📚 Solving Linear Equations Step by Step | 逐步解线性方程
nging formulas, sequences, inequalities and even the calculus you meet at A Level - assumes that you can solve an equation such as 2x + 3 = 11 quickly and accurately. In this guide we build one reliable, step-by-step method for solving equations in the form 2x + 3 = 11, 3(x − 2) = 9 and 5x − 2 = 3x + 4, and we explain every move so that you never have to guess what to do next.
线性方程是 KS3 和 GCSE 代数的核心。你之后遇到的每一个主题——联立方程、直线图像、公式变形、数列、不等式，甚至 A Level 的微积分——都假定你能快速准确地解出 2x + 3 = 11 这样的方程。本文将为你建立一套可靠的分步方法，用来解 2x + 3 = 11、3(x − 2) = 9 和 5x − 2 = 3x + 4 这类方程，并解释每一步的理由，让你不必再猜测下一步该做什么。
You need four tools: the balancing method, inverse operations, bracket expansion and fraction clearing. Each section below gives the English explanation first and then the same idea in Chinese, followed by worked examples, a list of common mistakes, and exam-style practice with full solutions.
你需要掌握四个工具：平衡法、逆运算、去括号和消去分数。下面每一节都先给出英文讲解，再给出对应的中文讲解，之后配上例题、常见错误清单，以及带完整解答的考试风格练习。
1. What Is a Linear Equation? | 什么是线性方程？
the unknown, no unknown in a denominator, and no product of two unknowns. So 2x + 3 = 11, 5x − 2 = 3x + 4, x/3 + 1 = 4 and 7 − 4x = 2 are all linear equations, because in each one the unknown is simply multiplied by a number and then added to or subtracted from another number.
线性方程是指未知字母（通常是 x）只出现一次方的方程。它不含平方、立方，不含未知数的平方根，未知数不在分母中，也不出现两个未知数相乘。因此 2x + 3 = 11、5x − 2 = 3x + 4、x/3 + 1 = 4 和 7 − 4x = 2 都是线性方程，因为在每一个方程中，未知数只是被某个数乘，然后与另一个数相加或相减。
− 2 = 10 are not linear, because the unknown has a power greater than 1. Likewise 1/x = 5, √x uts the unknown in a denominator, the second takes a root of the unknown, and the third multiplies two unknowns together. Spotting the difference in the first two seconds saves a lot of wasted effort in an exam.
像 x² + 1 = 5 或 3x³ − 2 = 10 这样的方程不是线性方程，因为未知数的次数大于 1。同样，1/x = 5、√x = 3 和 xy = 6 也不是线性方程：第一个把未知数放在分母里，第二个对未知数开方，第三个把两个未知数相乘。在考试中，用最初两秒判断出这个区别，可以避免大量无用功。
| Equation | Linear? | Reason |
|---|---|---|
| 2x + 3 = 11 | Yes | x appears to the power 1 only |
| 5x − 2 = 3x + 4 | Yes | x on both sides, still power 1 |
| x² + 1 = 5 | No | x is squared |
| 3x³ − 2 = 10 | No | x is cubed |
| 1/x = 5 | No | x in the denominator |
ement. For 2x + 3 = 11 the solution is x = 4, because 2(4) + 3 = 11. A linear equation in one unknown normally has exactly one solution; the exceptions are the special cases discussed in Section 10, where an equation can be true for every value of x or for none at all.
线性方程的解，就是使方程两边相等的未知数的值——也就是把方程变成一个正确等式的那个值。对于 2x + 3 = 11，解是 x = 4，因为 2(4) + 3 = 11。只含一个未知数的线性方程通常恰好有一个解；例外情况是第 10 节讨论的特殊情形，此时方程可能对任何 x 都成立，或者对任何 x 都不成立。
2. The Balancing Method | 平衡法
Think of an equation as a balanced see-saw with the equals sign as the pivot. The left-hand side and the right-hand side have exactly the same value, which is why the see-saw is level. If you add something to one side only, the see-saw tips over and the equation is no longer true. But if you do exactly the same thing to both sides, the balance is preserved and the new equation has the same solution as the old one.
