Solving Simultaneous Equations: A Complete Guide | 联立方程组的解法与应用全归纳

📚 Solving Simultaneous Equations: A Complete Guide | 联立方程组的解法与应用全归纳

Simultaneous equations are a set of equations that share the same unknown variables and must be solved together. Their solutions represent values that satisfy every equation in the system at the same time.

联立方程组是一组含有相同未知数的方程，必须放在一起求解。它们的解是同时满足方程组中每一个方程的未知数的值。

In A-Level Mathematics, simultaneous equations appear in algebra, coordinate geometry, matrices and real-world modelling. A confident grasp of the methods, the geometry, and the conditions for different numbers of solutions is essential for exam success.

在 A-Level 数学中，联立方程组出现在代数、坐标几何、矩阵和实际建模中。扎实掌握解法、对应的几何意义以及不同解个数的条件，是考试取得好成绩的关键。

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

A linear equation in two variables has the general form ax + by = c, where a, b and c are constants. When we have two or more such equations with the same variables, we call them simultaneous equations.

含有两个变量的一次方程一般可以写成 ax + by = c 的形式，其中 a、b、c 是常数。当两个或多个这样的方程含有相同的变量时，它们就构成联立方程组。

The simplest case is a pair of linear equations:

最简单的情形是一对线性方程：

2x + 3y = 7
x – y = 1

Here the solution is x = 2 and y = 1, because substituting these values makes both equations true.

这里的解是 x = 2，y = 1，因为代入后两个方程都成立。

Simultaneous equations can also involve one linear equation and one quadratic equation. These systems are common in coordinate geometry when finding the intersection of a straight line and a curve.

联立方程组也可以由一个线性方程和一个二次方程构成。这类方程组在坐标几何中经常用于求直线与曲线的交点。

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2. Key Methods: Substitution and Elimination | 关键方法：代入消元法与加减消元法

Substitution Method | 代入消元法

Rearrange one equation to make one variable the subject, then substitute this expression into the other equation.

先改写其中一个方程，把一个变量表示成另一个变量的表达式，再把这个表达式代入另一个方程。

Consider:

例如：

y = 2x – 1
3x + 2y = 12

Substitute y = 2x – 1 into 3x + 2y = 12:

把 y = 2x – 1 代入 3x + 2y = 12：

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

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

然后 y = 2(2) – 1 = 3。所以解为 x = 2，y = 3。

Elimination Method | 加减消元法

Multiply one or both equations so that the coefficients of one variable are opposites, then add the equations to eliminate that variable.

将一个或两个方程乘以适当的数，使某一变量的系数互为相反数，然后把两个方程相加，从而消去该变量。

For example:

例如：

2x + 3y = 7
x – y = 1

Multiply the second equation by 3:

把第二个方程两边乘以 3：

3x – 3y = 3

Now add the two equations:

然后把两个方程相加：

5x = 10
x = 2

Substitute x = 2 into x – y = 1, so y = 1.

把 x = 2 代入 x – y = 1，得到 y = 1。

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3. Solving Linear-Linear Systems | 线性-线性方程组的求解

For a pair of linear equations, the solution is the point where the two straight lines intersect.

对两个线性方程组成的方程组，其解就是两条直线的交点。

There are three possible cases:

一共有三种可能的情况：

Case | 情况	Condition | 条件	Number of Solutions | 解的个数
Intersecting lines | 两直线相交	a₁/a₂ ≠ b₁/b₂	One unique solution | 唯一解
Parallel and distinct lines | 两直线平行且不重合	a₁/a₂ = b₁/b₂ ≠ c₁/c₂	No solution | 无解
Same line | 两直线重合	a₁/a₂ = b₁/b₂ = c₁/c₂	Infinitely many solutions | 无穷多解

For example, the pair 2x + y = 5 and 4x + 2y = 10 are the same line, so every point on the line is a solution.

例如，2x + y = 5 与 4x + 2y = 10 表示同一条直线，因此直线上的每一个点都是方程组的解。

Always check whether the coefficient ratios are consistent before assuming there is a unique solution.

在假设方程组一定有唯一解之前，必须先检查系数之间的比值是否一致。

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4. Solving Linear-Quadratic Systems | 线性-二次方程组的求解

A common A-Level question involves one linear and one quadratic equation. The linear equation is usually substituted into the quadratic equation.

