📚 Solving Simultaneous Equations by Elimination | 使用消元法解联立方程组
Simultaneous equations are a pair (or more) of equations that share the same variables. The solution is the set of values that makes all equations true at the same time. In KS3 and Cambridge Checkpoint Mathematics, learning to solve them by elimination builds strong algebraic foundations for IGCSE and beyond.
联立方程组是一组共享相同变量的方程。所谓解，就是能够同时满足所有方程的那组值。在 KS3 和剑桥 Checkpoint 数学阶段，学习用消元法解联立方程组能为 IGCSE 及更高阶的代数学习打下扎实基础。
1. What Are Simultaneous Equations? | 什么是联立方程组？
b, c, d, e and f are constants. Graphically, each equation represents a straight line. The solution is the point where the lines intersect.
通常情况下，二元一次联立方程组形如 ax + by = c 和 dx + ey = f，其中 a、b、c、d、e、f 为常数。从几何角度看，每个方程代表一条直线；方程组的解就是这两条直线的交点坐标。
If the lines intersect at exactly one point, there is a unique solution. If they are parallel and distinct, there is no solution. If they are the same line, there are infinitely many solutions. At KS3, we mostly work with systems that have one unique solution.
如果两条直线相交于恰好一点，方程组有唯一解；如果两条直线平行但不重合，则无解；如果两条直线重合，就有无穷多解。在 KS3 阶段，我们主要处理具有唯一解的方程组。
2. The Idea Behind Elimination | 消元法的基本思想
Elimination means we combine the two equations to remove one of the variables. This is done by making the coefficients of either x or y equal in magnitude (but possibly opposite in sign), so that adding or subtracting the equations cancels that variable.
消元法是指我们将两个方程进行组合，使其中一个变量的系数绝对值相等（符号则可同可异），然后通过相加或相减把这一变量消去，从而得到一个只含单个变量的方程。
Once one variable is eliminated, we solve the resulting linear equation in one unknown. Then we substitute back to find the other variable. The elimination method is systematic and reduces the chance of arithmetic mistakes compared to guessing.
消去一个变量后，我们解出只含一个未知量的线性方程，再代回原方程求另一个变量。相比猜测法，消元法步骤清晰，能有效减少计算错误。
3. Same-Sign vs Opposite-Sign Coefficients | 同号系数与异号系数
When the coefficients of the variable you want to eliminate have the same sign, you subtract one equation from the other. When they have opposite signs, you add the equations. Getting this rule right is crucial.
如果要消去的变量在两个方程中的系数符号相同，就将两方程相减；如果符号相反，则将两方程相加。准确应用这一规则至关重要。
y. For 2x + y = 7 and 3x − y = 8, the y-coefficients have opposite signs, so we add them.
例如：对于 2x + y = 7 和 3x + y = 10，两个 y 的系数都为正，应相减以消去 y；而对于 2x + y = 7 和 3x − y = 8，y 的系数一正一负，应将两方程相加。
4. Step-by-Step Procedure | 逐步消元流程
Follow these steps every time: (1) Label the equations (1) and (2). (2) Decide which variable to eliminate. (3) Multiply one or both equations by suitable numbers so that the chosen variable has the same coefficient magnitude. (4) Add or subtract the equations to eliminate that variable. (5) Solve the resulting single-variable equation. (6) Substitute the found value back into one original equation to find the other variable. (7) Check your solution in both equations.
每次解题都遵循下列步骤：(1) 将两个方程分别标为 (1) 和 (2)。(2) 决定消去哪一個变量。(3) 选择一个或多个方程乘以适当的数，使要消去的变量系数绝对值相等。(4) 通过相加或相减消去该变量。(5) 解所得的一元一次方程。(6) 将求出的值代回原方程之一，解出另一个变量。(7) 将解代入原两个方程进行检验。
5. Example 1: Direct Elimination (Same Coefficient) | 例 1：直接消元（系数全等）
uation (2) from equation (1): (4x + 3y) − (2x + 3y) = 17 − 9. This gives 2x = 8, so x = 4.
解方程组：4x + 3y = 17 和 2x + 3y = 9。注意到 y 的系数都是 3，大小相等且同号。将方程 (1) 减去方程 (2)：(4x + 3y) − (2x + 3y) = 17 − 9，得到 2x = 8，因此 x = 4。
= 1/3. Check in equation (1): 4(4) + 3(1/3) = 16 + 1 = 17. The solution is x = 4, y = 1/3.
将 x = 4 代入方程 (2)：2(4) + 3y = 9 → 8 + 3y = 9 → 3y = 1 → y = 1/3。代回方程 (1) 检验：4(4) + 3(1/3) = 16 + 1 = 17，成立。解为 x = 4, y = 1/3。
6. Example 2: Multiplying One Equation | 例 2：乘上一个方程
magnitude. Simply add the two equations: (3x + 2y) + (x − 2y) = 12 + 4 → 4x = 16 → x = 4.
解方程组：3x + 2y = 12 和 x − 2y = 4。y 的系数分别为 +2 和 −2，绝对值相等符号相反。直接将两方程相加：(3x + 2y) + (x − 2y) = 12 + 4 → 4x = 16 → x = 4。
−2(0)=4. Note that no multiplication was needed here; it is a straightforward elimination.
