📚 Solving Simultaneous Equations: Elimination, Substitution and Graphs | 解联立方程组：消元法、代入法与图像法
In IGCSE Mathematics, simultaneous equations appear regularly in both core and extended papers. They require you to find values that satisfy two or more equations at the same time. This guide explains the main algebraic methods, graphical interpretation, special cases, and exam-style applications, with fully worked examples for every technique.
在 IGCSE 数学中，联立方程组是核心卷和扩展卷中的常见题型。题目要求你找到同时满足两个或更多方程的值。本指南将讲解主要的代数方法、图像含义、特殊情况以及考试型应用，并为每一种技巧配有完整例题。
1. What Are Simultaneous Equations? | 什么是联立方程组
A system of simultaneous equations consists of two or more equations involving the same unknown variables. A solution is a pair of values (x, y) that satisfies every equation in the system. For two linear equations, the solution represents the point where their straight-line graphs intersect.
联立方程组由两个或更多含有相同未知数的方程组成。方程组的解是一组数值 (x, y)，它满足系统中的每一个方程。对于两个线性方程，解表示两条直线图像的交点坐标。
Because both equations must be true at the same time, a candidate answer has to be checked in both of them. A pair of values that works in one equation but fails in the other is not a solution, no matter how tidy it looks.
由于两个方程必须同时成立，任何候选答案都必须在两个方程中检验。只满足其中一个方程、却在另一个方程中不成立的数值不是解，无论它看起来多么整齐。
x + y = 7
2x − y = 5
n true: 4 + 3 = 7 and 2(4) − 3 = 5. In coordinate form, the two lines intersect at (4, 3).
唯一解是 x = 4, y = 3，因为代入这两个值后每个方程都成立：4 + 3 = 7，2(4) − 3 = 5。在坐标形式中，两条直线相交于点 (4, 3)。
am, the answer is normally written as a coordinate pair in brackets, such as (4, 3), or as 'x = 4, y = 3' use for the final step, so that an arithmetic slip is caught before you hand in the paper.
考试中答案通常写成括号形式的坐标对，例如 (4, 3)，或写成 'x = 4, y = 3'。写完后一定要把结果代入你最后一步没有使用的那个方程，这样在交卷前就能发现计算失误。
2. The Elimination Method: The Core Idea | 消元法：核心思路
Elimination works best when the coefficients of one variable can be made equal or opposite. You add or subtract the two equations to remove that variable, then solve for the remaining unknown. The logic is simple: if two quantities are equal, adding equal amounts to both keeps them equal, so whole equations can be combined safely.
当某个变量的系数可以化为相同或相反数时，消元法最有效。你将两个方程相加或相减，消去该变量，然后解出剩下的未知数。其逻辑很简单：如果两个量相等，给两边加上相等的量仍然相等，所以整个方程可以安全地合并。
ions by a constant so that one variable has matching or opposite coefficients; (3) add (if the coefficients are opposite) or subtract (if they are the same) to eliminate that variable; (4) solve the resulting one-variable equation and back-substitute.
四个步骤是：（1）把两个方程写成同一形式，通常是 ax + by = c；（2）给一个或两个方程乘上适当的常数，使某个变量的系数相同或相反；（3）若系数相反则相加，若相同则相减，从而消去该变量；（4）解出剩下的一元方程，再回代求出另一个未知数。
If you choose to subtract, keep the order consistent: subtract the whole of the second equation from the whole of the first, including the constant term on the right-hand side. Most lost marks in this topic come from dropping a sign at this exact step.
如果你选择相减，一定要保持顺序一致：用第一个方程整体减去第二个方程，包括等号右边的常数项。这个题型中大多数丢分都发生在这关键的一步上，原因就是符号写错。
3. Elimination Worked Examples | 消元法例题精讲
Example 1 - coefficients already opposite. Adding the two equations removes y immediately, which is the fastest possible route to the answer.
例 1——系数已经相反。将两个方程相加可以立刻消去 y，这是最快的解题路径。
2x + y = 10
3x − y = 5
y = 4. The solution is (3, 4), and it checks in the second equation because 3(3) − 4 = 5.