把方程想象成一架平衡的跷跷板，等号就是支点。左边和右边的值完全相等，所以跷跷板保持水平。如果你只在一边加上某个东西，跷跷板就会倾斜，等式不再成立；但如果你对两边做完全相同的事，平衡就得以保持，新方程与原方程的解完全相同。
This single idea gives us four legal moves, and every linear equation is solved using only these moves:
这一个想法就给出了四种合法操作，而所有线性方程都只用这几种操作来解：
- Add the same number to both sides: if a = b then a + c = b + c.｜在两边加上同一个数：若 a = b，则 a + c = b + c。
- Subtract the same number from both sides: if a = b then a − c = b − c.｜在两边减去同一个数：若 a = b，则 a − c = b − c。
- Multiply both sides by the same non-zero number: if a = b then ac = bc.｜两边同乘一个非零数：若 a = b，则 ac = bc。
- Divide both sides by the same non-zero number: if a = b then a ÷ c = b ÷ c (c ≠ 0).｜两边同除一个非零数：若 a = b，则 a ÷ c = b ÷ c（c ≠ 0）。
mple shows the method in action. To solve x + 5 = 12, we want x on its own, so we undo the "+5" step, so the solution is x = 7, which we can check by substituting back: 7 + 5 = 12. True.
一个最简单的例子可以展示这个方法。解 x + 5 = 12 时，我们希望 x 单独留在一边，所以用两边同时减 5 来抵消 "+5"。左边的 +5 与 −5 抵消，右边 12 − 5 = 7。每一步方程都保持平衡，所以解是 x = 7，代回检验：7 + 5 = 12，成立。
x + 5 = 12 → x + 5 − 5 = 12 − 5 → x = 7
The golden rule of presentation is one operation per line. Write the equation, perform a single move, write the new equation underneath, and keep the equals signs vertically aligned. This makes your working easy for an examiner to follow and much easier for you to check.
书写时的黄金法则是：一行只做一个操作。写出方程，做一次变换，把新方程写在下面，并让所有等号竖直对齐。这样阅卷老师容易看懂你的过程，你自己检查起来也轻松得多。
3. Inverse Operations: The Engine of the Method | 逆运算：方法的核心
An inverse operation is an operation that undoes another one. Addition and subtraction are inverses of each other; multiplication and division are inverses of each other; squaring and square-rooting are inverses of each other. To free the unknown, you apply the inverse of whatever has been done to it, and you apply that inverse to both sides of the equation.
逆运算就是能抵消另一种运算的运算。加法与减法互为逆运算；乘法与除法互为逆运算；平方与开平方互为逆运算。要想把未知数"解放"出来，你就要对它被施加的运算使用逆运算，并且把这个逆运算同时施加到方程两边。
| Operation applied to x | Inverse operation | Example |
|---|---|---|
| + 7 | − 7 | x + 7 = 10 → x = 3 |
| − 4 | + 4 | x − 4 = 9 → x = 13 |
| × 6 | ÷ 6 | 6x = 42 → x = 7 |
| ÷ 5 | × 5 | x/5 = 3 → x = 15 |
| x² (not linear) | √ | x² = 49 → x = ±7 |
ed by 2 and then had 3 added. So we undo backwards: subtract 3 first, then divide by 2. Students who divide by 2 first end up with x + 1.5 = 5.5, which is still solvable but messier - and messier working means more chances to slip.
当有两个或更多运算时，顺序很重要。要按它们被施加顺序的相反次序来抵消。在 2x + 3 中，未知数 x 先被乘以 2，然后加上 3。所以我们要倒着来：先减 3，再除以 2。如果先除以 2，会得到 x + 1.5 = 5.5，虽然仍可解，但更繁琐——而过程越繁琐，出错的机会就越多。
A useful memory aid is BIDMAS in reverse: brackets, indices, division and multiplication, addition and subtraction were applied in that order, so we peel them off in the opposite order - deal with brackets last when expanding, but with fractions and roots first when they wrap the whole side.
一个实用的记忆方法是"倒过来的 BIDMAS"：括号、指数、除与乘、加与减是按这个顺序作用的，所以我们要按相反顺序逐层剥开——展开时括号最后处理，而当分数或根号包住整个一边时则优先处理。
4. Worked Example: 2x + 3 = 11 | 例题：2x + 3 = 11
This is the classic first linear equation, and it contains exactly the two operations we have just discussed: a multiplication and an addition. The unknown has been doubled and then had 3 added, giving 11. We undo those operations backwards, one line at a time, keeping the equals signs lined up.
这是最经典的第一个线性方程，它恰好包含我们刚讨论的两种运算：一次乘法和一次加法。未知数先被乘以 2，再加上 3，得到 11。我们倒序抵消这些运算，一行一步，并让等号对齐。
2x + 3 = 11
2x + 3 − 3 = 11 − 3 → 2x = 8
2x ÷ 2 = 8 ÷ 2 → x = 4
Step 1: subtract 3 from both sides to remove the "+3". Step 2: divide both sides by 2 to remove the "×2" it is not optional in an exam-style answer - it takes five seconds and catches most slips.
