = 8

x − y = −1

The solution is a pair (x, y) that satisfies both equations at the same time. For nonlinear simultaneous equations, one equation may be quadratic, such as y = x² − 3, and the other linear, such as y = 2x + 1.

解是一对 (x, y)，能够同时满足两个方程。对于非线性联立方程，一个方程可能是二次的，例如 y = x² − 3，另一个是线性的，例如 y = 2x + 1。

2. Solving Linear Equations by Elimination | 线性方程组：消元法

Elimination is a powerful method when coefficients of one variable are easy to align. The goal is to add or subtract the equations to cancel one variable.

当某个变量的系数很容易对齐时，消元法非常有效。目标是通过相加或相减两个方程，消去一个变量。

Example | 例子

3x + 2y = 12

5x − 2y = 4

Add the equations: (3x + 5x) + (2y − 2y) = 12 + 4, giving 8x = 16, so x = 2. Substitute x = 2 into the first equation: 3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3.

将两式相加：(3x + 5x) + (2y − 2y) = 12 + 4，得到 8x = 16，因此 x = 2。将 x = 2 代入第一个方程：3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3。

Key tip | 关键提示: If the coefficients are not opposites, multiply one or both equations by a suitable number before adding or subtracting.

如果系数不是相反数，在相加或相减之前，先将一个或两个方程乘以适当的数。

3. Solving Linear Equations by Substitution | 线性方程组：代入法

Substitution is essential when one equation is already solved for a variable, or when elimination is messy. Rearrange one equation to make x or y the subject, then substitute into the other.

当某个方程已经解出某个变量时，或使用消元法较麻烦时，代入法非常重要。先整理一个方程，将 x 或 y 表示出来，然后代入另一个方程。

Example | 例子

y = 2x + 1

x + y = 10

Substitute y = 2x + 1 into x + y = 10: x + (2x + 1) = 10 → 3x + 1 = 10 → 3x = 9 → x = 3. Then y = 2(3) + 1 = 7. So the solution is (3, 7).

将 y = 2x + 1 代入 x + y = 10：x + (2x + 1) = 10 → 3x + 1 = 10 → 3x = 9 → x = 3。然后 y = 2(3) + 1 = 7。因此解是 (3, 7)。

4. Graphical Meaning of Solutions | 解的几何意义

For a pair of linear equations, the solution is the point where the two straight lines intersect. If the lines are parallel, there is no solution. If the lines are identical, there are infinitely many solutions.

对于一对线性方程，解就是两条直线相交的点。如果两条线平行，则无解。如果两条线重合，则有无数个解。

For a linear and a quadratic equation, the graph of the quadratic is a curve (parabola). The solution points are the intersections of the line and the curve. There can be zero, one, or two solutions.

对于线性方程和二次方程的组合，二次方程的图像是一条曲线（抛物线）。解就是直线与曲线的交点。可能有零个、一个或两个解。

5. Nonlinear Simultaneous Equations: Substitution | 非线性联立方程：代入法

When one equation is quadratic and the other is linear, substitution is the standard method. Always substitute the linear equation into the quadratic equation.

当一个方程是二次的，另一个是线性的，代入法是标准解法。始终将线性方程代入二次方程。

Example | 例子

y = x² − 2

y = 3x − 2

Substitute y = 3x − 2 into the quadratic: 3x − 2 = x² − 2. Then x² − 3x = 0 → x(x − 3) = 0 → x = 0 or x = 3. For x = 0, y = −2; for x = 3, y = 7. Solutions are (0, −2) and (3, 7).

将 y = 3x − 2 代入二次方程：3x − 2 = x² − 2。整理得 x² − 3x = 0 → x(x − 3) = 0 → x = 0 或 x = 3。当 x = 0 时，y = −2；当 x = 3 时，y = 7。解为 (0, −2) 和 (3, 7)。

6. Handling Quadratic Terms: x², y², and xy | 处理二次项：x²、y² 和 xy

Nonlinear equations may include x², y², or even xy. When substituting a linear expression into a quadratic, remember to expand carefully. For example, if y = 2x + 1 and the equation is x² + y² = 25, substitute to get x² + (2x + 1)² = 25.

非线性方程可能包含 x²、y²，甚至 xy。当把线性表达式代入二次方程时，务必小心展开。例如，如果 y = 2x + 1，方程为 x² + y² = 25，代入后得到 x² + (2x + 1)² = 25。

Expand: x² + 4x² + 4x + 1 = 25 → 5x² + 4x − 24 = 0. Solve by factorising or using the quadratic formula. Never forget to find y for each x value.

