Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations are the foundation of algebra at KS3. They appear in almost every Cambridge Checkpoint test and are essential for later topics such as simultaneous equations, graphs and formulae. A strong grasp of solving linear equations will help you tackle more complex problems with confidence.

一元一次方程是 KS3 代数的基础。它们几乎出现在每一次剑桥 Checkpoint 考试中，也是后续联立方程、图像和公式等内容的重要基础。牢固掌握一元一次方程的解法，能帮助你更有信心地应对更复杂的问题。

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1. What Is a Linear Equation? | 什么是一元一次方程？

n − 2 = 10. Linear means the variable is not squared, cubed or raised to any higher power.

在 KS3 阶段，一元一次方程是指未知数的最高次数为 1 的代数等式。它用等号连接两个表达式，例如 x + 5 = 12 或 3n − 2 = 10。这里的“一次”表示未知数没有被平方、立方或进行更高次幂运算。

the solution or root. For example, x = 7 is the solution of x + 5 = 12 because 7 + 5 = 12.

解方程就是求出使等式两边相等的未知数的值，这个值叫做方程的解或根。例如，x = 7 是 x + 5 = 12 的解，因为 7 + 5 = 12。

Linear equations can have positive or negative coefficients, and the variable may appear on one side or on both sides. Recognising the type of equation helps you choose the best solving method.

一元一次方程的系数可以是正数或负数，未知数可以只出现在等式一边，也可以同时出现在两边。识别方程的类型有助于选择最合适的解法。

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2. The Balance Method | 天平法

The balance method treats an equation like a set of scales. Whatever you do to one side, you must do the same to the other side to keep it balanced. This is the golden rule of solving equations.

天平法把方程看作一台天平。无论你对一边做了什么运算，都必须对另一边做同样的运算，才能保持平衡。这是解方程的金科玉律。

• Add the same amount to both sides — 两边同加一个数；
• Subtract the same amount from both sides — 两边同减一个数；
• Multiply or divide both sides by the same non-zero number — 两边同乘或同除一个非零数。

These operations are called inverse operations because addition and subtraction undo each other, and multiplication and division undo each other. Using inverse operations is the key to isolating the unknown variable.

这些运算叫做逆运算，因为加法和减法互为逆运算，乘法和除法互为逆运算。使用逆运算是分离未知数的关键。

For example, to undo an addition of 6, you subtract 6 from both sides. To undo a multiplication by 4, you divide both sides by 4.

例如，要抵消加 6 的运算，你需要两边同减 6。要抵消乘 4 的运算，你需要两边同除以 4。

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3. Solving Simple Equations | 解简单方程

on its own and gives x = 8. Showing the operation on both sides makes your working clear.

要解 x + 7 = 15，两边同减 7。这样 x 就单独留在左边，得到 x = 8。把两边同时进行的运算写清楚，过程会非常清晰。

x + 7 − 7 = 15 − 7 → x = 8

For 4x = 28, divide both sides by 4. The solution is x = 7 because 4 × 7 = 28.

对于 4x = 28，两边同除以 4。解为 x = 7，因为 4 × 7 = 28。

4x ÷ 4 = 28 ÷ 4 → x = 7

first subtract 3 from both sides to get 2x = 8, then divide both sides by 2 to get x = 4.

两步方程也很常见。对于 2x + 3 = 11，先两边同减 3，得到 2x = 8，再两边同除以 2，得到 x = 4。

2x + 3 − 3 = 11 − 3 → 2x = 8 → x = 4

ue = x, to match examination style. This makes your final line easy for examiners to find.

最终答案应写成 x = 数值 的形式，而不是 数值 = x，以符合考试书写规范。这样阅卷老师也能更容易找到你的最终答案。

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4. Equations with Unknowns on Both Sides | 未知数在等式两边

llect like terms so that the variable appears on only one side. This is often called moving terms or collecting terms.

当未知数出现在等式两边时，例如 5x + 2 = 2x + 14，首先要移项合并同类项，使未知数只出现在一边。这一过程通常叫做移项或合并同类项。

: 3x + 2 = 14. Then subtract 2 from both sides: 3x = 12. Finally divide by 3 to get x = 4.

两边同减 2x：3x + 2 = 14。再两边同减 2：3x = 12。最后两边同除以 3，得到 x = 4。

5x + 2 − 2x = 2x + 14 − 2x → 3x + 2 = 14 → 3x = 12 → x = 4

It is usually easier to move the smaller variable term first. This avoids negative coefficients and reduces sign errors, especially when the variable is already accompanied by a negative sign.

通常建议先移较小的未知数项，这样可以避免出现负系数，减少符号错误，尤其是当未知数前已经有负号时。

x + 1. Add x to both sides first: 7 = 3x + 1. Then subtract 1: 6 = 3x. Divide by 3: x = 2.

例如：7 − x = 2x + 1。先两边同加 x：7 = 3x + 1。再两边同减 1：6 = 3x。两边同除以 3：x = 2。

7 − x + x = 2x + 1 + x → 7 = 3x + 1 → 6 = 3x → x = 2

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5. Equations with Brackets | 含括号的方程

law. For example, 3(x + 4) = 27 becomes 3x + 12 = 27 after multiplying 3 by both x and 4.

如果方程中有括号，应先用分配律展开。例如 3(x + 4) = 27 在将 3 分别乘以 x 和 4 后，化为 3x + 12 = 27。

e by 3 to get x = 5. Always check that the bracket has been fully expanded before solving.

然后两边同减 12：3x = 15，再两边同除以 3，得到 x = 5。解方程前一定要确保括号已经完全展开。

3(x + 4) = 27 → 3x + 12 = 27 → 3x = 15 → x = 5

ves 9 − 2x + 6 = 5, so 15 − 2x = 5. Then subtract 15 from both sides: −2x = −10, so x = 5.

如果括号前是负号，要把负号乘进括号内的每一项。例如 9 − 2(x − 3) = 5 化为 9 − 2x + 6 = 5，即 15 − 2x = 5。然后两边同减 15：−2x = −10，所以 x = 5。

9 − 2(x − 3) = 5 → 9 − 2x + 6 = 5 → 15 − 2x = 5 → −2x = −10 → x = 5

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6. Equations with Fractions | 含分数的方程

When an equation contains fractions, you can multiply every term by the lowest common denominator to clear the fractions. This makes the equation much easier to solve.

当方程含有分数时，可以在每一项上乘以最简公分母来去分母。这样方程会变得更容易求解。

Consider x/2 + 3 = 7. Multiply every term by 2: x + 6 = 14. Then subtract 6 to get x = 8.

例如 x/2 + 3 = 7。每一项乘以 2：x + 6 = 14。然后两边同减 6，得到 x = 8。

x/2 + 3 = 7 → x + 6 = 14 → x = 8

e fraction bar acts as a grouping symbol, so keep the numerator together when multiplying.

对于 (2x − 1)/3 = 5 这类方程，先两边同乘以 3，再求解：2x − 1 = 15，所以 2x = 16，x = 8。分数线起到括号一样的分组作用，所以乘以分母时要保持分子为一个整体。

(2x − 1)/3 × 3 = 5 × 3 → 2x − 1 = 15 → 2x = 16 → x = 8

ul example is 2x/5 + 1 = 3. Multiply every term by 5: 2x + 5 = 15. Then 2x = 10, so x = 5.

另一个有用的例子是 2

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