📚 Solving Linear Equations: From One-Step to Variables on Both Sides | 解一元一次方程：从一步到两边含变量
e style of Cambridge Lower Secondary Checkpoint questions such as those in worksheet p100_2.
在剑桥 KS3 数学中，一元一次方程是最重要的代数技能之一。本文讲解解方程所需的关键方法，包括一步方程、两步方程、括号以及两边含变量的方程。示例遵循剑桥 Lower Secondary Checkpoint 题目的风格，例如练习 p100_2 中的问题。
1. What Is a Linear Equation? | 什么是一元一次方程？
a linear equation. Solving it means finding the value of x that makes the statement true.
一元一次方程是指未知数（通常写作 x）的指数仅为 1 的方程。方程中不含 x² 项、不含 x 的平方根，也没有变量出现在分母中。例如，2x + 3 = 11 就是一元一次方程。解方程就是求出使等式成立的 x 值。
Linear equations appear throughout the Cambridge KS3 curriculum because they are used to express rules, solve problems, and prepare for graphs of straight lines. Mastering them early makes later algebra topics much easier.
一元一次方程贯穿剑桥 KS3 课程，因为它们是表达规则、解决问题以及学习直线图像的基础。尽早掌握这一内容会让后续的代数主题变得简单许多。
2. The Balance Method | 天平法
Think of an equation as a balance scale. The left side must equal the right side. Whatever you do to one side, you must do to the other side. If you add 5 to one side, add 5 to the other. If you divide one side by 3, divide the other side by 3 as well. This keeps the equation balanced while you isolate the unknown.
可以把方程想象成一架天平。左边必须等于右边。无论你对一边做什么运算，另一边也必须做相同运算。如果给一边加 5，另一边也要加 5。如果把一边除以 3，另一边也要除以 3。这样在分离未知数的过程中，方程始终保持平衡。
The balance method is especially useful because it gives a clear reason for every step. Instead of guessing values, you perform inverse operations on both sides until x is left on its own.
天平法特别有用，因为它为每一步都提供了清晰的理由。你不用猜测数值，而是对两边同时进行逆运算，直到 x 单独留在一边。
| Operation 运算 | Equation 方程 | Balanced Step 平衡步骤 |
|---|---|---|
| Add 3 加 3 | x − 3 = 7 | x − 3 + 3 = 7 + 3 |
| Subtract 2 减 2 | x + 2 = 9 | x + 2 − 2 = 9 − 2 |
| Divide by 4 除以 4 | 4x = 20 | 4x ÷ 4 = 20 ÷ 4 |
| Multiply by 5 乘以 5 | x ÷ 5 = 3 | x ÷ 5 × 5 = 3 × 5 |
3. Solving One-Step Equations | 解一步方程
ivide both sides by 5 to get x = 7. For x ÷ 4 = 3, multiply both sides by 4 to get x = 12.
一步方程只需要一次逆运算就能分离出 x。对于 x + 6 = 9，两边同时减去 6，因为减法是加法的逆运算，得到 x = 3。对于 5x = 35，两边同时除以 5，得到 x = 7。对于 x ÷ 4 = 3，两边同时乘以 4，得到 x = 12。
The key is to identify which operation is attached to the unknown and then do the opposite. Addition and subtraction undo each other, while multiplication and division undo each other.
关键是要找出未知数上附加了哪种运算，然后进行相反的运算。加法和减法互为逆运算，乘法和除法也互为逆运算。
x + 6 = 9 → x + 6 − 6 = 9 − 6 → x = 3
5x = 35 → 5x ÷ 5 = 35 ÷ 5 → x = 7
4. Solving Two-Step Equations | 解两步方程
e both sides by 3 to get x = 5. This order follows the reverse of the order of operations.
两步方程包含两种运算。对于 3x + 5 = 20，首先处理加法或减法，然后处理乘法或除法。两边先同时减去 5，得到 3x = 15，然后两边同时除以 3，得到 x = 5。这个顺序与运算顺序相反。
en divide by 4 to find x = 5. Always show each step clearly so you can check your working.
剑桥 KS3 练习中常见的问题是解诸如 4x − 7 = 13 的方程。两边先同时加 7，得到 4x = 20，然后两边除以 4，得到 x = 5。每一步都要写清楚，以便检查解题过程。
3x + 5 = 20 → 3x = 15 → x = 5
4x − 7 = 13 → 4x = 20 → x = 5
5. Expanding Brackets First | 先展开括号
2x + 6 = 14. Then subtract 6 from both sides to get 2x = 8, and divide by 2 to get x = 4.
如果方程含有括号，应在求解前先展开括号。对于 2(x + 3) = 14，将括号内两项都乘以 2，得到 2x + 6 = 14。然后两边同时减去 6，得到 2x = 8，再除以 2，得到 x = 4。
−3x + 6 = 9. Then subtract 6 from both sides: −3x = 3. Finally divide by −3 to get x = −1.
当括号前是负数时要特别小心。对于 −3(x − 2) = 9，展开得到 −3x + 6 = 9。然后两边同时减去 6，得到 −3x = 3。最后两边除以 −3，得到 x = −1。
2(x + 3) = 14 → 2x + 6 = 14 → 2x = 8 → x = 4
6. Variables on Both Sides | 两边都含变量
to get 3x − 3 = 9. Then add 3 to both sides to get 3x = 12, and divide by 3 to find x = 4.
有些方程两边都含有 x，例如 5x − 3 = 2x + 9。目标是所有含 x 的项集中到一边，所有数字项集中到另一边。两边同时减去 2x，得到 3x − 3 = 9。然后两边同时加 3，得到 3x = 12，再除以 3，得到 x = 4。
It does not matter whether you move x terms to the left or to the right, as long as you apply the same operation to both sides. Choose the side that keeps the coefficient of x positive when possible.
把含 x 的项移到左边还是右边都可以，只要对两边进行相同的运算即可。在可能的情况下，尽量选择让 x 的系数为正的一边。
5x − 3 = 2x + 9 → 5x − 2x − 3 = 9 → 3x = 12 → x = 4
7. Dealing with Negative Coefficients | 处理负系数
rom both sides to get x = 3. This avoids dividing by a negative number until the very end.
当未知数带负号时，通常把它移到另一边会更容易。对于 7 − x = 4，两边同时加 x，得到 7 = x + 4。然后两边同时减 4，得到 x = 3。这样可以避免在最后一步之前除以负数。
8. Always check that the sign is correct by substituting back into the original equation.
也可以两边同时乘以 −1 来改变未知数的符号。对于 −x = −8，两边乘以 −1 得到 x = 8。最后一定要代回原方程检验符号是否正确。
7 − x = 4 → 7 = x + 4 → x = 3
−x = −8 → x = 8
8. Checking Solutions | 检验解
x + 5 = 20, when x = 5, the left side becomes 3 × 5 + 5 = 20, which equals the right side.
务必把答案代回原方程检验。如果左边和右边得到相同的值，说明解是正确的。例如 3x + 5 = 20，当 x = 5 时，左边变为 3 × 5 + 5 = 20，与右边相等。
Checking is especially important in exam-style questions because it catches arithmetic mistakes and sign errors. It also helps you decide whether an answer is reasonable before you write it down.
在考试类题目中，检验尤其重要，因为它能发现算术错误和符号错误。它还能帮助你在写下答案之前判断答案是否合理。
| Equation 方程 | Solution 解 | Check 检验 |
|---|---|---|
| 2x − 4 = 10 | x = 7 | 2
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