展开：x² + 4x² + 4x + 1 = 25 → 5x² + 4x − 24 = 0。用因式分解或求根公式求解。千万不要忘记为每个 x 值求对应的 y 值。

Formula reminder | 公式提醒

x = (−b ± √(b² − 4ac)) / 2a

7. Common Pitfalls in Nonlinear Systems | 非线性方程组常见错误

Students often make mistakes in the following areas:

学生经常在以下方面犯错：

Only finding x, not y | 只求 x 不求 y

Always substitute back to find both coordinates. | 始终回代求出两个坐标。
Losing solutions when dividing by x | 除以 x 时丢失解

If you have x² − 3x = 0, never divide by x because x = 0 is a valid solution. Factorise instead. | 如果遇到 x² − 3x = 0，切勿除以 x，因为 x = 0 是一个合法解。应使用因式分解。
Making sign errors during expansion | 展开过程中出现正负号错误

When squaring (2x − 3), write (2x − 3)(2x − 3) and expand carefully. | 计算 (2x − 3)² 时，写成 (2x − 3)(2x − 3) 并仔细展开。

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

For linear systems, if the coefficients of x and y are proportional but the constant terms are not, the lines are parallel and there is no solution. For example:

对于线性方程组，如果 x 和 y 的系数成比例，但常数项不成比例，则两条线平行，无解。例如：

x + y = 3

2x + 2y = 7

Multiplying the first equation by 2 gives 2x + 2y = 6, which conflicts with 2x + 2y = 7. Hence no solution.

将第一个方程乘以 2 得到 2x + 2y = 6，与 2x + 2y = 7 矛盾，因此无解。

If the two equations are multiples of each other, they represent the same line, giving infinitely many solutions.

如果两个方程互为倍数，它们代表同一条直线，因此有无数个解。

9. Step-by-Step Problem-Solving Strategy | 分步解题策略

Follow this reliable sequence for any simultaneous equation problem:

对于任何联立方程问题，请遵循以下可靠步骤：

Identify the type | 判断类型

Is every equation linear? Is there a quadratic term? | 每个方程都是线性的吗？是否存在二次项？
Choose a method | 选择方法

For linear-linear, either elimination or substitution works. For nonlinear, use substitution. | 对于线性-线性，消元法或代入法均可。对于非线性，使用代入法。
Solve for one variable | 求出一个变量

Eliminate one variable or substitute to get a single equation in one variable. | 消去一个变量或代入，得到只含一个变量的方程。
Find the other variable | 求出另一个变量

Substitute back into any original equation. | 代回任意原始方程。
Check your answers | 检验答案

Plug the values into both original equations to verify. | 将数值代入两个原始方程进行验证。

10. Worked Exam Example | 考试真题练习

Question | 题目: Solve the simultaneous equations

y = x² − 4x + 1

y = x + 1

Solution | 解答

Set the two expressions for y equal: x² − 4x + 1 = x + 1 → x² − 5x = 0 → x(x − 5) = 0 → x = 0 or x = 5.

令两个 y 的表达式相等：x² − 4x + 1 = x + 1 → x² − 5x = 0 → x(x − 5) = 0 → x = 0 或 x = 5。

When x = 0, y = 1. When x = 5, y = 6. The solutions are (0, 1) and (5, 6).

当 x = 0 时，y = 1。当 x = 5 时，y = 6。解为 (0, 1) 和 (5, 6)。

Geometrically, the line y = x + 1 cuts the parabola at two distinct points. | 从几何角度看，直线 y = x + 1 与抛物线相交于两个不同的点。

11. Practice Questions | 练习题

Try these on your own. Solutions are shown below.

请自行尝试以下题目，解答如下。

Question \| 题目	Answer \| 答案
2x + y = 5, x − y = 1	x = 2, y = 1
y = x², y = 2x	(0, 0) and (2, 4)
x² + y² = 10, y = x + 2	(1, 3) and (−3, −1)

12. Final Summary | 最终总结

Linear simultaneous equations can be solved by elimination or substitution, while nonlinear systems require substitution. Always factorise quadratic equations fully, find both coordinates, and check your answers. With regular practice, solving simultaneous equations becomes a straightforward and reliable skill for your IGCSE exams.

线性联立方程组可以通过消元法或代入法求解，而非线性方程组则需要使用代入法。始终对二次方程进行完整因式分解，求出两个坐标，并检验答案。通过定期练习，解联立方程组将成为你在 IGCSE 考试中一项简单而可靠的技能。

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Solving Linear and Nonlinear Simultaneous Equations | 线性与非线性联立方程组的解法