📚 Solving Linear Equations with Brackets and Fractions | 解含括号和分数的一次方程
When you meet an equation such as 2(x + 3) = x + 8 or (x – 1)/3 = 4, you are being asked to find the value of the unknown that makes the statement true. This topic is central to the Cambridge KS3 mathematics curriculum and appears in many progression tests and Checkpoint papers. The skills covered here include expanding brackets, clearing fractions, collecting like terms and checking answers carefully.
当你遇到像 2(x + 3) = x + 8 或 (x – 1)/3 = 4 这样的方程时，你需要寻找使等式成立的未知数的值。这个主题是剑桥 KS3 数学课程的核心内容，在许多进阶测试和 Checkpoint 试卷中都会出现。这里涵盖的技能包括展开括号、去分母、合并同类项以及仔细检验答案。
1. What Is a Linear Equation? | 什么是一次方程？
A linear equation is an equation where the unknown appears only to the first power. This means you will never see x², x³, or any root of x in a linear equation. Common examples are 3x + 2 = 14, 5(y – 2) = 3y + 6, and (x + 4)/2 = 7. Linear equations are called ‘linear’ because their graphs are straight lines. At KS3 you usually solve for one unknown.
一次方程是指未知数只出现一次幂的方程。这意味着你永远不会在一次方程中看到 x²、x³ 或任何 x 的根式。常见的例子有 3x + 2 = 14、5(y – 2) = 3y + 6 和 (x + 4)/2 = 7。一次方程之所以称为 ‘linear’，是因为它们的图像是直线。在 KS3 阶段，你通常求解一个未知数。
2x + 3 = 11 is a linear equation; 2x² + 3 = 11 is not.
2. Balancing the Equation | 方程的天平原理
An equation is like a balance scale. The equals sign tells you that the left side and right side have exactly the same value. Whatever operation you do to one side, you must do to the other side. If you add 3 to the left, add 3 to the right. If you multiply the left by 2, multiply the right by 2. This rule keeps the equation true while you rearrange it.
方程就像一个天平。等号告诉你左边和右边的值完全相同。无论你对一边做什么运算，都必须对另一边做同样的运算。如果你在左边加 3，右边也要加 3。如果你将左边乘以 2，右边也要乘以 2。这条规则能保证你在变形方程时等式仍然成立。
x + 3 = 7 → x + 3 − 3 = 7 − 3 → x = 4
3. Solving One-Step Equations | 解一步方程
A one-step equation needs only one inverse operation. For x + 5 = 12, subtract 5 from both sides. For 3x = 18, divide both sides by 3. For x/4 = 2, multiply both sides by 4. Always choose the inverse operation that cancels the number next to x.
一步方程只需要一次逆运算。对于 x + 5 = 12，两边都减去 5。对于 3x = 18，两边都除以 3。对于 x/4 = 2，两边都乘以 4。始终选择能抵消 x 旁边数字的逆运算。
x + 5 = 12 → x = 12 − 5 → x = 7
3x = 18 → 3x ÷ 3 = 18 ÷ 3 → x = 6
x/4 = 2 → (x/4) × 4 = 2 × 4 → x = 8
4. Solving Two-Step Equations | 解两步方程
In a two-step equation, perform addition or subtraction first, then multiplication or division. For example, 2x + 3 = 11: first subtract 3 from both sides, then divide both sides by 2. This reverses the order of operations used to build the expression.
在两步方程中，先做加法或减法，再做乘法或除法。例如，2x + 3 = 11：先从两边减去 3，再将两边除以 2。这正好逆向了构建表达式时的运算顺序。
2x + 3 = 11 → 2x = 8 → x = 4
Remember that addition and subtraction are inverse operations, just as multiplication and division are inverse operations. Reversing the order correctly is the key to a smooth solution.
记住加法和减法互为逆运算，乘法和除法也互为逆运算。正确地逆转顺序是顺利解题的关键。
5. Expanding Brackets | 展开括号
When a bracket is multiplied by a number, multiply each term inside the bracket by that number. For example, 2(x + 3) = 2x + 6 and −3(y − 2) = −3y + 6. Be careful with negative signs: multiplying two negatives gives a positive.
当括号乘以一个数时，用这个数去乘括号内的每一项。例如，2(x + 3) = 2x + 6，−3(y − 2) = −3y + 6。注意负号：两个负数相乘得到正数。
2(x + 3) = 2 × x + 2 × 3 = 2x + 6
−3(y − 2) = −3 × y + (−3) × (−2) = −3y + 6
6. Equations with Brackets | 含括号的方程
To solve an equation with brackets, expand the brackets first, then collect like terms. For example, 3(x + 2) = 21 expands to 3x + 6 = 21, then subtract 6 and divide by 3 to get x = 5. If there are brackets on both sides, expand both before balancing.
解含括号的方程时，先展开括号，再合并同类项。例如，3(x + 2) = 21 展开为 3x + 6 = 21，然后减去 6 再除以 3，得到 x = 5。如果两边都有括号，先展开两边再进行平衡。
3(x + 2) = 21 → 3x + 6 = 21 → 3x = 15 → x = 5
For an equation like 2(x + 1) = 3(x − 2), expand both sides to get 2x + 2 = 3x − 6, then collect x terms on one side and numbers on the other.
