📚 Solving Linear Equations and Inequalities: A Complete IGCSE Guide | 解一元一次方程与不等式：IGCSE 完整指南
Linear equations and inequalities are the backbone of IGCSE Mathematics. They are tested directly in algebra questions, but they also appear inside coordinate geometry, sequences, ratio problems, function work and multi-step word problems. A confident command of balancing, expanding brackets, clearing fractions, collecting unknowns on both sides, and reversing inequality signs will secure you many marks across both the Core and Extended papers.
一元一次方程与不等式是 IGCSE 数学的骨架。它们不仅会在代数题中直接考查，还会隐藏在坐标几何、数列、比例问题、函数题以及多步骤文字题里。熟练掌握等量平衡、去括号、去分母、把两边未知数合并，以及改变不等号方向等技能，能让你在 Core 与 Extended 两份试卷中稳稳拿到大量分数。
The aim of this guide is simple: to turn a topic that students often rush through into a reliable, step-by-step routine that works every time. We start from the definition, build up through increasingly complex equations, then move to inequalities where the single extra rule - flip the sign when you multiply or divide by a negative number - catches out thousands of candidates each year.
本指南的目标很简单：把这个很多学生草草带过的主题，变成一套可靠、可复制的分步流程。我们先从定义出发，逐步过渡到更复杂的方程，再进入不等式；不等式只多了一条规则，即在乘以或除以负数时要反转不等号，而正是这一条规则每年让成千上万的考生丢分。
1. What Makes an Equation Linear | 什么是一元一次方程
A linear equation in one unknown can be written in the standard form below, where a and b are constants and a is not zero. The key feature is that the unknown appears only to the power 1, so terms such as x², x³, the square root of x, or 1 divided by x never appear.
含有一个未知数的一元一次方程可以写成下面的标准形式，其中 a、b 为常数，且 a 不等于 0。关键特征是未知数只以一次方出现，因此不会出现 x²、x³、√x 或 1/x 这样的项。
ax + b = c, a ≠ 0
l to the right-hand side. Because the graph of y = ax + b is a straight line, a linear equation always has exactly one solution - unlike a quadratic, which may have two, one, or none.
例如，3x + 5 = 20 是一元一次方程，而 x² + 2x = 8 不是。一元一次方程的解就是使方程左右两边相等的 x 值。由于 y = ax + b 的图像是一条直线，一元一次方程总有且仅有一个解，这与可能有两个、一个或没有解的一元二次方程不同。
reduces to the form ax + b = c after expanding and collecting like terms counts as linear.
有些式子乍看复杂，但化简后仍然是一次方程。凡是经过展开与合并同类项后能化成 ax + b = c 形式的，都属于一元一次方程。
- Linear: 5(x − 2) = 3x + 4 (expands to 5x − 10 = 3x + 4)
一次：5(x − 2) = 3x + 4（展开得 5x − 10 = 3x + 4） - Linear: x/3 + 2 = 7 (the unknown is still to the power 1)
一次：x/3 + 2 = 7（未知数仍是一次方） - Not linear: x² − 4 = 0 (unknown squared)
非一次：x² − 4 = 0（未知数被平方） - Not linear: 3/x + 1 = 5 (unknown in the denominator)
非一次：3/x + 1 = 5（未知数出现在分母）
2. The Balancing Method | 等量平衡法
Think of an equation as a balance scale in perfect equilibrium. Whatever you do to one side, you must do to the other side. This keeps the equation true and allows you to isolate the unknown step by step. Addition and subtraction undo each other, and multiplication and division undo each other.
把方程想象成一台完全平衡的天平。你对一边做的任何操作，都必须对另一边做同样的操作。这样才能保持等式成立，并逐步把未知数单独留在一边。加法与减法互为逆运算，乘法与除法互为逆运算。
To solve x + 7 = 15, subtract 7 from both sides. To solve 4x = 28, divide both sides by 4.
