📚 Solving Linear Equations | 解一次方程
Linear equations are a core part of the Cambridge KS3 Mathematics curriculum, and they are also the foundation of almost every algebra topic that follows, from straight-line graphs to simultaneous equations and quadratic work. In this article you will learn how to solve one-step, two-step and multi-step equations, how to handle brackets, fractions and unknowns on both sides, how to check your answers, and how to avoid the common mistakes that cost marks in tests and exams.
一次方程是剑桥 KS3 数学课程的核心内容，也是后续几乎所有代数主题的基础——从直线图像到联立方程，再到二次函数的内容，都离不开它。在本文中，你将学习如何解一步方程、两步方程和多步方程，如何处理括号、分数以及未知数出现在等号两边的情况，如何检验答案，以及如何避免在测验和考试中丢分的常见错误。
1. Understanding Equations | 理解方程
le 2x + 3 on its own is just an expression - it has a value that depends on x, but it makes no claim about being equal to anything.
方程是表示两个表达式相等的数学语句，它总是包含等号。例如，2x + 3 = 11 是一个方程，而单独写出的 2x + 3 只是一个表达式——它的值取决于 x，但并没有断言它等于任何东西。
1, substituting x = 4 gives 2 × 4 + 3 = 11, which is true, so x = 4 is the solution. Substituting x = 5 gives 13, which is not 11, so x = 5 is not a solution.
解方程就是找出所有能使该语句成立的未知数的值，这个值称为方程的解（或根）。对于 2x + 3 = 11，把 x = 4 代入得到 2 × 4 + 3 = 11，成立，所以 x = 4 是解；把 x = 5 代入得到 13，不等于 11，所以 x = 5 不是解。
. At KS3 the equations you meet almost always have exactly one solution, which is what makes the balance method so reliable.
有些方程对未知数的任意取值都成立（这类方程称为恒等式，例如 2(x + 3) = 2x + 6），也有些方程永远不成立（例如 x + 1 = x + 5）。在 KS3 阶段，你遇到的方程几乎都恰好只有一个解，这正是平衡法如此可靠的原因。
2. What Makes an Equation Linear? | 什么是一次方程
ighest power of the variable is 1. Typical KS3 examples include 5x − 2 = 13 and 7 = 2x + 1. Both are linear, because the only variable term in each is a number times x.
一次方程是指未知数只出现一次方、且没有与另一个未知数相乘的方程。换句话说，未知数的最高次数是 1。KS3 的典型例子包括 5x − 2 = 13 和 7 = 2x + 1，它们都是一次方程，因为其中唯一的未知项都是某个数乘以 x。
near, because the unknown sits in the denominator. You will solve those types of equation in later years, using different methods.
像 x² + 2 = 6 这样的方程不是一次方程，因为变量出现了平方。像 3 ÷ x = 6 这样的方程同样不是一次方程，因为未知数出现在分母上。这类方程要在以后的年级用不同的方法来解。
The graph of a linear equation is a straight line, which is exactly where the name comes from. Adding a constant such as + 3 or − 7 shifts the line up or down; changing the number multiplying x changes its steepness. Recognising this link between the algebra and the graph is a useful habit to build early.
一次方程的图像是一条直线，"一次方程"（直线方程）这个名字正来源于此。加上一个常数（如 + 3 或 − 7）会把直线上下平移；改变乘在 x 前面的数则会改变直线的陡峭程度。尽早建立代数与图像之间的这种联系，是一个很好的学习习惯。
3. Inverse Operations and the Balance Method | 逆运算与平衡法
To solve an equation, we need to isolate the unknown, which means getting the unknown on its own on one side of the equals sign. This is done by applying inverse operations. Addition and subtraction are inverses of each other, and multiplication and division are inverses of each other.
解方程时，我们需要把未知数单独留在等号一边，也就是让未知数"孤立"出来。这可以通过逆运算来实现。加法和减法互为逆运算，乘法和除法互为逆运算。
The balance method is the key idea behind every linear equation you will solve. Whatever operation you perform on one side of the equation, you must perform exactly the same operation on the other side. This keeps the equation balanced, just like keeping equal weights on the two pans of a balance scale.
平衡法是你要解的每一个一次方程背后的核心思想。无论你对等式的一边进行什么运算，都必须对另一边进行完全相同的运算。这样才能保持方程平衡，就像维持天平两端砝码等重一样。
Here is a quick reference table of operations and their inverses, with an example of each:
下面是一张运算与逆运算的速查表，每个都配有例子：
| Operation in the equation | 方程中的运算 | Inverse operation | 逆运算 | Example | 例子 |
|---|---|---|
| Add 7 (+ 7) | Subtract 7 (− 7) | x + 7 = 15 ⇒ x = 8 |
| Subtract 4 (− 4) | Add 4 (+ 4) | x − 4 = 9 ⇒ x = 13 |
| Multiply by 5 (× 5) | Divide by 5 (÷ 5) | 5x = 35 ⇒ x = 7 |
| Divide by 3 (÷ 3) | Multiply by 3 (× 3) | x ÷ 3 = 6 ⇒ x = 18 |
When an equation contains more than one operation, undo them in reverse order: deal with the addition or subtraction first, then the multiplication or division. This is the opposite order to the order of operations (BIDMAS), and Section 5 explains why.
