Solving Linear Equations: Brackets, Fractions and Word Problems | 解线性方程：括号、分数与文字题

📚 Solving Linear Equations: Brackets, Fractions and Word Problems | 解线性方程：括号、分数与文字题

Linear equations are the foundation of algebra at KS3. Whether you are solving a simple one-step equation or tackling a multi-step problem with brackets and fractions, the key is to keep the balance. This revision guide covers everything you need to master solving equations confidently — from expanding brackets to setting up equations from real-life problems.

线性方程是 KS3 代数的基石。无论你是在解简单的一步方程，还是处理带有括号和分数的多步问题，关键都是保持等式两边的平衡。本复习指南涵盖了你自信掌握方程求解所需的一切——从展开括号到根据现实问题建立方程。

1. What is an Equation? | 什么是方程？

An equation is a mathematical statement showing that two expressions are equal. It contains an equals sign ‘=’ and at least one unknown, usually represented by a letter such as x or y.

方程是一个数学陈述，表明两个表达式相等。它包含一个等号 “=” 和至少一个未知数，通常用 x 或 y 这样的字母表示。

The aim when solving an equation is to find the value of the unknown that makes the statement true. This process involves performing identical operations on both sides to keep the equation balanced.

解方程的目的是找出使陈述成立的未知数的值。这个过程需要在等式两边进行相同的操作，以保持方程平衡。

Think of an equation like a balance scale: whatever you do to one side, you must also do to the other, or the scale tips.

可以把方程想象成一个天平：无论你对一边做了什么，也必须对另一边做同样的操作，否则天平就会倾斜。

2. Solving One-Step Equations | 解一步方程

A one-step equation requires only one inverse operation to isolate the unknown. For example, x + 7 = 15 is solved by subtracting 7 from both sides.

一步方程只需要一个逆运算就能把未知数分离出来。例如，x + 7 = 15 可以通过两边同时减去 7 来求解。

If the unknown is multiplied by a number, we divide. For 4x = 20, dividing both sides by 4 gives x = 5. Always do the opposite operation to undo what is being done to x.

如果未知数乘以了一个数，我们就做除法。对于 4x = 20，两边同时除以 4 得到 x = 5。始终用相反的运算来撤消施加在 x 上的运算。

Even with subtraction or division, the same rule applies: for x − 3 = 8, add 3 to both sides; for x / 5 = 2, multiply both sides by 5. Keep the balance.

即使遇到减法或除法，同样的规则也适用：对于 x − 3 = 8，两边加 3；对于 x / 5 = 2，两边乘 5。保持平衡。

3. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations. For instance, 2x + 3 = 11. First, subtract 3 from both sides to get 2x = 8, then divide by 2 to find x = 4.

两步方程涉及两种运算。例如 2x + 3 = 11。首先两边减 3 得到 2x = 8，然后除以 2 得出 x = 4。

The order is crucial: undo addition or subtraction first, then multiplication or division. This reverses the normal order of operations (BIDMAS).

顺序至关重要：先撤消加法或减法，然后再撤消乘法或除法。这与通常的运算顺序 (BIDMAS) 相反。

Another example: (x/4) − 2 = 3. Add 2 to both sides to get x/4 = 5, then multiply by 4 to obtain x = 20.

另一个例子：(x/4) − 2 = 3。两边加 2 得到 x/4 = 5，然后乘以 4 得到 x = 20。

4. Equations with Brackets | 含括号的方程

Brackets group terms together. When solving an equation like 3(x + 2) = 21, you have two main approaches: expand the brackets first or treat the bracket as a single quantity.

括号将项组合在一起。当求解像 3(x + 2) = 21 这样的方程时，有两种主要方法：先展开括号，或者把括号当作一个整体来处理。

If you expand first, multiply the term outside by each term inside: 3x + 6 = 21. Then solve as a two-step equation: subtract 6, get 3x = 15, divide by 3, x = 5.

如果先展开，就把外面的项乘以括号内的每一项：3x + 6 = 21。然后按两步方程求解：减 6 得到 3x = 15，除以 3 得 x = 5。

Alternatively, divide both sides by 3 first: (x + 2) = 7, then subtract 2 to get x = 5. Both methods work; expanding is often safer when the coefficient is not a factor of the constant.