要解 x + 7 = 15，两边同时减去 7；要解 4x = 28，两边同时除以 4。
x + 7 − 7 = 15 − 7 → x = 8
4x ÷ 4 = 28 ÷ 4 → x = 7
, first add 3 to both sides, then divide by 2. The order matters: you are peeling away the layers closest to x last.
当方程涉及两种运算时，按相反顺序依次还原。对于 2x − 3 = 11，先两边同时加 3，再两边同时除以 2。顺序很重要：离 x 最近的那一层运算最后处理。
2x − 3 = 11 → 2x = 14 → x = 7
7: 2 × 7 − 3 = 14 − 3 = 11, which matches the right-hand side, so the solution is correct.
务必把答案代回原方程检验。当 x = 7 时：2 × 7 − 3 = 14 − 3 = 11，与右边相等，因此解是正确的。
3. Two-Step and Multi-Step Equations | 求解两步与多步方程
Most exam equations need more than one operation. A reliable routine is: simplify each side separately, collect all unknown terms on one side, collect all number terms on the other side, then divide by the coefficient of x.
考试中的大多数方程都需要不止一步运算。可靠的流程是：先把每一侧分别化简，再把所有含未知数的项集中到一侧，把常数项集中到另一侧，最后除以 x 的系数。
Worked example: solve 5x + 4 = 3x + 12.
例题：解 5x + 4 = 3x + 12。
5x + 4 = 3x + 12 → 5x − 3x = 12 − 4 → 2x = 8 → x = 4
A common shortcut is to move terms across the equals sign and change their sign. Moving +3x to the left makes it −3x; moving +4 to the right makes it −4. This is exactly the same as subtracting from both sides, just written more quickly.
常用的快捷做法是把项移到等号另一边并改变符号。把 +3x 移到左边就变成 −3x；把 +4 移到右边就变成 −4。这与两边同时相减本质上完全一样，只是书写更快。
- If x has a fraction as a coefficient: multiply both sides by the reciprocal. For (2/3)x = 8, multiply by 3/2 to get x = 12.
若 x 的系数是分数：两边乘以它的倒数。对于 (2/3)x = 8，两边乘以 3/2 得 x = 12。 - If the same term appears on both sides: subtract it from both sides first to simplify.
若两边出现相同的项：先两边同时减去它来化简。 - If there are like terms on the same side: combine them before moving anything.
若同一侧有同类项：先合并，再移项。
4. Expanding Brackets Before Solving | 先展开括号再求解
When a bracket is multiplied by a number, you may either expand first or divide both sides first. Both are valid; expanding is safer when there are brackets on both sides or when the division would create awkward fractions.
当括号外乘有一个数时，你可以先展开，也可以先两边同时相除。两种方法都正确；当两边都有括号，或相除会产生难看的分数时，先展开更稳妥。
= 9, so x = 5. Notice that dividing first is quicker here because 27 divides exactly by 3.
解 3(x + 4) = 27。展开得 3x + 12 = 27，所以 3x = 15，x = 5。若先相除，得 x + 4 = 9，同样 x = 5。这里先相除更快，因为 27 能被 3 整除。
Negative multipliers are where most marks are lost. Every term inside the bracket must change sign.
括号前是负数乘数时，最容易丢分。括号内的每一项都必须变号。
−2(x − 5) = 14 → −2x + 10 = 14 → −2x = 4 → x = −2
A more demanding example combines two brackets: solve 2(3x − 1) − 3(x + 2) = 5.
再看一个更复杂的双括号例子：解 2(3x − 1) − 3(x + 2) = 5。
6x − 2 − 3x − 6 = 5 → 3x − 8 = 5 → 3x = 13 → x = 13/3
The answer does not have to be an integer. Leaving it as an exact fraction, or as 4.33 to three significant figures if the question asks for a decimal, is perfectly acceptable.
