📚 Simultaneous Equations: Elimination, Substitution and Graphs | 联立方程：消元法、代入法与图像法
In IGCSE and A-Level mathematics, simultaneous equations are one of the core algebra topics. A system of simultaneous equations is a set of two or more equations that share the same variables, and the solution is the pair of values that makes all the equations true at the same time. This topic matters because it appears not only as direct algebra questions but also inside word problems, graphical problems and higher-tier questions involving non-linear equations.
在 IGCSE 与 A-Level 数学中，联立方程是核心代数主题之一。联立方程组是由两个或更多共享相同变量的方程组成的集合，它的解就是能同时使所有方程成立的一组数值。这个主题非常重要，因为它不仅以直接代数题的形式出现，还会出现在文字应用题、图像题以及提高卷中涉及非线性方程的问题里。
1. What Are Simultaneous Equations? | 什么是联立方程？
quations in two unknowns, usually written x and y. The general form of a linear equation in two variables is ax + by = c, where a, b and c are constants and a and b are not both zero. Because there are two unknowns, a single equation is never enough - you need two independent equations to pin down one pair of values.
联立方程是指必须由相同的未知数值同时满足的方程。在 IGCSE 中，通常需要处理含有两个未知数的两个方程，未知数通常写作 x 和 y。二元一次方程的一般形式是 ax + by = c，其中 a、b 和 c 是常数，且 a、b 不同时为零。因为有两个未知数，单独一个方程永远不够——需要两个相互独立的方程才能确定唯一的一组解。
nd algebraically by elimination or substitution, or graphically by finding the point where the two lines intersect.
例如，x + y = 5 和 x - y = 1 构成一个线性方程组。有序数对 (3, 2) 同时满足两个方程，因为 3 + 2 = 5 且 3 - 2 = 1。解可以通过消元法或代入法进行代数求解，也可以通过找出两条直线交点的图像方法求得。
x + y = 5 x - y = 1 ⇒ x = 3, y = 2
The word 'simultaneous' simply means 'at the same time'. A common exam trap is to solve one equation and stop, forgetting that the answer must be checked in both equations. Always write your final answer as a coordinate pair (x, y) unless the question asks for the values separately.
"Simultaneous（联立）"这个词的意思就是"同时"。一个常见的考试陷阱是只解出一个方程就停下，忘记答案必须在两个方程中都成立。除非题目分别要求 x 和 y 的值，否则最终答案应写成坐标对 (x, y) 的形式。
2. The Graphical Meaning: Intersection of Two Lines | 图像意义：两条直线的交点
Every linear equation in two variables represents a straight line on the coordinate plane, and every point on that line is a solution of that single equation. The solution of the simultaneous system is therefore the point that lies on both lines at once - the point of intersection. This is why a pair of linear equations usually has exactly one solution.
二元一次方程在坐标平面上表示一条直线，直线上的每一个点都是这个单一方程的解。因此，联立方程组的解就是同时落在两条直线上的那个点——也就是交点。这就是为什么一对线性方程通常恰好只有一个解。
There are three possible cases. If the two lines meet at a single point, there is one unique solution and the equations are consistent and independent. If the two lines are parallel with different intercepts, there is no solution. If the two equations describe the same line, there are infinitely many solutions.
存在三种可能情形。若两条直线交于一点，则有唯一解，方程组相容且独立；若两条直线平行但截距不同，则无解；若两个方程描述的是同一条直线，则有无穷多组解。
| Case | 情形 | Lines | 直线关系 | Solutions | 解的个数 |
| Gradients different | 斜率不同 | Intersect once | 相交于一点 | One solution | 唯一解 |
| Gradients equal, intercepts differ | 斜率相同、截距不同 | Parallel | 平行 | No solution | 无解 |
| Same gradient and intercept | 斜率与截距都相同 | Coincident | 重合 | Infinitely many | 无穷多解 |
: m is the gradient and c is the y-intercept. If two lines have the same gradient but different values of c, they will never meet.
把每个方程整理成 y = mx + c 的形式便于比较：m 是斜率，c 是 y 轴截距。若两条直线斜率相同但 c 值不同，它们永远不会相交。
3. Method 1: Elimination by Adding or Subtracting | 方法一：加减消元法
Elimination works by adding or subtracting the two equations so that one variable cancels. If the coefficients of one variable are the same, subtract the equations; if the coefficients are opposite (for example +y and -y), add them. After one variable disappears, solve the remaining one-variable equation, then substitute back to find the other variable.