A-Level 常见题型包含一个线性方程和一个二次方程。通常的做法是把线性方程代入二次方程。

Example:

例如：

y = x² – 3x + 2
y = 2x – 1

Substitute y = 2x – 1 into the quadratic:

把 y = 2x – 1 代入二次方程：

2x – 1 = x² – 3x + 2
x² – 5x + 3 = 0

Use the quadratic formula:

利用求根公式：

x = (5 ± √(25 – 12)) / 2
x = (5 ± √13) / 2

Now substitute each x-value back into y = 2x – 1:

然后把每一个 x 值代回 y = 2x – 1：

x₁ = (5 + √13) / 2, y₁ = 4 + √13
x₂ = (5 – √13) / 2, y₂ = 4 – √13

The two intersection points are the solutions of the system.

这两个交点坐标就是方程组的解。

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5. The Discriminant and Number of Solutions | 判别式与解的个数

After substituting a linear equation into a quadratic equation, you usually obtain a quadratic equation of the form ax² + bx + c = 0. The number of real solutions is determined by the discriminant:

把线性方程代入二次方程后，通常会得到一个形如 ax² + bx + c = 0 的二次方程。实数解的个数由判别式决定：

Δ = b² – 4ac

Discriminant | 判别式	Number of Intersections | 交点个数	Geometric Meaning | 几何意义
Δ > 0	Two solutions | 两个解	Line intersects curve at two points | 直线与曲线相交于两点
Δ = 0	One solution | 一个解	Line is tangent to the curve | 直线与曲线相切
Δ < 0	No real solution | 无实数解	Line does not meet the curve | 直线与曲线不相交

This discriminant test is especially important for tangent problems. If a question says a line is tangent to a curve, set Δ = 0 and solve for the unknown parameter.

判别式检验在切线问题中尤其重要。如果题目说一条直线与曲线相切，就令 Δ = 0，然后求解未知参数。

For example, if y = mx + 2 is tangent to y = x² + 1, then x² + 1 = mx + 2 gives x² – mx – 1 = 0. Here Δ = m² + 4. Since m² + 4 > 0 for all real m, no tangent exists for this line family. Always check whether the required condition is actually possible.

例如，若 y = mx + 2 与 y = x² + 1 相切，则 x² + 1 = mx + 2，得到 x² – mx – 1 = 0。此时 Δ = m² + 4。因为对任意实数 m，m² + 4 > 0，所以这一族直线中不存在切线。记得检查所需条件是否真的可能成立。

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6. Using Graphs | 图形法

Graphical methods help visualise the system. The solution is the intersection point of the graphs of the equations.

图形法有助于理解方程组的几何意义。方程组的解就是各方程图像的交点。

For two straight lines, draw both lines on the same set of axes. If they meet at one point, that point is the solution. If they are parallel, there is no solution. If one line lies exactly on the other, there are infinitely many solutions.

对于两条直线，在同一个坐标系中画出两条直线。若它们交于一点，该点就是解；若两条直线平行，则无解；若一条直线完全重合在另一条上，则有无穷多解。

For a line and a curve, sketch the curve and the straight line. The number of intersection points matches the number of real solutions.

对于直线和曲线，画出曲线和直线。交点的个数与实数解的个数一致。

Graphs are excellent for checking answers, but in an exam you should still solve algebraically unless the question explicitly asks for a graphical estimate.

图像非常适合用来检验答案，但在考试中，除非题目明确要求作图估计，否则仍然应该用代数方法求解。

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7. Matrices and Determinants | 矩阵与行列式

A pair of simultaneous linear equations can be written in matrix form:

两个线性方程组可以写成矩阵形式：

[a₁ b₁] [x] [c₁]
[a₂ b₂] [y] = [c₂]

The determinant of the coefficient matrix is:

系数矩阵的行列式为：

D = a₁b₂ – a₂b₁

If D ≠ 0, the system has a unique solution. Cramer’s rule gives:

如果 D ≠ 0，方程组有唯一解。克莱姆法则给出：

x = (c₁b₂ – c₂b₁) / D
y = (a₁c₂ – a₂c₁) / D

If D = 0, the system is either inconsistent, giving no solution, or dependent, giving infinitely many solutions.

如果 D = 0，则方程组要么不相容，即无解；要么相关，即有无穷多解。

In A-Level Further Mathematics, the inverse matrix method is often written as X = A⁻¹B. Provided that A is non-singular, this gives the unique solution of the system.

在进阶数学中，逆矩阵法通常写成 X = A⁻¹B。只要矩阵 A 非奇异，就可以求出方程组的唯一解。

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8. Applications in Coordinate Geometry | 坐标几何中的应用

Simultaneous equations are used to find the intersection of lines and curves, including tangents and normals.

联立方程组常用于求直线与曲线的交点，包括切线和法线相关的问题。

To find where two lines meet, solve their equations together. For example:

要求两条直线的交点，只需要联立求解它们的方程。例如：

y = 2x + 1
y = -x + 7

Equating gives 2x + 1 = -x + 7, so 3x = 6, hence x = 2 and y = 5. The two lines intersect at (2, 5).