答案不一定非得是整数。保留精确分数形式，或在题目要求小数时写成三位有效数字 4.33，都是完全可以接受的。
5. Clearing Fractions and Decimals | 去分母与去小数
Fractional equations become much easier once you multiply every term by the lowest common multiple of the denominators. The golden rule is that every single term must be multiplied - including terms that do not contain a fraction.
只要把每一项都乘以各分母的最小公倍数，分数方程就会简单很多。黄金法则是：每一项都必须乘，包括那些不含分数的项。
Solve x/2 + x/3 = 5. The lowest common multiple of 2 and 3 is 6.
解 x/2 + x/3 = 5。2 和 3 的最小公倍数是 6。
6 × (x/2) + 6 × (x/3) = 6 × 5 → 3x + 2x = 30 → 5x = 30 → x = 6
3)/3, cross-multiplication is the quickest route: multiply the numerator of one side by the denominator of the other.
对于 (x + 1)/4 = (2x − 3)/3 这类方程，交叉相乘是最快的路径：用一侧的分子乘以另一侧的分母。
3(x + 1) = 4(2x − 3) → 3x + 3 = 8x − 12 → 15 = 5x → x = 3
3x + 5 = 14, so 3x = 9 and x = 3. This avoids careless decimal slips under exam pressure.
小数也可以用同样的思路清除。对于 0.3x + 0.5 = 1.4，把每一项乘以 10 得 3x + 5 = 14，所以 3x = 9，x = 3。这样可以避免考试压力下的小数运算失误。
6. Unknowns on Both Sides and Rearranging Formulas | 未知数在两边与公式变形
Equations with the unknown on both sides follow the same logic. Decide which side will keep the unknown - usually the side with the larger coefficient, so the result stays positive - then eliminate the other.
未知数出现在两边的方程遵循同样逻辑。先决定让哪一边保留未知数，通常选择系数较大的一侧，这样结果保持为正，然后消去另一侧。
Solve 7 − 2x = 4x − 5. Add 2x to both sides to keep the coefficient positive.
解 7 − 2x = 4x − 5。两边同时加 2x，使系数保持为正。
7 = 6x − 5 → 12 = 6x → x = 2
erent letter the subject. To make x the subject of y = 3x + 2, subtract 2 and divide by 3.
同样的平衡技巧也能用于公式变形，把另一个字母变成主项。要使 x 成为 y = 3x + 2 的主项，先减 2 再除以 3。
y − 2 = 3x → x = (y − 2)/3
ke x as a common factor to get x(a + b) = c, so x = c/(a + b), provided a + b is not zero.
若未知数出现两次，就把它提取成公因式。由 ax + bx = c，提取 x 得 x(a + b) = c，因此 x = c/(a + b)，前提是 a + b 不为零。
7. Solving Linear Inequalities | 解一元一次不等式
An inequality compares two expressions using one of four symbols: less than, greater than, less than or equal to, or greater than or equal to. Solving a linear inequality uses exactly the same steps as solving a linear equation, with one crucial extra rule introduced in the next section.
不等式用四种符号之一来比较两个式子：小于、大于、小于或等于、大于或等于。解一元一次不等式的步骤与解一元一次方程完全相同，只是多了一条关键规则，下一节会专门讲。
| Symbol | Meaning | 符号 | 含义 |
| < | is less than | < | 小于 |
| > | is greater than | > | 大于 |
| ≤ | is less than or equal to | ≤ | 小于或等于 |
| ≥ | is greater than or equal to | ≥ | 大于或等于 |
Solve 3x + 4 < 19. Subtract 4 from both sides, then divide by 3.
解 3x + 4 < 19。两边同时减 4，再除以 3。
3x + 4 < 19 → 3x < 15 → x < 5
heck with x = 4, which gives 16 < 19 (true), and x = 6, which gives 22 < 19 (false).
与方程不同，不等式有无穷多个解：所有小于 5 的数都成立。用 x = 4 检验得 16 < 19（成立），用 x = 6 检验得 22 < 19（不成立）。
8. Reversing the Inequality Sign | 不等号方向的反转
Here is the one rule that separates equations from inequalities: when you multiply or divide both sides by a negative number, the inequality sign must be reversed. Adding or subtracting a negative number does not require any change.