当方程包含不止一种运算时，要按相反的顺序逐步消去：先处理加法或减法，再处理乘法或除法。这与运算顺序（BIDMAS）恰好相反，第 5 节会解释原因。
4. Solving One-Step Equations | 解一步方程
A one-step equation needs only a single inverse operation to solve it. There are four basic forms: the unknown has a number added to it, a number subtracted from it, a number multiplied by it, or it is divided by a number.
一步方程只需要一次逆运算就能解出。它有以下四种基本形式：未知数被加上一个数、被减去一个数、被乘以一个数，或者被除以一个数。
Example: Solve x + 7 = 15. To undo the addition of 7, subtract 7 from both sides:
例题：解方程 x + 7 = 15。为了消去加 7，两边同时减去 7：
x + 7 − 7 = 15 − 7 ⇒ x = 8
Check: 8 + 7 = 15 ✓
检验：8 + 7 = 15 ✓
Example: Solve 4x = 20. To undo multiplication by 4, divide both sides by 4:
例题：解方程 4x = 20。为了消去乘以 4，两边同时除以 4：
4x ÷ 4 = 20 ÷ 4 ⇒ x = 5
Check: 4 × 5 = 20 ✓
检验：4 × 5 = 20 ✓
Example: Solve x ÷ 3 = 6. To undo division by 3, multiply both sides by 3:
例题：解方程 x ÷ 3 = 6。为了消去除以 3，两边同时乘以 3：
x ÷ 3 × 3 = 6 × 3 ⇒ x = 18
Check: 18 ÷ 3 = 6 ✓
检验：18 ÷ 3 = 6 ✓
ction equation is solved by adding, and a division equation is solved by multiplying - the inverse is always the "opposite" operation, not the same one.
例题：解方程 x − 4 = 9。这里的逆运算是加法，所以两边同时加 4：x = 13。注意，减法方程要用加法来解，除法方程要用乘法来解——逆运算永远是"相反"的运算，而不是相同的运算。
5. Solving Two-Step Equations | 解两步方程
Two-step equations need two inverse operations, applied one after the other. The standard order is to undo the addition or subtraction first, and then undo the multiplication or division.
两步方程需要两次逆运算，一次接一次地进行。标准顺序是先消去加法或减法，然后再消去乘法或除法。
lve 2x + 3 = 11. First subtract 3 from both sides because addition is the outer operation:
例题：解方程 2x + 3 = 11。首先两边同时减去 3，因为加法是外层的运算：
2x + 3 − 3 = 11 − 3 ⇒ 2x = 8
Then divide both sides by 2:
然后两边同时除以 2：
2x ÷ 2 = 8 ÷ 2 ⇒ x = 4
Check: 2 × 4 + 3 = 11 ✓
检验：2 × 4 + 3 = 11 ✓
Why this order? rst, so you remove the + 3 before dividing by 2. If you divided by 2 first you would get x + 1.5 = 5.5, which is still correct but requires an extra fraction step.
为什么是这个顺序？想一想表达式 2x + 3 是怎样"构造"出来的：先把 x 乘以 2，然后再加上 3。要还原这个过程，就必须先拆掉最后一步，所以在除以 2 之前先去掉 + 3。如果你先除以 2，会得到 x + 1.5 = 5.5，虽然结果依然正确，但会多出一个分数步骤。
. Add 2 to both sides to get 5x = 15, then divide by 5 to get x = 3. Check: 5 × 3 − 2 = 13 ✓.
第二个例题：解方程 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 = 13 ✓。
÷ 2 + 5 = 9. Subtract 5 first: x ÷ 2 = 4. Then multiply by 2: x = 8. Check: 8 ÷ 2 + 5 = 9 ✓.
第三个含除法的例题：解方程 x ÷ 2 + 5 = 9。先减 5：x ÷ 2 = 4；再乘以 2：x = 8。检验：8 ÷ 2 + 5 = 9 ✓。
6. Solving Multi-Step Equations | 解多步方程
Multi-step equations involve more than two operations. The strategy is always the same: simplify each side first, then apply inverse operations in reverse order until the unknown stands alone.
多步方程涉及两次以上的运算。策略始终相同：先把等号两边分别化简，再按相反顺序运用逆运算，直到未知数单独留在一边。
Step 1: Simplify each side separately. Collect like terms on the left and on the right, but do not move anything across the equals sign yet.
第一步：分别化简等号两边。把左边和右边的同类项分别合并，但此时还不要把任何项移到等号另一边。
Step 2: Remove the constant term from the side containing the unknown, using addition or subtraction.
第二步：用加法或减法，把含有未知数那一侧的常数项消掉。
Step 3: Divide (or multiply) both sides so that the coefficient of the unknown becomes 1.
第三步：两边同时除以（或乘以）某个数，使未知数的系数变为 1。
+ 7 = 19. Subtract 7: 3x = 12. Divide by 3: x = 4. Check: 4 × 4 + 7 − 4 = 16 + 7 − 4 = 19 ✓.
