📚 Solving Simultaneous Equations | 联立方程组求解
t satisfy every given equation at the same time. A single equation such as x + y = 10 has infinitely many solutions, but adding a second condition such as x − y = 4 pins the answer down to one unique ordered pair. This revision guide explains how to solve linear systems and linear-quadratic systems, how to avoid the errors that cost marks, and how to present your working so that an examiner can follow it.
联立方程组是 IGCSE 与 A Level 数学的核心内容之一。它要求你找出同时满足所有给定方程的两个或多个未知数的值。单独一个方程如 x + y = 10 有无数个解，但增加第二个条件如 x − y = 4 后就能确定唯一解。本复习指南将讲解如何求解线性方程组以及线性与二次联立方程组，如何避开那些会丢分的常见错误，以及如何书写解题过程让阅卷人一目了然。
1. What Are Simultaneous Equations? | 什么是联立方程组
A simultaneous equation system is a set of two or more equations that share the same variables. A valid solution must make every equation true at the same time - not just one of them. The solution of a system in two unknowns is usually written as an ordered pair, such as (7, 3), where the first value is x and the second is y.
联立方程组是由两个或多个共享相同变量的方程组成的系统。一个有效解必须使每一个方程都同时成立，而不仅仅是满足其中一个。含两个未知数的方程组，其解通常写成一个有序数对，例如 (7, 3)，其中第一个值是 x，第二个值是 y。
The number of equations you need depends on the number of unknowns. For two unknowns you normally need two independent equations; for three unknowns you need three independent equations, and so on. "Independent" means that no equation is simply a rearranged copy of another - if two equations say exactly the same thing, they give you no new information.
你需要多少个方程取决于有多少个未知数。对于两个未知数，通常需要两个独立方程；对于三个未知数，则需要三个独立方程，依此类推。"独立"意味着没有任何一个方程只是另一个方程的变形复制——如果两个方程说的完全是同一件事，它们就提供不了新信息。
uations gives 10 = 10 and 4 = 4, so the ordered pair (7, 3) is the solution of the system.
看下面的例子：x + y = 10 与 x − y = 4。将 x = 7、y = 3 代入，两个方程分别得到 10 = 10 和 4 = 4，因此有序数对 (7, 3) 就是该方程组的解。
x + y = 10 and x − y = 4 → x = 7, y = 3
2. Why Simultaneous Equations Matter | 为什么联立方程组重要
Simultaneous equations appear in almost every mathematics syllabus and in papers at every level. At IGCSE they are usually worth four to six marks; at A Level they are the foundation for linear algebra, matrices, vectors and the intersection of curves. If you cannot solve a two-by-two system reliably, later topics become much harder.
联立方程组几乎出现在所有数学考纲中，且各阶段考试都会考。在 IGCSE 阶段它们通常占 4 到 6 分；在 A Level 阶段，它们是线性代数、矩阵、向量以及曲线交点问题的基础。如果你不能稳定地解出二元一次方程组，后面的内容会变得困难得多。
They also model real situations. Two phone contracts with different fixed fees and different monthly rates, two cyclists travelling towards each other at different speeds, or two ingredients being mixed to a target cost - all of these reduce to a pair of equations that must both be satisfied. Learning the technique therefore pays off well beyond the exam hall.
它们同样可以描述真实情境。两种月租不同、资费不同的手机套餐，两名以不同速度相向而行的骑行者，或者两种原料混合以达到目标成本——这些都能化为需要同时满足的两个方程。因此掌握这套方法的价值远不止于考场。
3. The Graphical Meaning of a Solution | 解的图像意义
Every linear equation in two variables draws a straight line. Solving a linear system is exactly the same as finding the point where the two lines cross. Reading a graph is often the fastest way to check whether your algebraic answer is sensible.
每个二元一次方程都对应一条直线。解线性方程组，本质上就是找出这两条直线的交点。读图往往是检验代数答案是否合理的最快方法。
- If the two lines meet at a single point, the system has exactly one solution. | 若两条直线相交于一点，方程组有唯一解。
- If the two lines are parallel and distinct, there is no solution. | 若两条直线平行且不重合，则方程组无解。
- If the two lines are the same line, there are infinitely many solutions. | 若两条直线完全重合，则方程组有无穷多解。
For a linear-quadratic system, the line meets a curve (a parabola, circle or hyperbola) and there may be zero, one or two intersection points. The graph tells you how many answers to expect before you start solving.