或者，先两边除以 3：x + 2 = 7，然后减 2 得到 x = 5。两种方法都可行；当括号前的系数不是常数的因数时，展开通常更稳妥。

5. Expanding Brackets First | 先展开括号

When an equation has more than one set of brackets, always expand first. For example, 2(3x − 1) + 5(x + 4) = 40.

当一个方程有多个括号时，一定要先展开。例如 2(3x − 1) + 5(x + 4) = 40。

Expand to get 6x − 2 + 5x + 20 = 40. Collect like terms: 11x + 18 = 40. Subtract 18: 11x = 22. Divide by 11: x = 2.

展开得到 6x − 2 + 5x + 20 = 40。合并同类项：11x + 18 = 40。减 18：11x = 22。除以 11：x = 2。

Be especially careful with negative signs. For 4 − 2(x − 3) = 10, expand −2(x − 3) to get −2x + 6. The equation becomes 4 − 2x + 6 = 10, then 10 − 2x = 10, so x = 0.

要格外注意负号。对于 4 − 2(x − 3) = 10，将 −2(x − 3) 展开得到 −2x + 6。方程变为 4 − 2x + 6 = 10，即 10 − 2x = 10，所以 x = 0。

6. Unknowns on Both Sides | 未知数在两侧的方程

When the unknown appears on both sides of the equation, the first step is to collect all the variable terms on one side and all the constant terms on the other.

当方程两边都出现未知数时，第一步是把所有含未知数的项移到一边，把所有常数项移到另一边。

For 5x − 3 = 2x + 9, subtract 2x from both sides: 3x − 3 = 9. Then add 3: 3x = 12, so x = 4.

对于 5x − 3 = 2x + 9，两边减 2x：3x − 3 = 9。然后加 3：3x = 12，所以 x = 4。

If the variable terms are on the right, you can still work the same way. For 7 + 4x = 10x − 5, subtract 4x from both sides to get 7 = 6x − 5, add 5: 12 = 6x, divide by 6: x = 2.

如果含未知数的项在右边，仍然可以用相同的方法处理。对于 7 + 4x = 10x − 5，两边减 4x 得到 7 = 6x − 5，加 5：12 = 6x，除以 6：x = 2。

Always aim to keep the unknown positive to reduce sign errors — move the smaller variable term to the other side.

始终尽量让未知数的系数为正，以减少符号错误——把较小的含未知数的项移到另一边。

7. Dealing with Fractions | 处理分数方程

Equations with fractions can look tricky, but the trick is to eliminate the fractions by multiplying every term by the lowest common multiple (LCM) of the denominators.

含有分数的方程看起来棘手，但技巧是通过将每一项乘以分母的最小公倍数 (LCM) 来消去分数。

For (x + 1)/3 = (x − 2)/4, the LCM of 3 and 4 is 12. Multiply each term: 12×(x + 1)/3 = 12×(x − 2)/4. This simplifies to 4(x + 1) = 3(x − 2).

对于 (x + 1)/3 = (x − 2)/4，3 和 4 的 LCM 是 12。每项乘以 12：12×(x + 1)/3 = 12×(x − 2)/4。化简为 4(x + 1) = 3(x − 2)。

Then expand: 4x + 4 = 3x − 6. Subtract 3x: x + 4 = −6, so x = −10. Always check your solution in the original equation.

然后展开：4x + 4 = 3x − 6。减 3x：x + 4 = −6，所以 x = −10。一定要把解代入原方程检验。

If an equation contains a single fraction like (2x)/5 + 3 = 7, you can either multiply through by 5 or subtract 3 first, then multiply. Both paths work.

如果一个方程只含有一个分数，如 (2x)/5 + 3 = 7，你可以先全式乘以 5，也可以先减 3 再乘。两条路径都可行。

8. Checking Your Solution | 检验解

Substituting your answer back into the original equation is the only way to be sure you have not made a mistake. Replace the unknown with your value and simplify both sides.

把答案代回原方程是确保没有犯错的唯一方法。用你求出的值替换未知数，并化简两边。

If the left-hand side equals the right-hand side, your solution is correct. If not, retrace your steps — often a sign error or an incomplete expansion is the culprit.

如果左边等于右边，你的解就是正确的。如果不相等，重新检查步骤——通常符号错误或不完整的展开是罪魁祸首。

Checking also helps you build confidence and catch mistakes early, which is especially useful in exam conditions.

检验还能帮助你建立信心，及早发现错误，这在考试情境下尤其有用。

9. Common Mistakes to Avoid | 常见错误避免

Forgetting to multiply all terms on both sides by the same number.

