📚 Simultaneous Equations: Elimination, Substitution and Graphs | 联立方程组：消元法、代入法与图像法
In IGCSE Mathematics, a simultaneous equation is a pair (or system) of equations that share the same two unknowns. The solution is the pair of values that makes all equations true at the same time. This article reviews the main algebraic methods - substitution and elimination - as well as the graphical interpretation, special cases, and common exam-style applications.
在 IGCSE 数学中，联立方程组是指含有相同两个未知数的一组方程（通常为两个方程）。它的解就是同时使所有方程成立的一组未知数值。本文将复习主要代数方法——代入法和消元法，以及图像意义、特殊情形和常见考试应用题。
1. Introduction to Simultaneous Equations | 联立方程组简介
e solution is an ordered pair (x, y) that satisfies every equation in the system at the same time.
联立方程组是共用相同未知数的一组方程。在 IGCSE 中，通常求解两个含有两个变量的方程，常写作 ax + by = c 和 dx + ey = f。方程组的解就是同时满足组内每个方程的坐标对 (x, y)。
= 5 and 2x − y = 1 have the solution x = 2 and y = 3 because 2 + 3 = 5 and 2 × 2 − 3 = 1.
例如，方程 x + y = 5 与 2x − y = 1 的解为 x = 2，y = 3，因为 2 + 3 = 5 且 2 × 2 − 3 = 1。
x + y = 5
2x − y = 1
2. Graphical Interpretation and Solving by Graphs | 图像意义与图像法
If you draw each equation as a straight line on the same axes, the solution is the point where the two lines cross. This is because the intersection point lies on both lines, so its coordinates satisfy both equations.
如果在同一坐标系中把每个方程画成直线，方程组的解就是两条直线的交点。因为交点同时在两条直线上，所以它的坐标同时满足两个方程。
Graphical methods are useful for checking answers, but they can be inaccurate when the solution is not an integer. On graph paper, plot at least two points for each line, then read off the coordinates of the intersection.
图像法适合用来检验答案，但当解不是整数时可能不够精确。在方格纸上作图时，每条直线至少描两个点，然后读出交点坐标。
es through (0,−1) and (1,1). The intersection is (2,3), confirming the algebraic solution.
在上面的例子中，直线 x + y = 5 经过点 (0,5) 和 (5,0)，直线 2x − y = 1 经过点 (0,−1) 和 (1,1)。两条直线的交点是 (2,3)，这与代数解一致。
3. Solving by Substitution Method | 代入法求解
This is efficient when one equation already has y or x as the subject, such as y = 2x + 1.
代入法是指先把一个方程变形，将其中一个变量表示出来，再把这个表达式代入另一个方程。当某个方程已经是 y 或 x 为被表达对象时，例如 y = 2x + 1，代入法就非常高效。
tute this expression into the other equation. Step 3: solve the resulting one-variable equation. Step 4: substitute back to find the other variable.
第一步：把一个方程整理成 y = ... 或 x = ... 的形式。第二步：将这个表达式代入另一个方程。第三步：解得到的一元方程。第四步：代回原式求出另一个变量。
+ 1 = 16, so x = 3. Substitute x = 3 into y = 2x + 1 to get y = 7. The solution is (3,7).
例题：解方程组 y = 2x + 1 与 3x + y = 16。将第二个方程中的 y 替换：3x + (2x + 1) = 16。化简得 5x + 1 = 16，所以 x = 3。把 x = 3 代入 y = 2x + 1，得 y = 7。方程组的解为 (3,7)。
4. Solving by Elimination Method | 消元法求解
ning up like terms. If the coefficients of one variable are the same or opposites, add or subtract the equations directly. If not, multiply one or both equations by suitable constants first.
消元法通过相加或相减消去一个未知数。先把两个方程都写成 ax + by = c 的形式，并对齐同类项。如果某个未知数的系数相同或互为相反数，就可以直接相加或相减。如果系数不同，则需要先给一个或两个方程乘以适当的常数。
12, so x = 2. Substitute into 2x + 3y = 13 to get 4 + 3y = 13, so y = 3. Solution: (2,3).
例 1：2x + 3y = 13 与 4x − 3y = −1。y
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