📚 Solving Two-Step Linear Equations | 解两步线性方程
Solving two-step linear equations is one of the most important algebra skills in the Cambridge KS3 mathematics course. A two-step equation contains a variable, one addition or subtraction, and one multiplication or division. To find the value of the unknown, you must reverse the order of operations carefully. This revision article gives you clear methods, worked examples, common mistakes and exam-style practice so you can solve these equations with confidence.
解两步线性方程是剑桥 KS3 数学课程中最重要的代数技能之一。两步方程包含一个变量、一次加法或减法，以及一次乘法或除法。为了求出未知数的值，你必须仔细逆转运算顺序。这篇复习文章为你提供清晰的方法、例题、常见错误和考试型练习，帮助你自信地解出这类方程。
1. What Is a Two-Step Equation? | 什么是两步方程
ions such as 4n - 7 = 13 or x/3 + 5 = 9. The variable n or x is the unknown you must find.
两步线性方程是恰好需要两次逆运算才能隔离变量的方程。它通常具有 ax + b = c 或 x/a + b = c 的形式，其中 a、b、c 是数且 a ≠ 0。剑桥 KS3 题目经常要求你解诸如 4n - 7 = 13 或 x/3 + 5 = 9 这样的方程。变量 n 或 x 就是你要找的未知数。
quation exactly. For example, x = 5 is the solution of 2x + 3 = 13 because 2 × 5 + 3 = 13.
解这样一个方程意味着求出使两边相等的变量值。这个解必须完全满足原方程。例如，x = 5 是 2x + 3 = 13 的解，因为 2 × 5 + 3 = 13。
2. The Balancing Method | 平衡法
Think of an equation as a balance scale. The left side and the right side must always stay equal. Whatever operation you do to one side, you must do exactly the same to the other side. This is called the balancing method, and it keeps the equation true while you work towards the solution.
把方程想象成一架天平。左边和右边必须始终保持相等。你对一边进行什么运算，就必须对另一边做完全相同的运算。这就是平衡法，它能保持方程成立，同时帮助你逐步求出解。
d because the same value was added to both sides. This first step removes the subtraction.
例如，在方程 4n - 7 = 13 中，左边是 4n - 7，右边是 13。两边同时加 7 得到 4n = 20。方程仍然保持平衡，因为两边加上了相同的数。这一步就去掉了减法。
3. Inverse Operations in Order | 逆运算顺序
When solving an equation, you must undo operations in the reverse order to how they were applied. Normally in an expression such as 4n - 7, n is first multiplied by 4 and then 7 is subtracted. To solve, undo the subtraction first, then undo the multiplication. This is often remembered as 'undo the addition or subtraction first, then undo the multiplication or division'.
解方程时，你必须按照运算应用的相反顺序来逆运算。在 4n - 7 这样的表达式中，n 先乘以 4，然后减去 7。求解时要先消去减法，再消去乘法。这通常可以记为“先消去加减法，再消去乘除法”。
The table below shows the main operation and its inverse operation. Use it whenever you are not sure which operation to apply next.
下表列出了主要运算及其逆运算。当你不太确定下一步该用哪个运算时，可以参考它。
| Operation | Inverse operation |
| + b | − b |
| − b | + b |
| × a | ÷ a |
| ÷ a | × a |
4. Worked Example 1: ax + b = c | 例题 1：ax + b = c
s by 6 to undo the multiplication. This gives x = 3. The solution can be written as x = 3.
解方程 6x + 4 = 22。首先，两边同时减去 4，消去加法。得到 6x = 18。其次，两边同时除以 6，消去乘法。得到 x = 3。解可以写成 x = 3。
6x + 4 = 22 → 6x = 18 → x = 3
m x = value. Do not stop at 6x = 18, because the question asks for the value of x, not 6x.
始终把最终答案写成 x = 数值的形式。不要停在 6x = 18，因为题目要求的是 x 的值，而不是 6x 的值。
5. Worked Example 2: x/a + b = c | 例题 2：x/a + b = c
s gives x/3 = 7. Then undo the division by multiplying both sides by 3. This gives x = 21.
解方程 x/3 - 2 = 5。变量 x 先除以 3，再减去 2。求解时先消去减法，两边同时加 2。得到 x/3 = 7。然后消去除法，两边同时乘以 3。得到 x = 21。
x/3 − 2 = 5 → x/3 = 7 → x = 21
e same logic works for equations such as x/5 + 1 = 4, where multiplying by 5 gives x = 15.
注意，当 x/3 = 7 时，乘以 3 会抵消除以 3 的运算。同样的逻辑也适用于 x/5 + 1 = 4 这样的方程，乘以 5 后得到 x = 15。
6. Equations with Negative Coefficients | 负系数方程
5. Subtract 3 from both sides to get -4x = 12. Then divide both sides by -4 to get x = -3.
当 x 的系数为负数时，你必须两边同时除以这个负数。负号要跟着系数。例如，解 -4x + 3 = 15。两边同时减去 3，得到 -4x = 12。然后两边同时除以 -4，得到 x = -3。
−4x + 3 = 15 → −4x = 12 → x = 12 ÷ (−4) = −3
swer must be negative when the right side is positive. Check: -4 × (-3) + 3 = 12 + 3 = 15.
一个常见的符号错误是用 12 除以 4 而忘记负号。如果系数是 -4，当右边为正数时，最终答案一定是负数。检验：-4 × (-3) + 3 = 12 + 3 = 15。
7. Equations with Variables on Both Sides | 两边含变量的方程
= 8. Then add 2 to both sides to get 2m = 10. Finally divide both sides by 2 to get m = 5.
如果变量同时出现在方程的两边，要先把含变量的项合并到一边。例如，解 5m - 2 = 3m + 8。两边同时减去 3m，得到 2m - 2 = 8。然后两边同时加 2，得到 2m = 10。最后两边同时除以 2，得到 m = 5。
5m − 2 = 3m + 8 → 2m − 2 = 8 → 2m = 10 → m = 5
When choosing which side to keep the variable on, it is often easier to subtract the smaller coefficient. In this example, subtracting 3m from both sides avoids a negative variable term.
在选择把变量保留在哪一边时，通常减去较小的系数会更简单。在这个例子中，两边同时减去 3m 可以避免出现负的变量项。
8. Solving Equations with Brackets | 解含括号的方程
et is multiplied by 3. This gives y + 4 = 9. Then subtract 4 from both sides to get y = 5.
有些两步方程含有括号。例如，解 3(y + 4) = 27。你可以先两边同时除以 3，因为整个括号都被乘以 3。得到 y + 4 = 9。然后两边同时减去 4，得到 y = 5。
3(y + 4) = 27 → y + 4 = 9 → y = 5
then subtract 12 to get 3y = 15, then divide by 3 to get y = 5. Both methods are correct.
你也可以先展开括号：3y + 12 = 27，然后减去 12 得到 3y = 15，再除以 3 得到 y = 5。这两种方法都是正确的。
9. Checking Your Solution | 检验你的解
6 × 3 + 4 = 18 + 4 = 22. The left side equals the right side, so the solution is correct.
始终把你求出的最终答案代回原方程进行检验。对于方程 6x + 4 = 22，代入 x = 3：6 × 3 +
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