这就是方程与不等式的唯一分水岭：当两边同时乘以或除以一个负数时，不等号必须反转。而加上或减去一个负数则不需要改变方向。
Solve −2x > 8. Divide both sides by −2 and flip the sign.
解 −2x > 8。两边同时除以 −2，并反转不等号。
−2x > 8 → x < −4
Why does this happen? Multiplying by a negative number reflects every point on the number line about zero, so the order of the numbers is turned upside down. Since 2 < 3 but −2 > −3, any comparison must flip along with it.
为什么会这样？乘以负数相当于把数轴上的每个点关于原点做对称，数的顺序被完全颠倒。因为 2 < 3 而 −2 > −3，所以任何比较关系都必须跟着反转。
a fractional coefficient: solve −x/3 ≤ 2. Multiply both sides by −3 and reverse the sign.
再看一个系数为分数的例子：解 −x/3 ≤ 2。两边同时乘以 −3，并反转不等号。
−x/3 ≤ 2 → x ≥ −6
A useful safety check: pick a test value from your final answer and substitute it into the original inequality. If it satisfies the original statement, your sign change was correct.
一个实用的自查方法：从最终答案中取一个检验值，代回原不等式。如果它满足原式，说明你的符号反转是正确的。
9. Number Lines and Interval Notation | 数轴与区间表示法
Answers to inequalities are often shown on a number line. Use an open circle for strict inequalities (less than or greater than) and a filled circle for inclusive ones (less than or equal to, or greater than or equal to). The arrow shows which direction the solution continues.
不等式的答案常用数轴表示。严格不等式（小于或大于）用空心圆，含等号的不等式（小于或等于、大于或等于）用实心圆。箭头表示解继续延伸的方向。
- x < 5: open circle at 5, arrow pointing left.
x < 5：在 5 处画空心圆，箭头向左。 - x ≥ −6: filled circle at −6, arrow pointing right.
x ≥ −6：在 −6 处画实心圆，箭头向右。 - 2 ≤ x < 7: filled circle at 2 and open circle at 7, with a line joining them.
2 ≤ x < 7：在 2 处画实心圆、在 7 处画空心圆，中间用线段连接。
Interval notation is a compact alternative favoured in Extended papers. A round bracket means the endpoint is excluded; a square bracket means it is included. Infinity is always written with a round bracket because it is not a number you can reach.
区间表示法是 Extended 试卷中常用的简洁写法。圆括号表示端点不包含，方括号表示端点包含。无穷大永远用圆括号，因为它不是一个可以取到的数。
x < 5 → (−∞, 5) 2 ≤ x < 7 → [2, 7) x ≥ −6 → [−6, ∞)
are 2, 3, 4, 5 and 6 - note that 7 is excluded. Reading the question carefully for words such as integer, whole number, or positive is worth an easy mark.
有些题目只要求写出整数解。对于 2 ≤ x < 7，整数解是 2、3、4、5、6，注意 7 不包含在内。仔细读出题目中的整数、自然数或正数等字眼，往往能轻松拿分。
10. Forming Equations from Word Problems | 由文字题列出方程
Word problems test whether you can translate English into algebra. Use a four-step routine: define the unknown clearly with a letter, write an expression for every other quantity in terms of that letter, form the equation, then solve and interpret the answer in context.
文字题考查的是把语言翻译成代数的能力。使用四步流程：先用字母清楚地定义未知量，再用这个字母表示其他所有量，接着列出方程，最后求解并结合情境解释答案。
Example: the length of a rectangle is 3 cm more than its width, and the perimeter is 34 cm. Let the width be w, so the length is w + 3.
例题：一个长方形的长比宽多 3 厘米，周长是 34 厘米。设宽为 w，则长为 w + 3。
2(w + w + 3) = 34 → 4w +
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