📚 Solving Simultaneous Equations | 解联立方程组
Simultaneous equations are among the most frequently tested topics in IGCSE Mathematics. Whenever two unknown quantities are linked by two independent equations, you are dealing with a system of simultaneous equations. Mastery of this topic builds confidence across algebra, coordinate geometry, and real-world problem solving.
联立方程组是 IGCSE 数学中考查频率最高的内容之一。只要两个未知量由两个独立的方程联系，你就遇到了联立方程组。掌握这一主题能提升你在代数、坐标几何以及实际应用问题中的综合解题能力。
1. What Are Simultaneous Equations? | 什么是联立方程组？
xample, in the system below, only x = 3 and y = 1 make both equations true simultaneously.
联立方程组包含两个或多个共享相同变量的方程。该方程组的一个解是同时满足所有方程的有序数对 (x, y)。例如，在下面的方程组中，只有 x = 3 且 y = 1 时，两个方程才同时成立。
x + y = 4
x − y = 2
s both equations. The key idea is that two distinct lines in a plane intersect at exactly one point — that intersection point is the solution.
这里，当 x = 3、y = 1 时，3 + 1 = 4 且 3 − 1 = 2。没有任何其他数值对能同时满足两个方程。其核心思想是：平面内两条不同的直线恰好相交于一点——那个交点就是方程组的解。
2. The Elimination Method | 消元法
The elimination method involves adding or subtracting the two equations to cancel out one variable. This works best when the coefficients of one variable are equal in magnitude. Let us consider the system:
消元法通过将两个方程相加或相减来消去一个变量。当一个变量的系数绝对值相等时，这种方法最有效。我们来看这个方程组：
2x + y = 7
x − y = 2
Notice that the y-term has coefficient +1 in the first equation and −1 in the second. If we add the two equations, the y terms cancel:
注意，第一个方程中 y 的系数是 +1，第二个方程中 y 的系数是 −1。将两个方程相加，y 项就被消去了：
(2x + y) + (x − y) = 7 + 2
3x = 9
x = 3
Always check your answer by substituting both values into the other equation: 2(3) + 1 = 7 ✓.
现在将 x = 3 代回任意一个原方程。利用 x − y = 2，得到 3 − y = 2，所以 y = 1。务必代入另一个方程检验答案：2(3) + 1 = 7 ✓。
If neither variable has equal coefficients, you must multiply one or both equations by a constant first. For instance:
如果两个方程中没有一个变量的系数相等，你需要先将一个或两个方程乘以适当的常数。例如：
3x + 2y = 12
2x − y = 1
Multiply the second equation by 2 to make the y coefficients match in magnitude (2 and −2):
将第二个方程乘以 2，使 y 的系数绝对值相等（2 和 −2）：
3x + 2y = 12
4x − 2y = 2
. Substituting back, 3(2) + 2y = 12, which gives 2y = 6, so y = 3. The solution is (2, 3).
相加得到 7x = 14，因此 x = 2。代回原方程，3(2) + 2y = 12，得到 2y = 6，所以 y = 3。解为 (2, 3)。
3. The Substitution Method | 代入法
Substitution is the preferred method when one equation is already solved for a variable, or when one variable has a coefficient of 1. The procedure is: rearrange one equation to make one variable the subject, then substitute that expression into the other equation.
当其中一个方程已经解出某个变量，或某个变量的系数为 1 时，代入法是最优选择。步骤是：先从其中一个方程中解出一个变量，然后将该表达式代入另一个方程。
Consider the system:
考虑以下方程组：
y = 2x − 1
3x + 2y = 12
st equation already gives y in terms of x, substitute y = 2x − 1 into the second equation:
由于第一个方程已经用 x 表示 y，将 y = 2x − 1 代入第二个方程：
3x + 2(2x − 1) = 12
3x + 4x − 2 = 12
7x = 14
x = 2
3). Note that this is the same answer we found using elimination for a differently posed problem — different methods must always agree.