忘记两边所有项都乘以同一个数。
Incorrectly expanding brackets, especially when a negative sign is outside the bracket.

错误地展开括号，特别是括号外有负号时。
Adding instead of subtracting when moving terms, or vice versa.

移项时该减却加，或相反。
Not collecting like terms before solving.

求解前没有合并同类项。
Dividing before subtracting when solving two-step equations in the wrong order.

解两步方程时顺序错误，先除以系数再减。

Write each step clearly and avoid taking shortcuts until you are very confident. A neat layout reduces errors.

每一步都写清楚，在足够自信之前避免跳步。整洁的书写能减少错误。

10. Word Problems: Forming Equations | 文字题：列方程

Many real-life situations can be modelled with linear equations. The key is to define a variable clearly and translate words into mathematical expressions.

许多现实情境都可以用线性方程来建模。关键是要清晰地定义变量，并将文字转化为数学表达式。

For example: ‘Three times a number plus 7 is 28.’ Let the number be x. Then 3x + 7 = 28. Solve: 3x = 21, so x = 7.

例如：“一个数的三倍加 7 等于 28。” 设这个数为 x，则 3x + 7 = 28。求解：3x = 21，所以 x = 7。

‘The sum of two consecutive numbers is 45.’ Let the first number be n, then the next is n + 1. Equation: n + (n + 1) = 45, so 2n + 1 = 45, 2n = 44, n = 22; the numbers are 22 and 23.

“两个连续数的和是 45。”设第一个数为 n，则下一个是 n + 1。方程：n + (n + 1) = 45，所以 2n + 1 = 45，2n = 44，n = 22；这两个数是 22 和 23。

Read the problem several times and underline key quantities before writing an equation. Always state what your variable represents.

在写方程之前多读几遍题目，并在关键量下划线。一定要说明你的变量代表什么。

11. Word Problems: Solving Strategically | 文字题：巧妙求解

Once you have formed the equation, solve it step by step using the techniques above. Then interpret your answer in the context of the question.

一旦列出方程，就用上述技巧逐步求解。然后根据题目背景解释你的答案。

Example: ‘A rectangle has length (x + 5) cm and width 3 cm. Its perimeter is 34 cm. Find x.’ Perimeter = 2(length + width), so 2((x + 5) + 3) = 34. Simplify inside: x + 8; equation: 2(x + 8) = 34. Divide by 2: x + 8 = 17, x = 9.

例子：“一个矩形的长是 (x + 5) cm，宽是 3 cm。它的周长是 34 cm。求 x。”周长 = 2(长 + 宽)，所以 2((x + 5) + 3) = 34。括号内化简为 x + 8；方程：2(x + 8) = 34。除以 2：x + 8 = 17，x = 9。

Always check that your solution makes sense. If length becomes negative or zero for a geometric problem, you may have made a mistake.

始终检查你的解是否合理。如果几何问题的长度变成负数或零，你很可能犯了错误。

Write the final answer as a sentence when the problem asks for a specific quantity, e.g. ‘The number is 9’ or ‘The length is 14 cm.’

当题目要求某个具体的量时，把最终答案写成一句话，如“这个数是 9”或“长度为 14 cm”。

12. Summary and Key Points | 小结与关键点

An equation is a balance; do the same to both sides.

方程是一座天平；对两边做同样的操作。
Follow the reverse of BIDMAS to undo operations: subtract/add first, then multiply/divide.

按 BIDMAS 的逆顺序撤消运算：先加减，后乘除。
For brackets, expand carefully or divide first if possible.

对于括号，仔细展开，如果可能也可先除。
Eliminate fractions by multiplying by the LCM of denominators.

乘以分母的 LCM 来消去分数。
Collect variable terms on one side and constants on the other when unknowns appear on both sides.

当未知数出现在两边时，把含未知数的项移到一边，常数移到另一边。
Always substitute your answer back to check.

总要把答案代回去检验。
In word problems, define your variable, write an equation, solve, and interpret the answer.

在文字题中，定义变量，列出方程，求解，并解释答案。

Mastering linear equations now will give you a strong base for more advanced algebra. Practise daily, and soon you will solve them with speed and accuracy.

现在就掌握线性方程，将为你学好更高阶的代数打下坚实基础。每天练习，很快你就能快速而准确地解出它们。

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Solving Linear Equations: Brackets, Fractions and Word Problems | 解线性方程：括号、分数与文字题