消元法的思路是把两个方程相加或相减，使其中一个未知数被消去。若某个未知数的系数相同，就相减；若系数互为相反数（例如 +y 与 -y），就相加。一个未知数消失后，解剩下的一元一次方程，再代回求另一个未知数。
2x + y = 11 x - y = 1
and x = 4. Substituting x = 4 into the second equation gives 4 - y = 1, so y = 3. The solution is (4, 3).
这里 y 项分别是 +y 与 -y，所以两式相加可直接消去 y。左边得 3x，右边得 12，因此 3x = 12，x = 4。把 x = 4 代入第二个方程得 4 - y = 1，所以 y = 3。解为 (4, 3)。
h is also correct. Checking is not optional - it is the quickest way to catch a sign slip.
把两个方程都验证一下：2(4) + 3 = 8 + 3 = 11，正确；4 - 3 = 1，也正确。检验不是可有可无的步骤——它是发现正负号错误最快的方法。
- Step 1: Write both equations in the form ax + by = c.
第一步：把两个方程都整理成 ax + by = c 的形式。 - Step 2: Make the coefficients of one variable equal or opposite.
第二步：使其中一个未知数的系数相等或互为相反数。 - Step 3: Add or subtract to eliminate that variable.
第三步：相加或相减，消去该未知数。 - Step 4: Solve, substitute back, and check.
第四步：求解、代回，并检验。
4. Elimination with Multiplication First | 先乘再消元：系数不匹配的处理
Most exam questions do not have matching coefficients straight away. In that case you multiply one or both equations by a suitable constant first. Choose the multiplier so that the coefficients of one variable become equal or opposite, then proceed as in Section 3.
大多数考题的系数并不会正好匹配。这时需要先把其中一个或两个方程整体乘以适当的常数。选择合适的倍数，使某个未知数的系数变得相等或互为相反数，然后按第 3 节的方法继续。
3x + 2y = 16 2x - 3y = -11
s y: 13x = 26, so x = 2. Substituting into the first equation gives 3(2) + 2y = 16, so 2y = 10 and y = 5.
把第一个方程乘以 3、第二个方程乘以 2，使 y 的系数变成 +6 和 -6，得到 9x + 6y = 48 与 4x - 6y = -22。两式相加消去 y：13x = 26，所以 x = 2。代入第一个方程得 3(2) + 2y = 16，所以 2y = 10，y = 5。
correct. Notice that when you multiply an equation, every term on both sides must be multiplied, including the constant term.
用第二个方程检验：2(2) - 3(5) = 4 - 15 = -11，解 (2, 5) 正确。注意：方程两边相乘时，每一项都必须乘，包括常数项。
A useful shortcut is to use the lowest common multiple of the two coefficients. For coefficients 2 and 3, the lowest common multiple is 6, so you multiply by 3 and 2 respectively. Working with the smallest possible numbers reduces arithmetic errors.
一个实用技巧是取两个系数的最小公倍数。系数为 2 和 3 时，最小公倍数是 6，因此分别乘以 3 和 2。使用尽可能小的数可以降低计算错误率。
5. Method 2: Substitution | 方法二：代入法
Substitution means rearranging one equation to make one variable the subject, then replacing that variable in the other equation. It is the natural choice when one equation already gives y in terms of x, or x in terms of y, because no rearrangement is needed.
代入法是先把其中一个方程整理成某个未知数为主题的形式，再把这个表达式代入另一个方程。当某个方程已经以 y = ... 或 x = ... 的形式给出时，代入法最自然，因为不需要额外变形。
y = 2x - 1 3x + y = 9
and x = 2. Then y = 2(2) - 1 = 3. The solution is (2, 3), which checks in both equations.
把 y = 2x - 1 代入第二个方程得 3x + (2x - 1) = 9。化简得 5x - 1 = 9，所以 5x = 10，x = 2。再由 y = 2(2) - 1 = 3。解为 (2, 3)，在两个方程中都成立。
r equation. Always substitute into the equation you have not yet used, otherwise you simply get an identity such as 5 = 5.
如果两个方程都没有整理好，也可以自行变形。由 x + y = 5 可得 y = 5 - x，再代入另一个方程。切记要代入"还没有用过"的那个方程，否则只会得到像 5 = 5 这样的恒等式。
- Best for: one variable already isolated, or one coefficient equal to 1.
最适合：某个未知数已被单独表示，或某个系数等于 1。 - Avoid when: both equations are in standard form with large coefficients.
不适合：两个方程都是标准形式且系数较大时。 - Always: expand brackets carefully and collect like terms.
务必：小心展开括号并合并同类项。
6. Choosing the Most Efficient Method | 如何选择最合适的方法
ion is often quicker. If one equation is already solved for one variable, substitution is faster. Both methods always give the same answer, so the choice is about speed and accuracy, not correctness.