再将 x = 2 代回 y = 2x − 1，得到 y = 2(2) − 1 = 3。解为 (2, 3)。请注意，这与我们之前用消元法解决不同问题得到的答案一致——不同的方法得出的结果必须相同。
4. The Graphical Method | 图像法
range each equation into the form y = mx + c before plotting, or use the intercept method.
从图像上看，每个线性方程代表一条直线。两个线性方程组的解就是两条直线交点的坐标。画图前应将每个方程化为 y = mx + c 的形式，或使用截距法。
Take the system:
以这个方程组为例：
y = x + 1
y = −2x + 7
Plot both lines on the same axes. The first line has gradient 1 and y-intercept 1. The second has gradient −2 and y-intercept 7. By drawing them carefully, you will find they intersect at (2, 3).
在同一坐标轴上画出两条直线。第一条线斜率为 1，y 截距为 1；第二条线斜率为 −2，y 截距为 7。仔细作图后，你会发现它们相交于点 (2, 3)。
The graphical method is useful for estimating solutions and for checking answers, but it is rarely precise enough for exact values in an exam. Always prefer an algebraic method for final answers, and use a graph only to verify.
图像法适合估算解和检验答案，但在考试中很少足够精确。务必优先使用代数方法得出最终答案，用图像仅作验证。
5. Forming Equations from Word Problems | 从应用题中建立方程
Many IGCSE questions present a real-world scenario and require you to translate the words into two equations. The key is to define your variables clearly at the start. Let the unknown quantities be x and y, then read each sentence carefully and convert it into an equation.
许多 IGCSE 题目给出实际情境，要求你将文字翻译成两个方程。关键在于一开始就明确设定变量。设未知量为 x 和 y，然后仔细阅读每句话并转化为方程。
Example: "Two pens and one notebook cost $7. One pen and three notebooks cost $11." Let p be the price of a pen and n the price of a notebook. The equations are:
例："两支笔和一个笔记本共 7 美元。一支笔和三个笔记本共 11 美元。"设 p 为笔的单价，n 为笔记本的单价，则方程为：
2p + n = 7
p + 3n = 11
hich simplifies to 5p = 10, so p = 2. Substitute back: 2(2) + n = 7, so n = 3. A pen costs $2 and a notebook costs $3.
用消元法：第一个方程乘以 3 得到 6p + 3n = 21。用它减去第二个方程：(6p + 3n) − (p + 3n) = 21 − 11，化简得 5p = 10，所以 p = 2。代回：2(2) + n = 7，所以 n = 3。一支笔 2 美元，一个笔记本 3 美元。
When forming equations, always pay attention to units and whether the question asks for "each" item or "total" — these detail points are a common source of lost marks.
建立方程时，始终注意单位以及问题是问"每件"物品还是"总计"——这些细节是常见的失分点。
6. One Linear and One Quadratic | 线性与二次方程的联立
In extended IGCSE Mathematics, you must also solve systems where one equation is linear and the other is quadratic. The method is always substitution: substitute the linear equation into the quadratic, producing a quadratic equation in one variable, which you then solve by factorising or using the quadratic formula.
在 IGCSE 扩展数学中，你还需要求解一个线性方程与一个二次方程组成的方程组。方法始终是代入：将线性方程代入二次方程，得到一个关于一个变量的二次方程，然后通过因式分解或二次公式求解。
Example: solve the system
例：解方程组
y = x + 1
y = x² − 2x + 1
Substitute y = x + 1 into the quadratic:
将 y = x + 1 代入二次方程：
x + 1 = x² − 2x + 1
0 = x² − 3x
x(x − 3) = 0
4. The solutions are (0, 1) and (3, 4). Graphically, these are the two intersection points of the line with the parabola.
因此 x = 0 或 x = 3。对每个 x，用线性方程求出对应的 y：当 x = 0 时，y = 1；当 x = 3 时，y = 4。解为 (0, 1) 和 (3, 4)。在图像上，这是直线与抛物线的两个交点。
Always collect all terms on one side and set the quadratic equal to 0 before solving. If the quadratic does not factorise, use the quadratic formula:
求解前务必将所有项移到一边，使二次方程等于 0。如果二次式无法因式分解，则用二次公式：
x = (−b ± √b² − 4ac) ÷ 2a
Make sure you show the substitution clearly — examiners award method marks even if the final arithmetic goes wrong.
