📚 Solving Linear Equations | 解一元一次方程
Linear equations are one of the most important building blocks in KS3 mathematics. They appear when two algebraic expressions are joined by an equals sign, and the goal is to find the unknown value that makes the statement true. Being confident with linear equations helps you solve real-life problems and prepares you for harder topics such as simultaneous equations and quadratic equations.
一元一次方程是 KS3 数学中最重要的基础模块之一。当两个代数式用等号连接时，就出现了方程，我们的目标是求出使等式成立的未知数的值。熟练掌握一元一次方程有助于解决实际问题，并为你后续学习联立方程和二次方程等更难的内容做好准备。
1. What is a Linear Equation? | 什么是一元一次方程？
n often be written in the form ax + b = c, where a, b and c are numbers and a is not zero.
一元一次方程是指未知数（通常写作 x）的指数仅为 1 的方程。例如，2x + 3 = 7 就是一元一次方程，因为 x 的最高次数是 1。这类方程通常可以写成 ax + b = c 的形式，其中 a、b 和 c 是已知数，且 a 不等于 0。
ne. At KS3 level, solving a linear equation means finding the value of x that makes the left-hand side equal to the right-hand side.
“线性”一词来源于 y = ax + b 的图像是一条直线。在 KS3 阶段，解一元一次方程就是求出使等号左边等于右边的 x 值。
2x + 3 = 7 → x = 2
2. The Balance Method | 天平法
Think of an equation as a balance scale. The equals sign is the pivot or centre. If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the other side to keep the balance. This rule is called the balance method, and it is the golden rule for solving equations.
把方程想象成一台天平。等号就是支点或中心。如果你对一边进行加、减、乘、除，必须对另一边做完全相同的运算，才能保持平衡。这条规则叫做天平法，也是解方程的黄金法则。
o x = 7. The equation stays balanced because the same operation was applied to both sides.
例如，如果 x + 3 = 10，两边同时减去 3，得到 x + 3 − 3 = 10 − 3。左边化简为 x，右边化简为 7，因此 x = 7。因为两边做了相同的运算，方程仍然平衡。
x + 3 = 10 → x + 3 − 3 = 10 − 3 → x = 7
Always remember that whatever you do to one side, you must do to the other. This is the most important idea in this entire topic.
始终记住：你对一边做了什么，就必须对另一边做同样的事。这是本主题中最核心的思想。
3. Solving One-Step Equations | 解一步方程
, add a to both sides. Addition and subtraction are inverse operations, so they undo each other.
一步方程只需一步逆运算就能分离出未知数。如果方程是 x + a = b，就在两边同时减去 a；如果是 x − a = b，就在两边同时加上 a。加法和减法是互逆运算，因此可以相互抵消。
x + 5 = 12 → x = 12 − 5 → x = 7
x − 4 = 9 → x = 9 + 4 → x = 13
or x/2 = 5, multiply both sides by 2. Multiplication and division are inverse operations.
对于乘法和除法，要使用相反的运算。如果 3x = 18，两边同时除以 3；如果 x ÷ 2 = 5 或 x/2 = 5，两边同时乘以 2。乘法和除法是互逆运算。
3x = 18 → x = 18 ÷ 3 → x = 6
x ÷ 2 = 5 → x = 5 × 2 → x = 10
4. Solving Two-Step Equations | 解两步方程
oing addition or subtraction first, then undo multiplication or division. This order follows the reverse of the usual order of operations, which is sometimes called BODMAS or BIDMAS.
两步方程包含两种运算，例如 2x + 3 = 11。首先消去加法或减法，再消去乘法或除法。这个顺序与通常的运算顺序（有时称为 BODMAS 或 BIDMAS）相反。
nstant term before the coefficient of x, because dividing too early can create unnecessary fractions.
解 2x + 3 = 11 时，两边先同时减去 3，得到 2x = 8。然后两边同时除以 2，得到 x = 4。一定要先消去常数项，再处理 x 的系数，因为过早除以系数会产生不必要的分数。
2x + 3 = 11 → 2x = 11 − 3 → 2x = 8 → x = 4
x = 15, then divide by 5 to get x = 3. Checking: 5(3) − 2 = 15 − 2 = 13, which is correct.
另一个例子是 5x − 2 = 13。两边同时加 2 得到 5x = 15，再除以 5 得到 x = 3。检验：5(3) − 2 = 15 − 2 = 13，结果正确。
5x − 2 = 13 → 5x = 15 → x = 3
5. Equations with Brackets | 含括号的方程
mple, 2(x + 3) = 14 becomes 2x + 6 = 14, and then you can solve it as a two-step equation.
如果方程含有括号，先用分配律展开。用括号外的项乘以括号内的每一项。例如，2(x + 3) = 14 展开为 2x + 6 = 14，然后就可以按两步方程求解。
2(x + 3) = 14 → 2x + 6 = 14 → 2x = 8 → x = 4
Be very careful with negative signs outside brackets. The expression −(x + 2) means −1 multiplied by the whole bracket, giving −x − 2. A common error is to write −x + 2, which is incorrect because the negative applies to both terms inside the bracket.
特别注意括号外的负号。表达式 −(x + 2) 表示 −1 乘以整个括号，展开后得到 −x − 2。一个常见错误是写成 −x + 2，这是不正确的，因为负号要作用于括号内的每一项。
lve 3(2x − 1) = 21. Expand: 6x − 3 = 21. Add 3 to both sides: 6x = 24. Divide by 6: x = 4.
例如，解 3(2x − 1) = 21。展开：6x − 3 = 21。两边加 3：6x = 24。除以 6：x = 4。
3(2x − 1) =
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