📚 Solving Linear Equations with Brackets and Fractions | 解含括号和分数的线性方程
Linear equations are the foundation of Key Stage 3 algebra. Being able to solve equations with brackets and fractions confidently will help you with graphs, sequences, and later topics such as simultaneous equations.
线性方程是 KS3 代数的基础。熟练掌握含括号和分数的方程求解，将帮助你学习图像、数列以及后续的联立方程等主题。
1. What is a linear equation? | 什么是线性方程？
ower of 1. It can usually be written in the form ax + b = c, where a, b and c are numbers.
线性方程是未知变量的最高次数为 1 的方程。它通常可以写成 ax + b = c 的形式，其中 a、b 和 c 是数。
n Key Stage 3, you will meet equations such as 2x + 3 = 11, 5(x - 2) = 20 and x/4 + 1 = 3.
在 KS3 阶段，你会遇到 2x + 3 = 11、5(x - 2) = 20 和 x/4 + 1 = 3 这样的方程。
2x + 3 = 11
2. The balancing method | 天平法
Think of an equation as a balanced scale. Whatever you do to one side, you must do to the other side. This is called the balancing method and it is the key rule for solving equations.
把方程想象成一架平衡的天平。你对一边做什么，另一边也必须做同样的事情。这叫做天平法，也是解方程的核心规则。
Use inverse operations to undo addition, subtraction, multiplication and division. For example, to undo + 5, subtract 5 from both sides.
使用逆运算来撤销加、减、乘、除。例如，要撤销 + 5，就在两边同时减去 5。
Example: Solve 3x + 5 = 20.
3x + 5 = 20
Subtract 5 from both sides to get 3x = 15. Then divide both sides by 3 to get x = 5.
两边同时减去 5，得到 3x = 15。然后两边同时除以 3，得到 x = 5。
3. Solving two-step equations | 解两步方程
A two-step equation needs two inverse operations. Always undo addition or subtraction before multiplication or division.
两步方程需要两次逆运算。一定要先撤销加法或减法，再撤销乘法或除法。
Example: Solve x/4 - 2 = 3.
x/4 - 2 = 3
Add 2 to both sides: x/4 = 5. Then multiply both sides by 4: x = 20.
两边加 2：x/4 = 5。然后两边乘以 4：x = 20。
This order prevents negative coefficients and makes the working clearer.
这样的顺序可以避免负系数，也使过程更清晰。
4. Expanding brackets | 去括号
law first: a(b + c) = ab + ac. Multiply the term outside by every term inside the bracket.
当方程含有括号时，先使用分配律：a(b + c) = ab + ac。把括号外的项与括号内的每一项相乘。
| Expression | Expanded form |
|---|---|
| 3(x + 4) | 3x + 12 |
| 2(x - 3) | 2x - 6 |
| -2(x + 1) | -2x - 2 |
Example: Solve 3(x + 4) = 27.
例题：解 3(x + 4) = 27。
Expand first: 3x + 12 = 27. Then subtract 12: 3x = 15. Finally divide by 3: x = 5.
先展开：3x + 12 = 27。然后减 12：3x = 15。最后除以 3：x = 5。
5. Equations with negative signs and brackets | 含负号和括号的方程
A negative sign outside a bracket means you must multiply every term inside by -1. Many students forget to change the sign of the second term.
括号外有负号时，必须把括号内的每一项都乘以 -1。很多学生忘记改变第二项的符号。
Example: Solve 5 - 2(x + 1) = 1.
5 - 2(x + 1) = 1
Expand -2(x + 1) as -2x - 2. The equation becomes 5 - 2x - 2 = 1, so 3 - 2x = 1.
把 -2(x + 1) 展开为 -2x - 2。方程变为 5 - 2x - 2 = 1，即 3 - 2x = 1。
Subtract 3 from both sides: -2x = -2. Divide by -2: x = 1.
两边减 3：-2x = -2。除以 -2：x = 1。
6. Equations with variables on both sides | 变量在两侧的方程
When the variable appears on both sides, collect like terms first. Move the smaller variable term to the other side so the coefficient stays positive.
当变量同时出现在两边时，先合并同类项。把较小的变量项移到另一边，这样系数保持为正。
Example: Solve 5x + 2 = 3x + 10.
5x + 2 = 3x + 10
ract 3x from both sides: 2x + 2 = 10. Then subtract 2: 2x = 8. Finally divide by 2: x = 4.
两边减 3x：2x + 2 = 10。再减 2：2x = 8。最后除以 2：x = 4。
Always check by substituting x = 4 back into the original equation.
一定要把 x = 4 代回原方程检验。
7. Equations with fractions | 含分数的方程
x/3 + 1 = 5, you can treat x/3 as x divided by 3. Undo the addition first, then multiply.
分数线表示除法。在 x/3 + 1 = 5 这样的方程中，你可以把 x/3 看作 x 除以 3。先撤销加法，再乘以分母。
Example: Solve x/3 + 1 = 5.
x/3 + 1 = 5
Subtract 1: x/3 = 4. Multiply both sides by 3: x = 12.
减 1：x/3 = 4。两边乘以 3：x = 12。
8. Clearing denominators | 去分母
If there are two or more fractions, find the lowest common denominator and multiply every term by it. This clears the fractions in one step.
如果有两个或更多分数，找出最小公分母，并把每一项都乘以它。这样就能一步去掉所有分数。
Example: Solve x/2 + x/3 = 5.
x/2 + x/3 = 5
The lowest common denominator of 2 and 3 is 6. Multiply every term by 6:
6(x/2) + 6(x/3) = 6 × 5
This gives 3x + 2x
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图