第一步：两边同时减 3，去掉 "+3"。第二步：两边同时除以 2，去掉 "×2"。未知数现在单独留在左边，所以解是 x = 4。第三步是检验，在考试风格的解答中它不是可有可无的——它只需五秒，却能抓住大多数笔误。
Check: 2(4) + 3 = 8 + 3 = 11 ✓
ided a ≠ 0. Recognising this pattern means that a two-mark equation really does take you a few seconds.
同样的结构可以解所有 ax + b = c 形式的方程：两边先减 b，再除以 a。用符号表示，若 ax + b = c，则 x = (c − b)/a，只要 a ≠ 0。认出这个模式，意味着两分的方程题真的只要几秒钟。
ives gives a positive; dividing by a negative flips the sign of the other side, so keep a careful eye on the minus signs.
注意负系数。在 7 − 4x = 2 中，我们先两边减 7 得到 −4x = −5，再除以 −4 得到 x = 1.25。负负相除得正；除以负数会改变另一边的符号，所以一定要仔细盯住负号。
5. Equations with Brackets: 3(x − 2) = 9 | 含括号的方程：3(x − 2) = 9
Brackets are simply a package: 3(x − 2) means three copies of the whole bracket added together, which is why expanding gives 3x − 6 and not 3x − 2. There are two perfectly valid routes to the solution, and choosing the quicker one is a genuine exam skill.
括号只是一个"包裹"：3(x − 2) 表示三个完整的括号相加，所以展开后是 3x − 6，而不是 3x − 2。有两条完全可行的路径可以求解，选择更快的那条是真正的考试技巧。
Route A - expand first. Multiply out the bracket, then solve the resulting two-step equation. Route B - divide first. If the number outside the bracket divides the number on the other side exactly, divide both sides by it straight away and the bracket disappears.
路径 A——先展开。把括号乘开，再解得到的两步方程。路径 B——先除。如果括号外的数能整除另一边的数，就先把两边都除以它，括号自然消失。
Route A: 3(x − 2) = 9 → 3x − 6 = 9 → 3x = 15 → x = 5
Route B: 3(x − 2) = 9 → x − 2 = 3 → x = 5
Both routes give x = 5, and the check confirms it: 3(5 − 2) = 3 × 3 = 9 ✓. When the outside number also divides the bracket' tly, Route B is faster; when it does not, Route A always works, so it is the safe default.
两条路径都得到 x = 5，检验也确认了：3(5 − 2) = 3 × 3 = 9 ✓。当括号外的数也能整除括号内的各项时，路径 B 更快；当不能时，路径 A 总是有效，因此它是稳妥的默认选择。
vide both sides by −2 at the start: x + 4 = −5, so x = −9. Check: −2(−9 + 4) = −2(−5) = 10 ✓ sign: writing −2x + 8 instead of −2x − 8, because a negative times a positive is negative.
再看一个带负乘数的例子：解 −2(x + 4) = 10。展开得 −2x − 8 = 10，所以 −2x = 18，x = −9。也可以一开始两边除以 −2：x + 4 = −5，所以 x = −9。检验：−2(−9 + 4) = −2(−5) = 10 ✓。这里最常见的错误是漏掉符号：把 −2x − 8 写成 −2x + 8，因为负数乘正数得负数。
+ 6 + 3x − 3 = 18, then 5x + 3 = 18, then 5x = 15, so x = 3. Collecting like terms first keeps the numbers small and the working short.
当同一边出现两个括号时，先把两个括号都展开并合并同类项，再做别的。例如 2(x + 3) + 3(x − 1) = 18 变成 2x + 6 + 3x − 3 = 18，即 5x + 3 = 18，于是 5x = 15，x = 3。先合并同类项能让数字更小、过程更短。
6. Unknowns on Both Sides: 5x − 2 = 3x + 4 | 未知数在两边：5x − 2 = 3x + 4
When the unknown appears on both sides, you first gather all the x terms on one side and all the numbers on the other. The balancing rule still applies: whatever you subtract from one side you must subtract from the other, so subtracting 3x from both sides is perfectly legal and removes the x on the right.
当未知数出现在两边时，要先把所有含 x 的项集中到一边，把所有数字集中到另一边。平衡法则依然适用：你在一边减去什么，就必须在另一边也减去什么，所以两边同时减 3x 完全合法，这样右边的 x 就被消掉了。
5x − 2 = 3x
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