确保清晰地展示代入过程——即使最终计算出错，阅卷官也会给方法分。
7. Common Mistakes and Exam Tips | 常见错误与考试技巧
Below are the most common errors IGCSE candidates make with simultaneous equations, along with advice to avoid them:
以下是 IGCSE 考生在联立方程组中最常犯的错误，以及避免这些错误的建议：
Forgetting to check the solution in both equations. Always substitute your values back into one of the original equations to verify.
忘记在两个方程中检验答案。始终将数值代回某个原方程进行验证。
Sign errors when subtracting equations. If you subtract one equation from another, change every sign of the subtracted equation first.
相减方程时出现符号错误。从另一个方程中减去一个方程时，先改变被减方程的所有正负号。
Mixing up x and y values in the final answer. Give your answer as an ordered pair (x, y) in the correct order.
在最终答案中搞混 x 和 y 的值。以有序数对 (x, y) 的形式给出答案，注意顺序。
iply every term. For 2x − y = 1 multiplied by 2, you must get 4x − 2y = 2, not 4x − y = 1.
方程乘以常数时，没有乘以每一项。2x − y = 1 乘以 2 必须得到 4x − 2y = 2，而不是 4x − y = 1。
Using a graphical solution when an exact algebraic answer is required. Examiners accept graphs only when the question specifically says "by drawing".
当题目要求精确代数答案时却使用图像法。只有当题目明确说"通过作图"时，阅卷官才接受图像解法。
Dropping one solution when solving a linear-quadratic system. Since a line can intersect a curve twice, always record both ordered pairs.
求解线性二次方程组时漏掉一个解。由于直线与曲线可能有两个交点，务必写出两个有序数对。
8. Practice Questions | 练习题
Try these problems on your own before checking the solutions. They reflect the standard difficulty of IGCSE exam questions.
请先独立完成以下题目，再核对答案。这些题的难度与 IGCSE 考试相当。
Question 1: Solve 3x + 2y = 8 and x − 2y = 0.
Question 2: Solve y = 5 − x and y = x² + 3x + 1.
Question 3: The sum of two numbers is 12 and their difference is 4. Find the numbers.
Solutions:
解答：
4x = 8, so x = 2. Substituting into x − 2y = 0 gives 2 − 2y = 0, so y = 1. Answer: (2, 1).
第 1 题：两式相加得 4x = 8，所以 x = 2。代入 x − 2y = 0 得 2 − 2y = 0，所以 y = 1。答案：(2, 1)。
+ 3x + 1: 5 − x = x² + 3x + 1, so 0 = x² + 4x − 4. Using the quadratic formula, x = (−4 ± √32) ÷ 2 = −2 ± √8. Then y = 5 − x = 7 ∓ √8. Answers: (−2 + √8, 7 − √8) and (−2 − √8, 7 + √8).
第 2 题：将 y = 5 − x 代入 y = x² + 3x + 1：5 − x = x² + 3x + 1，所以 0 = x² + 4x − 4。用二次公式，x = (−4 ± √32) ÷ 2 = −2 ± √8。则 y = 5 − x = 7 ∓ √8。答案：(−2 + √8, 7 − √8) 和 (−2 − √8, 7 + √8)。
a + b = 12 and a − b = 4. Adding: 2a = 16, so a = 8. Then b = 4. The numbers are 8 and 4.
第 3 题：设两个数为 a 和 b。则 a + b = 12，a − b = 4。相加：2a = 16，所以 a = 8，则 b = 4。两个数为 8 和 4。
Once you feel comfortable with these methods, practise past-paper questions under timed conditions. Consistent practice is the most reliable way to secure full marks on this topic.
当你熟练掌握这些方法后，请在限时条件下练习历年真题。持续练习是在这一主题上获得满分的可靠途径。
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