将两个方程相加得到 5x = 15，所以 x = 3。代回原方程得 2(3) + y = 10，因此 y = 4。解为 (3, 4)，并且它在第二个方程中成立，因为 3(3) − 4 = 5。
e, multiplying 2x + 3y = 8 by 2 and keeping 4x − y = 5 unchanged makes the x terms 4x and 4x, ready for subtraction.
例 2——需要先配平系数。如果系数不匹配，就给一个或两个方程乘上合适的常数。例如，将 2x + 3y = 8 乘以 2，而 4x − y = 5 保持不变，就可使 x 项都变为 4x，便于相减消元。
2x + 3y = 8 → ×2 → 4x + 6y = 16
4x − y = 5
so x = 23/14. Checking in the original first equation: 2(23/14) + 3(11/7) = 23/7 + 33/7 = 56/7 = 8, which is correct.
用加倍后的第一个方程减去第二个方程，即可消去 x：7y = 11，所以 y = 11/7。代入 4x − y = 5 得 4x = 5 + 11/7 = 46/7，因此 x = 23/14。代回原第一个方程检验：2(23/14) + 3(11/7) = 23/7 + 33/7 = 56/7 = 8，成立。
s 10x + 4y = 22, while the second stays as 3x + 4y = 13. Subtracting eliminates y and leaves 7x = 9.
例 3——两个方程都需要调整。另一个非常常见的组合是 5x + 2y = 11 与 3x + 4y = 13。将第一个方程乘以 2 得到 10x + 4y = 22，而第二个方程仍为 3x + 4y = 13。相减后消去 y，得到 7x = 9。
4/7 = 91/7 = 13. Fraction answers are perfectly acceptable unless the question asks for decimals to a given accuracy.
因此 x = 9/7。代回 5x + 2y = 11 得 45/7 + 2y = 11，所以 2y = 32/7，y = 16/7。在第二个方程中检验：3(9/7) + 4(16/7) = 27/7 + 64/7 = 91/7 = 13。除非题目要求保留到指定精度的近似值，否则分数答案完全可接受。
4. The Substitution Method | 代入法
Substitution is useful when one equation has an isolated variable or can be rearranged easily. You substitute that expression into the other equation to create a one-variable equation, solve it, and then substitute the value back to find the second unknown.
当某个方程中已经有一个变量被单独表示出来，或者可以轻松变形时，代入法就很有用。你把这个表达式代入另一个方程，得到一个一元方程，解出它，再把结果代回求出第二个未知数。
The three steps are: (1) make x or y the subject of one equation, ideally the one with coefficient 1 or −1; (2) replace that variable in the other equation with the expression and simplify; (3) solve, then substitute back into the rearranged equation, not into the original messy one, to keep the arithmetic light.
三个步骤是：（1）把某个方程变形为 x 或 y 的表达式，最好选择系数为 1 或 −1 的那个方程；（2）在另一个方程中用该表达式替换这个变量并化简；（3）求解，然后代回到你变形后的式子中（而不是回到原来那个较复杂的方程）求另一个值，这样计算更轻便。
Substitution is also the only practical algebraic route for systems where one equation is a curve, such as a circle or a parabola, because you cannot simply add or subtract a linear equation and a quadratic one to cancel a variable.
当方程组中有一个方程表示曲线（如圆或抛物线）时，代入法是唯一实用的代数途径，因为你无法简单地通过相加减来消去线性方程与二次方程中的变量。
5. Substitution Worked Examples | 代入法例题精讲
Example 1 - one variable already isolated. Here y is already the subject of the first equation, so the substitution is immediate.
例 1——一个变量已经单独表示。这里 y 已经是第一个方程的主元，代入可以立刻进行。
y = 3x − 2
x + y = 10
nd x = 3. Then y = 3(3) − 2 = 7. The solution is (3, 7), and it checks because 3 + 7 = 10.
把 y 的表达式代入第二个方程得 x + 3x − 2 = 10，所以 4x = 12，x = 3。于是 y = 3(3) − 2 = 7。解为 (3, 7)，并因 3 + 7 = 10 而成立。