例题：解方程 4x + 7 − x = 19。先合并同类项：4x − x = 3x，于是方程变为 3x + 7 = 19。两边减 7：3x = 12；再除以 3：x = 4。检验：4 × 4 + 7 − 4 = 16 + 7 − 4 = 19 ✓。
btract 20 from both sides: −3x = −12. Divide by −3: x = 4. Check: 20 − 3 × 4 = 20 − 12 = 8 ✓.
含负系数的例题：解方程 20 − 3x = 8。这里未知项是负的，这是常见的出错点。两边同时减 20：−3x = −12；再除以 −3：x = 4。检验：20 − 3 × 4 = 20 − 12 = 8 ✓。
A useful habit is to write one line per operation, keeping the equals signs aligned vertically. This makes arithmetic slips easy to spot and shows the examiner exactly what you did.
一个很有用的习惯是每进行一步运算就另起一行，并让等号上下对齐。这样既容易发现计算失误，也让阅卷老师一眼看清你的解题步骤。
7. Equations with Brackets | 带括号的方程
When an equation contains brackets, you have two valid routes. You can expand the brackets first and then solve, or you can divide both sides by the number in front of the bracket if it divides exactly. Both methods earn full marks; choosing the quicker one saves time.
当方程中含有括号时，你有两条可行的路径：可以先展开括号再解方程，也可以在两边的数能被括号前的数整除时，直接两边同时除以这个数。两种方法都能得满分，选择更快的一种可以节省时间。
d: 3x + 12 = 27. Subtract 12: 3x = 15. Divide by 3: x = 5. Check: 3 × (5 + 4) = 3 × 9 = 27 ✓.
方法 A——先展开。解方程 3(x + 4) = 27。展开得 3x + 12 = 27；两边减 12 得 3x = 15；再除以 3 得 x = 5。检验：3 × (5 + 4) = 3 × 9 = 27 ✓。
= 9. Subtract 4: x = 5. The same answer, in fewer steps, because 27 ÷ 3 is a whole number.
方法 B——先除。解方程 3(x + 4) = 27。两边同时除以 3：x + 4 = 9；再减 4：x = 5。答案相同，但步骤更少，因为 27 ÷ 3 是整数。
rnatively divide by −2 first: x − 3 = −5, so x = −2. Check: −2 × (−2 − 3) = −2 × (−5) = 10 ✓. Notice how a negative number multiplied by a negative number is a positive number.
展开时要注意符号。解方程 −2(x − 3) = 10。展开得 −2x + 6 = 10，所以 −2x = 4，x = −2。也可以先两边同时除以 −2：x − 3 = −5，所以 x = −2。检验：−2 × (−2 − 3) = −2 × (−5) = 10 ✓。注意负数乘负数得正数。
t 4(x + 3) = 30 does not, so expansion is the safer route: 4x + 12 = 30, 4x = 18, x = 4.5.
如果括号不能整除，就展开。例如 5(2x − 1) = 35 可以整除（x = 4），但 4(x + 3) = 30 不能整除，所以展开更稳妥：4x + 12 = 30，4x = 18，x = 4.5。
8. Equations with the Unknown on Both Sides | 未知数在等号两边的方程
Many exam questions place the unknown on both sides of the equals sign. The method is to collect all the unknown terms on one side and all the number terms on the other side, then solve as normal.
很多考题会把未知数放在等号的两边。方法是把所有含未知数的项集中到一边，把所有常数项集中到另一边，然后按常规方法求解。
tract 2: 2x = 8. Divide by 2: x = 4. Check: LHS = 5 × 4 + 2 = 22 and RHS = 3 × 4 + 10 = 22 ✓.
例题：解方程 5x + 2 = 3x + 10。两边同时减去 3x，先消掉较小的未知项：2x + 2 = 10；两边减 2：2x = 8；再除以 2：x = 4。检验：左边 = 5 × 4 + 2 = 22，右边 = 3 × 4 + 10 = 22 ✓。
5x − 4 = 11. Add 4: 5x = 15. Divide by 5: x = 3. Check: 7 × 3 − 4 = 17 and 2 × 3 + 11 = 17 ✓.
结果为正的另一个例题（结果含负数的情况）：解方程 7x − 4 = 2x + 11。两边减 2x：5x − 4 = 11；两边加 4：5x = 15；再除以 5：x = 3。检验：7 × 3 − 4 = 17，2 × 3 + 11 = 17 ✓。
ee a negative answer, solve 2x + 7 = 5x + 19: subtract 2x to get 7 = 3x + 19, subtract 19 to get −12 = 3x, so x = −4. Check: 2 × (−4) + 7 = −1 and 5 × (−4) + 19 = −1 ✓.
产生负数解的例题：解方程 4x + 9 = 6x + 1。两边同时减 4x：9 = 2x + 1；再减 1：8 = 2x；除以 2 得 4 = x，即 x = 4。要看负数解，可以解 2x + 7 = 5x + 19：两边减 2x 得 7 = 3x + 19，
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