📚 Solving Linear Equations | 解一次方程
Solving linear equations is a fundamental skill in Key Stage 3 mathematics. These equations involve an unknown variable, usually represented by a letter such as x, and the goal is to find the value that makes the equation true. By applying inverse operations, we can isolate the variable step by step. Understanding this process builds a strong foundation for more advanced algebraic topics.
解一次方程是中学初级阶段数学的一项基本技能。这类方程含有一个未知数，通常用字母如 x 表示，目标是求出使等式成立的值。通过运用逆运算，我们可以逐步把未知数单独留在等号的一边。掌握这一过程为更高级的代数内容打下坚实的基础。
1. What is a Linear Equation? | 什么是一次方程？
is a linear equation because x is raised to the power of one. The graph of a linear equation is a straight line, which is why it is called 'linear'. In Key Stage 3, we focus on equations with one variable and simple coefficients.
一次方程是指未知数的最高次数为 1 的代数等式。例如，2x + 3 = 7 是一次方程，因为 x 的指数是 1。一次方程的图像是一条直线，因此被称为“线性”方程。在中学初级阶段，我们主要学习含一个未知数且系数简单的方程。
2. Understanding the Balance Method | 理解天平法
Imagine an equation as a balanced scale. Whatever you do to one side, you must do exactly the same to the other side to keep it balanced. This is the core idea behind solving equations. If you add 4 to the left side, you must add 4 to the right side. The balance method helps us perform operations systematically without breaking the equality.
我们可以把等式想象成一架平衡的天平。你对一边做的任何操作，都必须对另一边做完全相同的操作，才能保持平衡。这就是解方程的核心思想。如果你在左边加 4，就必须在右边也加 4。天平法帮助我们有条不紊地进行运算，而不破坏等式的平衡。
3. Solving Equations by Adding or Subtracting | 通过加减解方程
t side. The inverse of adding 5 is subtracting 5. So, we subtract 5 from both sides: x + 5 – 5 = 11 – 5, which simplifies to x = 6. Similarly, for x – 3 = 9, we add 3 to both sides because addition is the inverse of subtraction: x – 3 + 3 = 9 + 3, giving x = 12.
要解方程 x + 5 = 11，我们需要把左边的 +5 去掉。加 5 的逆运算是减 5。因此，两边同时减去 5：x + 5 – 5 = 11 – 5，化简得 x = 6。同理，对于 x – 3 = 9，我们在两边同时加 3，因为加法是减法的逆运算：x – 3 + 3 = 9 + 3，得到 x = 12。
4. Solving Equations by Multiplying or Dividing | 通过乘除解方程
multiply both sides by 4: (x ÷ 4) × 4 = 2 × 4, so x = 8. Always use the inverse operation.
当未知数乘以一个数时，例如 3x = 15，我们用除法来分离 x。两边同时除以 3：3x ÷ 3 = 15 ÷ 3，得到 x = 5。如果未知数除以一个数，如 x ÷ 4 = 2，我们就在两边同时乘以 4：(x ÷ 4) × 4 = 2 × 4，所以 x = 8。始终使用逆运算。
5. Two-Step Equations | 两步方程
is important: undo addition or subtraction before multiplication or division. This follows the reverse of the order of operations (BIDMAS/BODMAS).
许多方程需要两步才能解出。例如 2x + 3 = 9。首先，两边减 3：2x = 6。然后，两边除以 2：x = 3。运算顺序很重要：先处理加减，再处理乘除。这正好是运算顺序（BIDMAS/BODMAS）的逆过程。
6. Equations with Negative Coefficients | 含负系数的方程
Consider 5 – get 5 = 2 + x, then subtract 2 from both sides to find x = 3. Alternatively, you can treat –x as –1x and divide by –1, but the first method avoids negative division. For –2x = 8, divide both sides by –2 to get x = –4.
考虑方程 5 – x = 2。一种方法是两边同时加 x，得到 5 = 2 + x，然后两边减 2，求出 x = 3。另一种方法可以把 –x 看作 –1x，然后除以 –1，但前一种方法避免了负数的除法。对于 –2x = 8，两边除以 –2 得到 x = –4。
7. Equations with the Variable on Both Sides | 未知数在等号两边的方程
t 2: 2x = 8, and divide by 2: x = 4. Always aim to reduce the equation to a familiar form.
当方程两边都含有未知数时，如 3x + 2 = x + 10，我们首先把含未知数的项移到一边。两边同时减去 x：2x + 2 = 10。再减 2：2x = 8，然后除以 2：x = 4。始终要把方程化简成我们熟悉的形式。
8. Equations Involving Fractions | 含分数的方程
x/2) + (x/3) = 5 can be solved by multiplying by 6 to get 3x + 2x = 30, so 5x = 30, x = 6.
解方程 x/3 + 1 = 5 时，先两边减 1：x/3 = 4。然后两边乘以 3：x = 12。如果有多个分数，找出最小公分母，并把每一项都乘以它来去分母。例如 (x/2) + (x/3) = 5，可以两边乘以 6，得 3x + 2x = 30，即 5x = 30，x = 6。
9. Checking Your Solution | 检验答案
ight side. This step helps catch mistakes and reinforces your understanding. If the check fails, retrace your steps carefully.
一定要把你的解代回原方程进行验证。对于 2x + 5 = 13，我们求得 x = 4。检验：2(4) + 5 = 8 + 5 = 13，与右边相等。这一步有助于发现错误并加深理解。如果检验失败，请仔细回溯你的解题步骤。
10. Common Misconceptions | 常见错误观念
e terms to the other side without changing the sign, e.g., turning 3x + 2 = 5x into 2 = 5x – lance method: subtract 3x from both sides to get 2 = 2x. Careful notation prevents errors.
一个常见错误是忘记对等号两边同时操作。另一个错误是在逆运算时错误地使用 BIDMAS 规则。还有的学生在移项时不改变符号，例如错误地将 3x + 2 = 5x 变为 2 = 5x – 3x。一定要使用天平法：两边同时减去 3x，得到 2 = 2x。细致的书写可以避免错误。
11. Real-Life Applications | 实际应用
ituations, such as calculating costs, distances, or ages. For example, a plumber charges a £30 call‑out fee plus £20 per hour. The total cost C for h hours is given by C = 20h + 30. If the bill is £ + 30 = 90 to find h = 3 hours. Equations help us make predictions and informed decisions.
一次方程可以模拟许多日常情景，比如计算费用、距离或年龄。例如，一位水管工收取 30 英镑的上门费加每小时 20 英镑的工时费。h 小时的总费用 C = 20h + 30。如果账单是 90 英镑，我们可以解方程 20h + 30 = 90，得出 h = 3 小时。方程帮助我们做出预测和明智的决定。
12. Summary | 总结
Solving linear equations relies on performing inverse operations on both sides of the equation. Mastery of one‑step, two‑step, and multi‑step equations builds algebraic fluency. Remember to check your solutions and practise regularly. These skills are not only essential for examinations but also for logical reasoning in everyday life.
解一次方程的关键在于对等号两边同时进行逆运算。掌握一步、两步和多步方程的解法有助于提高代数流畅度。记住要检验答案，并坚持练习。这些技能不仅对应考至关重要，对日常生活中的逻辑推理也大有裨益。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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