将 x = 4 代入 x − 2y = 4：4 − 2y = 4 → −2y = 0 → y = 0。解 (4,0) 满足两个方程。此题不需要乘以任何数，是直接的消元。
7. Example 3: Scaling Both Equations | 例 3：两个方程同时缩放
6. Multiply equation (1) by 2 and equation (2) by 3 to get 6x + 8y = 20 and 6x + 15y = 27.
解方程组：3x + 4y = 10 和 2x + 5y = 9。x 和 y 的系数都不相等。选择消去 x，系数分别是 3 和 2，最小公倍数为 6。将方程 (1)×2，方程 (2)×3，得到 6x + 8y = 20 和 6x + 15y = 27。
=27−20 → 7y=7 → y=1. Substitute y=1 into (1): 3x+4(1)=10 → 3x=6 → x=2. Solution: x=2, y=1.
新方程中 x 系数均为 +6（同号），因此相减：(6x+15y)−(6x+8y)=27−20 → 7y=7 → y=1。将 y=1 代入 (1)：3x+4=10 → 3x=6 → x=2。解为 x=2, y=1。
8. Checking Your Solution Effectively | 有效检验你的解
Always substitute your found values into both original equations, not just the one used for back-substitution. This catches any mistake made earlier. Write your check clearly in your working.
一定要将求得的解同时代入原两个方程，而不仅仅是那个用于回代的方程。这样可以及时发现之前的错误。在解题步骤中清晰地写出检验过程。
h the right-hand sides. If either fails, re-examine your elimination steps and arithmetic.
以上题为例，x=2, y=1：3(2)+4(1)=6+4=10，2(2)+5(1)=4+5=9，都与右端相等。若任一方程不成立，就需要重新检查消元过程和计算。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One frequent mistake is forgetting to multiply the constant term when scaling an equation. Every term in the equation must be multiplied by the same factor, including the number on the right-hand side.
常见错误之一是在对方程乘以某个数时，忘记乘常数项。方程中的每一项，包括右边的常数，都必须乘上同一个因子。
s signs: (3x − 5y) − (2x − 3y) = 3x − 5y − 2x + 3y. Careful expansion prevents sign slips.
另一个错误是在方程相减时混淆符号。借助括号可避免丢失负号：(3x − 5y) − (2x − 3y) = 3x − 5y − 2x + 3y。细心展开就能防止符号出错。
hen coefficients have the same sign. Remember: same sign → subtract; opposite signs → add.
学生有时会在系数同号时错误地相加。记住：同号相减，异号相加。
10. Real-World Application Problem | 实际应用题
Two stadium seat tickets and three programme booklets cost £13. One stadium ticket and two booklets cost £8. Find the price of one ticket and one booklet. Let ticket price be £t and booklet price be £b.
两张体育场门票和三份节目册共花费 13 英镑，一张门票和两份节目册花费 8 英镑。求一张门票和一份节目册的价格。设门票价格为 £t，节目册为 £b。
2t+4b)−(2t+3b)=16−13 → b=3. Substituting b=3 into t+2b=8 gives t+6=8 → t=2. A ticket costs £2 and a booklet £3.
列方程：2t + 3b = 13，t + 2b = 8。将第二个方程乘以 2 使 t 系数相等：2t + 4b = 16。新方程减去第一个方程：(2t+4b)−(2t+3b)=16−13 → b=3。将 b=3 代入 t+2b=8 得 t+6=8 → t=2。门票 £2，节目册 £3。
11. Practice Questions to Build Confidence | 巩固练习
= 9 and 2x + 3y = 12. (c) 4x + 3y = 14 and 7x − 2y = 10. (d) 2x + 5y = 16 and 3x − 4y = 1.
尝试用消元法解下列方程组，并一步一步写出过程：(a) 5x + 2y = 11 和 3x − 2y = 5。(b) x + 3y = 9 和 2x + 3y = 12。(c) 4x + 3y = 14 和 7x − 2y = 10。(d) 2x + 5y = 16 和 3x − 4y = 1。
a different answer, review the scaling and sign rules carefully before reading the next section.
参考答案：(a) x=2, y=0.5；(b) x=3, y=2；(c) x=2, y=2；(d) x=3, y=2。如果答案不一致，请仔细复习缩放和符号规则再继续。
12. Summary and Key Takeaways | 总结与核心要点
The elimination method turns a system of two equations into a single equation by removing one variable. Its power lies in the careful handling of coefficients through multiplication, addition, and subtraction. Always aim to make the coefficient magnitudes equal for the targeted variable.
消元法通过去除一个变量，将两个方程构成的系统转化为单个方程。其威力在于通过乘、加、减，巧妙地处理系数。始终以“使目标变量的系数绝对值相等”为目标。
Remember the golden rules: multiply all terms, use brackets when subtracting, check both original equations, and never mix up the same-sign and opposite-sign cases. With consistent practice, elimination becomes a swift and reliable algebraic tool for KS3 and beyond.
记住黄金法则：乘时所有项都要乘；相减时使用括号以避免符号错误；同时检验两个原方程；切勿混淆同号和异号的情况。通过持续练习，消元法将成为 KS3 及以后迅速而可靠的代数工具。
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