对于线性与二次联立方程组，直线与曲线（抛物线、圆或双曲线）相交，交点可能有零个、一个或两个。图像能让你在动手求解之前就知道应该得到几个答案。
4. Method 1: Elimination | 方法一：消元法
Elimination works by adding or subtracting the two equations so that one variable disappears. The two coefficients of the variable you want to remove must have equal size; if they do not, multiply one or both equations by a suitable number first. The golden rule is that whatever you multiply an equation by, you must multiply every term, including the constant on the right-hand side.
消元法的思路是把两个方程相加或相减，使其中一个未知数消失。你想消去的那个未知数，在两个方程中的系数绝对值必须相等；如果不等，就先给一个或两个方程乘以适当的数。黄金法则是：方程乘以某个数时，必须乘以每一项，包括等号右边的常数项。
6. The y-coefficients are +3 and −3, so adding the two equations eliminates y immediately:
设方程组为 2x + 3y = 12 与 4x − 3y = 6。y 的系数分别是 +3 和 −3，因此把两个方程直接相加即可消去 y：
3y) + (4x − 3y) = 12 + 6 → 6x = 18 → x = 3
titute into the simpler of the two original equations, and always substitute your final pair back into the equation you did not use, as a free check.
把 x = 3 代回第一个方程：2(3) + 3y = 12，得 3y = 6，y = 2。解为 x = 3，y = 2。请尽量代入两个原方程中较简单的那个，并且最后一定要把这组解代回你没用过的那一个方程，作为免费的验算。
obtain 6x + 4y = 32; then adding gives 11x = 42. Notice that you must also double the 16.
当系数大小不相等时，先放大。对于 3x + 2y = 16 与 5x − 4y = 10，将第一个方程乘以 2 得 6x + 4y = 32，再相加得 11x = 42。注意等号右边的 16 也必须同时乘以 2。
5. Method 2: Substitution | 方法二：代入法
rranged into that form in one easy step. You replace that variable in the other equation with its expression, producing a single equation in one unknown.
当其中一个方程已经写成 x = …… 或 y = …… 的形式，或者只需一步就能整理成这种形式时，代入法就是首选。你把另一个方程中的这个未知数换成这个表达式，就得到只含一个未知数的方程。
ake y = 2x + 1 together with 3x + y = 11. Substitute the first expression into the second:
以 y = 2x + 1 与 3x + y = 11 为例，把第一个表达式代入第二个方程：
3x + (2x + 1) = 11 → 5x = 10 → x = 2
ssion rather than a number, keep the brackets until you have expanded - this is where most sign errors begin.
然后 y = 2(2) + 1 = 5，所以解为 x = 2，y = 5。当你代入的是一个表达式而不是一个数时，先保留括号，直到展开完成为止——大多数符号错误都是从这里开始的。
Substitution is also the standard route for linear-quadratic systems, because it turns two equations into one quadratic that you already know how to solve. For that reason alone it is worth practising until it feels automatic.
代入法也是解线性与二次联立方程组的标准路径，因为它把两个方程转化为一个你已经会解的二次方程。仅凭这一点，就值得把它练到不假思索的程度。
6. Choosing the Faster Method | 如何选择更快的方法
Both methods always work, so the choice is about speed and about reducing the chance of an arithmetic slip. Use the table below as a quick decision guide in the exam.
两种方法永远都行得通，所以选择的关键在于速度，以及降低算错的风险。考试时可用下表快速判断。
| Form of the system | 方程组形式 | Best method | 推荐方法 |
| Both equations in the form ax + by = c | 两个方程都是 ax + by = c 形式 | Elimination | 消元法 |
| One equation already gives x = ... or y = ... | 一个方程已给出 x = …… 或 y = …… | Substitution | 代入法 |
| One equation is linear, the other is quadratic | 一个方程是一次，另一个是二次 | Substitution | 代入法 |
| Coefficients are large or fractional | 系数很大或含分数 | Elimination after clearing fractions | 先去分母再消元 |
Whatever you choose, write each step on its own line. Method marks are awarded for a clearly shown elimination or substitution, even if the final arithmetic goes wrong.