对于像 2(x + 1) = 3(x − 2) 这样的方程，先展开两边得到 2x + 2 = 3x − 6，然后把含 x 的项移到一边，数字移到另一边。
2x + 2 = 3x − 6 → 2 + 6 = 3x − 2x → 8 = x → x = 8
7. Equations with Fractions | 含分数的方程
To clear a fraction, multiply every term on both sides by the denominator. For example, (x + 2)/3 = 5 can be multiplied by 3 to get x + 2 = 15, so x = 13. This works because multiplying by the denominator reverses division by that denominator.
要消去分数，将两边每一项都乘以分母。例如，(x + 2)/3 = 5 可以两边乘以 3，得到 x + 2 = 15，所以 x = 13。这是因为乘以分母正好抵消了除以该分母的运算。
(x + 2)/3 = 5 → (x + 2)/3 × 3 = 5 × 3 → x + 2 = 15 → x = 13
Always multiply the entire side, not just the fraction. For example, in x/2 + 1 = 4, multiply both terms on the left by 2: x + 2 = 8.
始终乘以整个一边，而不仅仅是分数部分。例如，在 x/2 + 1 = 4 中，将左边两项都乘以 2：x + 2 = 8。
8. Using the Lowest Common Denominator | 使用最小公分母
If an equation has more than one fraction, multiply every term by the lowest common denominator (LCD) of all fractions. For example, in x/2 + 1/3 = 5, the LCD of 2 and 3 is 6. Multiplying every term by 6 clears all fractions in one step: 3x + 2 = 30.
如果一个方程有多个分数，用所有分数的最小公分母 (LCD) 乘以每一项。例如，在 x/2 + 1/3 = 5 中，2 和 3 的最小公分母是 6。将每一项乘以 6 可以一步消去所有分数：3x + 2 = 30。
x/2 + 1/3 = 5 → 6 × (x/2) + 6 × (1/3) = 6 × 5 → 3x + 2 = 30 → 3x = 28 → x = 28/3
The LCD is the smallest number that all denominators divide into exactly. Using it keeps the numbers smaller and easier to manage than multiplying by the product of all denominators.
最小公分母是所有分母都能整除的最小数字。使用它比将所有分母相乘得到的数字更小，也更容易计算。
9. Variables on Both Sides | 变量在两边
When x appears on both sides, collect the variable terms on one side and the number terms on the other. For example, 5x + 2 = 2x + 11: subtract 2x from both sides to get 3x + 2 = 11, then solve as a two-step equation.
当 x 出现在两边时，把含有变量的项移到一边，数字项移到另一边。例如，5x + 2 = 2x + 11：从两边减去 2x，得到 3x + 2 = 11，然后按两步方程求解。
5x + 2 = 2x + 11 → 5x − 2x + 2 = 11 → 3x + 2 = 11 → 3x = 9 → x = 3
If the variable terms are negative, you can add the variable term to both sides instead. For example, 7 − x = 2x + 1 becomes 7 − x + x = 2x + 1 + x, giving 7 = 3x + 1.
如果含变量的项是负数，你可以改为在两边加上这个含变量的项。例如，7 − x = 2x + 1 变为 7 − x + x = 2x + 1 + x，得到 7 = 3x + 1。
10. Checking Your Solution | 检验你的解
Always substitute your answer back into the original equation. For x = 4 in 2(x + 3) = 14, the left side is 2(4 + 3) = 2 × 7 = 14, which matches the right side. Checking helps you catch arithmetic slips and sign errors before moving on.
始终将你的答案代回原方程检验。对于 2(x + 3) = 14 中的 x = 4，左边为 2(4 + 3) = 2 × 7 = 14，与右边相等。检验可以帮助你在继续之前发现计算失误和符号错误。
Solve 4(x − 2) = 2x + 6 → 4x − 8 = 2x + 6 → 2x = 14 → x = 7
Check: left side = 4(7 − 2) = 4 × 5 = 20; right side = 2(7) + 6 = 14 + 6 = 20. Both sides match, so x = 7 is correct.
检验：左边 = 4(7 − 2) = 4 × 5 = 20；右边 = 2(7) + 6 = 14 + 6 = 20。两边相等，所以 x = 7 正确。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Students often forget to multiply all terms inside a bracket, or they change a sign incorrectly when subtracting. Another common error is multiplying only one term by the LCD when clearing fractions. Write each step on a new line, keep the equals signs aligned, and check your final answer to reduce these mistakes.
学生经常忘记乘以括号内的所有项，或者在减法时错误地改变符号。另一个常见错误是在去分母时只将一项乘以最小公分母。每一步都换行书写，保持等号对齐，并检验最终答案，可以减少这些错误。
| Mistake | 错误 | Example | 示例 | Correct | 正确做法 |
|---|---|---|
| Forgetting to expand all terms | 忘记展开所有项 | 2(x + 3) = 2x + 3 | 2(x + 3) = 2x + 6 |
| Wrong sign when subtracting | 减法时符号错误 | −2(x − 3) = −2x − 6 |
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