理解方程的结构有助于选择最高效的方法。如果两个方程都是 ax + by = c 的标准形式，消元法通常更快；如果某个方程已经把某个未知数孤立出来，代入法更快。两种方法得到的答案永远相同，所以选择只关乎速度与准确度，与对错无关。
| Situation | 情形 | Recommended method | 推荐方法 |
| Matching or opposite coefficients | 系数相同或相反 | Add or subtract directly | 直接相加或相减 |
| Both in standard form, no match | 都是标准形式且不匹配 | Elimination with multipliers | 先乘再消元 |
| One variable isolated | 某未知数已孤立 | Substitution | 代入法 |
| One linear, one quadratic | 一个一次、一个二次 | Substitution (essential) | 必须用代入法 |
A good habit is to scan the coefficients for one second before you start writing. Two or three seconds of planning saves minutes of messy algebra, especially in an exam where time pressure is high.
一个好习惯是动笔前先花一秒扫一眼系数。两三秒的规划可以省下几分钟的繁琐运算，在时间紧张的考试中尤其重要。
7. Checking Your Answer and Common Errors | 检验答案与常见错误
Checking takes ten seconds and protects marks. Substitute your values of x and y into the equation you did not use for the final substitution. If both sides are equal, the answer is correct. This is especially valuable in elimination questions where a single sign error can change the answer completely.
检验只需要十秒，却能保住分数。把求得的 x 和 y 代入"最后一步没有用到"的那个方程。如果两边相等，答案就是正确的。在消元法题目中这一点尤其重要，因为一个正负号错误就足以让答案完全错误。
- Sign error when subtracting: subtract the whole equation, not just the first term.
相减时的符号错误：要减去整个方程，而不只是第一项。 - Forgetting to multiply everything: when scaling an equation, multiply the constant too.
忘记整体相乘：缩放方程时，常数项也要乘。 - Solving for only one variable: the question usually wants both x and y.
只求出一个未知数：题目通常要求同时给出 x 和 y。 - Substituting into the wrong equation: this produces a true but useless identity.
代入错误的方程：会得到成立但没有用处的恒等式。 - Losing one solution: in quadratic systems there may be two answers.
漏解：在二次方程组中可能存在两组答案。
Write your working in clear steps: label the equations (1) and (2), show the elimination or substitution, and finish with a boxed answer. Examiners award method marks, so a clear layout can earn credit even when the final number is wrong.
书写要清晰、分步：给方程标上 (1)、(2)，写出消元或代入过程，最后把答案框起来。考官会给出过程分，因此即使最终数字有误，清晰的排版依然可能得分。
8. Word Problems: Forming the Equations | 文字应用题：如何建立方程
Word problems test whether you can translate English into algebra before solving. Define your variables clearly first - write 'let x be ...' and 'let y be ...' - then build two equations from the two pieces of information given. Units must be consistent throughout.
文字应用题考查的是在求解之前把中文或英文描述翻译成代数的能力。先清楚地定义变量——写出"设 x 为……"、"设 y 为……"——再根据题中给出的两条信息建立两个方程。单位必须前后一致。
Consider this example: two coffees and three teas cost £11.40, while three coffees and two teas cost £12.60. Let c be the price of a coffee and t the price of a tea, both in pounds.
看一个例子：两杯咖啡加三杯茶共 11.40 英镑，三杯咖啡加两杯茶共 12.60 英镑。设 c 为咖啡单价、t 为茶单价，单位均为英镑。
2c + 3t = 11.40 3c + 2t = 12.60
5t = 9.00, so t = 1.80. Then 2c + 5.40 = 11.40, so 2c = 6.00 and c = 3.00. A coffee costs £3.00 and a tea costs £1.80.
把第一个方程乘以 3、第二个方程乘以 2：6c + 9t = 34.20 与 6c + 4t = 25.20。相减得 5t = 9.00，所以 t = 1.80。再由 2c + 5.40 = 11.40 得 2c = 6.00，c = 3.00。咖啡单价 3.00 英镑，茶单价 1.80 英镑。
Other classic contexts include age problems ('in five years' time the father will be three times as old as his son'), number problems involving digits, and mixture or speed problems. In each case the structure is the same: two unknowns, two equations, then your chosen method.
其他经典情境包括年龄问题（"五年后父亲的年龄是儿子的三倍"）、涉及数字的整数问题，以及混合或行程问题。每种情况的结构都相同：两个未知数、两个方程，然后选择方法求解。
9. Solving Graphically: Plotting and Reading Off | 图像法：作图与读点
line you need only two or three points, so build a small table of values for each equation. Rearranging into y = mx + c form makes plotting much easier.
图像
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