无论选哪种方法，每一步都单独写一行。只要消元或代入的过程清楚可见，即使最后算错，方法分依然能给到。
7. Linear-Quadratic Systems | 线性与二次联立方程组
to two solutions, each written as a full ordered pair. One equation is usually already in the form y = ... or x = ..., which makes substitution the natural approach.
线性与二次联立方程组，是把一条直线与一条曲线配对，例如 y = x² + 3x − 1 或 x² + y² = 25。由于直线可能与曲线相交两次，你应该预期最多有两个解，且每个解都要写成完整的有序数对。通常其中一个方程已经是 y = …… 或 x = …… 的形式，代入法自然成为首选。
The routine is fixed and worth memorising:
解题流程是固定的，值得记牢：
- Step 1: rearrange the linear equation to make x or y the subject. | 第一步：把一次方程整理成 x 或 y 为主元的形式。
- Step 2: substitute it into the quadratic equation. | 第二步：代入另一个二次方程。
- Step 3: expand and collect everything on one side, giving ... = 0. | 第三步：展开并把所有项移到一边，得到 …… = 0。
- Step 4: solve by factorising, completing the square or the quadratic formula. | 第四步：用因式分解、配方或求根公式求解。
- Step 5: substitute each x-value back to find the matching y-value. | 第五步：把每个 x 值代回，求出对应的 y 值。
The quadratic formula, stated here in Unicode form, is the reliable fallback whenever factorising is not obvious:
当因式分解不明显时，下面用 Unicode 形式写出的求根公式就是可靠的后备方案：
x = (−b ± √b² − 4ac) ÷ 2a
8. Worked Example: Line and Parabola | 例题：直线与抛物线
Since both equations are already in the form y = ..., set the two right-hand sides equal.
求解方程组 y = x² − 2x − 3 与 y = x + 1。由于两个方程都已是 y = …… 的形式，令两边相等即可。
x² − 2x − 3 = x + 1 → x² − 3x − 4 = 0
ch value into the linear equation y = x + 1, because it is far simpler than the quadratic.
因式分解得 (x − 4)(x + 1) = 0，所以 x = 4 或 x = −1。接着把每个值代入一次方程 y = x + 1，因为它比二次方程简单得多。
- When x = 4: y = 4 + 1 = 5, giving the point (4, 5). | 当 x = 4 时：y = 4 + 1 = 5，得到点 (4, 5)。
- When x = −1: y = −1 + 1 = 0, giving the point (−1, 0). | 当 x = −1 时：y = −1 + 1 = 0，得到点 (−1, 0)。
−1) − 3 = 0. Always state both coordinates - a bare x = 4 with no y-value can lose a mark.
两个解为 (4, 5) 和 (−1, 0)。把它们代回二次方程做最后检验：4² − 2(4) − 3 = 5 与 (−1)² − 2(−1) − 3 = 0，均成立。务必写出两个坐标——只写 x = 4 而不写 y 值可能会丢分。
9. Special Cases: No Solution or Infinitely Many | 特殊情况：无解或无穷多解
Not every system has exactly one solution. For linear systems, compare the gradients. If the gradients are equal but the intercepts differ, the lines are parallel and there is no solution. If both the gradient and the intercept match, the equations describe the same line and there are infinitely many solutions.
并非每个方程组都恰好有一个解。对线性方程组而言，比较斜率即可。若斜率相等但截距不同，两直线平行，方程组无解；若斜率与截距都相同，两个方程描述的是同一条直线，方程组有无穷多解。
or linear-quadratic systems, use the discriminant Δ = b² − 4ac of the resulting quadratic:
对线性与二次联立方程组，可用化简后二次方程的判别式 Δ = b² − 4ac 来判断：
- Δ > 0: two distinct intersection points, two solutions. | Δ > 0：两个不同交点，两个解。
- Δ = 0: the line is a tangent, one repeated solution. | Δ = 0：直线相切，一个重根解。
- Δ < 0: no real intersection, no real solutions. | Δ < 0：没有实交点，无实数解。
Examiners frequently ask you to "show that the line is a tangent to the curve" n show that Δ = 0, ideally with the intermediate line b² − 4ac = 0 written out explicitly.