令两边相等，得到 2x + 1 = -x + 7，所以 3x = 6，因此 x = 2，y = 5。两条直线相交于点 (2, 5)。

For tangent questions, start by writing the equation of the straight line, then substitute it into the curve equation. Use the discriminant to impose the tangent condition.

对于切线问题，先写出直线方程，然后代入曲线方程，再利用判别式建立相切条件。

Example: Find the value of k for which y = kx – 2 is tangent to y = x² – 4x + 2.

例如：求 k 的值，使得 y = kx – 2 与 y = x² – 4x + 2 相切。

kx – 2 = x² – 4x + 2
x² – (k + 4)x + 4 = 0

Δ = (k + 4)² – 16 = 0
k + 4 = ±4
k = 0 or k = -8

The line is tangent when k = 0 or k = -8.

当 k = 0 或 k = -8 时，该直线与抛物线相切。

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9. Real-World Applications | 实际应用

Simultaneous equations model situations where two conditions must be satisfied at the same time.

联立方程组可以用来建模那些需要同时满足两个条件的实际情境。

In economics, market equilibrium occurs where demand equals supply.

在经济学中，市场均衡出现在需求等于供给的位置。

Suppose demand is P = -2Q + 10 and supply is P = Q + 1. At equilibrium:

假设需求函数为 P = -2Q + 10，供给函数为 P = Q + 1。在均衡点：

-2Q + 10 = Q + 1
3Q = 9
Q = 3, P = 4

So the equilibrium quantity is 3 units and the equilibrium price is 4.

因此均衡数量为 3 个单位，均衡价格为 4。

In business, break-even analysis compares total cost and total revenue.

在商业中，盈亏平衡分析比较总成本与总收入。

If C = 100 + 5x and R = 9x, then break-even occurs when:

如果 C = 100 + 5x，R = 9x，那么盈亏平衡发生在：

100 + 5x = 9x
4x = 100
x = 25

The company breaks even after selling 25 units.

该公司销售 25 个单位后达到盈亏平衡。

In physics, simultaneous equations are used when two unknown quantities are linked by two independent relationships, such as resolving forces or solving kinematic problems.

在物理中，当两个未知量由两个独立关系联系时，例如力的分解或运动学问题，也会使用联立方程组。

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10. Common Mistakes and Exam Tips | 常见错误与考试技巧

• Always substitute the x-values back into the simpler equation, not the quadratic, to avoid unnecessary arithmetic errors.

  求出一个变量的值后，一定要代回较简单的那个方程，而不是代回二次方程，这样可以避免不必要的计算错误。

• Do not stop at x. For linear-quadratic systems, every x-value must be paired with its corresponding y-value.

  不要只求出 x 就结束。在线性-二次方程组中，每一个 x 值都必须配一个对应的 y 值。

• Check the discriminant before deciding how many solutions exist. A quadratic equation may have two, one, or no real roots.

  在判断解的个数之前，先计算判别式。二次方程可能有两个、一个或没有实数根。

• When using elimination, multiply all terms in the equation by the same number, not just one term.

  使用加减消元法时，方程两边所有项都必须乘以同一个数，不能只乘其中一个项。

• If the determinant of a linear system is zero, explain whether the lines are parallel or identical before concluding no solution or infinite solutions.

  如果线性方程组的行列式为 0，要先判断两直线是平行还是重合，再下结论说无解或有无穷多解。

• Do not use rounded decimals unless the question asks for an approximate answer. Exact surd forms such as (5 ± √13) / 2 are preferred in A-Level.

  除非题目要求近似答案，否则不要使用四舍五入后的小数。A-Level 中更倾向于保留 (5 ± √13) / 2 这类精确根式形式。

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11. Practice Questions | 巩固练习

Question | 题目	Solution Hint | 解答提示
1. Solve 3x + 2y = 8 and x – y = 1.	Substitute x = y + 1, or eliminate y. Answer: x = 2, y = 1.
2. Solve y = x² – 2x and y = x + 4.	Set x² – 2x = x + 4, so x² – 3x – 4 = 0. Answer: (4, 8) and (-1, 3).
3. Find k if y = kx + 1 is tangent to y = x² – x.	Substitute to obtain x² – (k + 1)x – 1 = 0. Since Δ = (k + 1)² + 4 > 0, no such real k exists.

Work through every step clearly. In simultaneous equation questions, clarity is just as important as the final answer.

每一步都要写清楚。在联立方程组题目中，过程的清晰程度和最终答案一样重要。

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Published by TutorHao | Mathematics Revision Series | aleveler.com

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