考题中经常出现"证明该直线与曲线相切"的要求。这句话就是一个信号：先列出二次方程，再证明 Δ = 0，最好把 b² − 4ac = 0 这一步明确写出来。
10. Word Problems and Modelling | 应用题与建模
Word problems follow a predictable four-step pattern: define the variables in words, write two equations from the information, solve them, then answer the question that was actually asked. The last step matters - many candidates find x correctly and then forget to convert it back into the context, such as the price of one ticket or the speed of one cyclist.
应用题遵循可预测的四步模式：用文字定义变量、根据题意写出两个方程、求解、然后回答题目真正问的问题。最后一步很关键——很多考生正确求出了 x，却忘记把它还原到情境中，例如一张票的价格或一名骑行者的速度。
Consider a typical example. A theatre sells 200 tickets. Adult tickets cost £12 and child tickets cost £7, and the total income is £1,890. Let a be the number of adult tickets and c the number of child tickets:
看一个典型例子。剧场售出 200 张票，成人票每张 12 英镑，儿童票每张 7 英镑，总收入为 1890 英镑。设成人票为 a 张，儿童票为 c 张：
a + c = 200 and 12a + 7c = 1890
= 102. Notice that the answer required is two numbers in context, not just the value of a.
把第一个方程乘以 7 得 7a + 7c = 1400，与第二个方程相减得 5a = 490，所以 a = 98，c = 102。注意题目要求的是情境中的两个数字，而不只是 a 的值。
11. Common Mistakes and Exam Technique | 常见错误与应试技巧
Most lost marks in this topic come from a small set of recurring slips rather than from not knowing the method. Read through the list below before your next mock paper.
这个专题的失分，大多来自少数几个反复出现的疏漏，而不是不懂方法。下一次模考前请通读下面的清单。
- Forgetting to multiply the constant term when scaling an equation. | 给方程乘以倍数时忘记同时乘常数项。
- Losing a minus sign when subtracting one equation from another. | 两式相减时丢掉负号。
- Substituting a value into the equation you have just used instead of the other one. | 把求出的值代回刚用过的方程，而不是另一个方程。
- Finding x but never calculating the matching y. | 求出 x 却没有算出对应的 y。
- Giving only one solution for a linear-quadratic system. | 线性与二次方程组只给出一个解。
- Rounding decimals too early and carrying the error into later steps. | 过早对小数四舍五入，把误差带入后续步骤。
For presentation, write one equation above the other, label them (1) and (2), and show the addition or subtraction as a separate line. If the question says "use an algebraic method", a graphical reading alone will not earn the method marks. If a question asks for answers to a given number of decimal places, keep full accuracy until the final line.
书写方面，把两个方程上下排列并标注 (1)、(2)，把相加或相减单独写成一行。若题目写明"用代数方法求解"，只靠读图是拿不到方法分的。若题目要求答案保留若干位小数，请在最后一行之前一直保持完整精度。
12. Quick Revision Checklist | 快速复习清单
Use this checklist in the final week of revision. If you can tick every line without notes, you are ready for this topic in the exam.
在复习的最后一周使用这份清单。如果你能不看书就勾选每一项，说明这个专题你已经准备好了。
- I can explain what makes a solution valid for every equation. | 我能解释为什么一个解必须满足每一个方程。
- I can solve a linear system by elimination, including scaling. | 我能用消元法解线性方程组，包括先放大系数。
- I can solve a linear system by substitution. | 我能用代入法解线性方程组。
- I can link a solution to the intersection point of two graphs. | 我能把解与两个图像的交点联系起来。
- I can turn a linear-quadratic system into one quadratic and solve it. | 我能把线性与二次方程组化为一个二次方程并求解。
- I can use Δ = b² − 4ac to decide how many solutions exist. | 我能用 Δ = b² − 4ac 判断解的个数。
- I can set up equations from a word problem and answer in context. | 我能从应用题中建立方程并在情境中作答。
- I always check my final pair in both original equations. | 我总是把最终解代回两个原方程进行验算。
With both methods fluent and the checklist complete, simultaneous equations become one of the most reliable sources of marks in the whole paper. Practise a mixed set of linear and linear-quadratic questions under timed conditions, and keep your working lines tidy - clarity earns marks just as much as correct arithmetic does.
当两种方法都已熟练、清单全部勾选完毕，联立方程组就会成为整份试卷中最稳定的得分点之一。请在限时条件下练习一组线性与线性-二次混合题，并保持书写整洁——过程清楚与算得正确同样能得分。
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