📚 Mastering Inequalities for GCSE OCR Maths | GCSE OCR 数学不等式考点精讲
Inequalities are a fundamental topic in GCSE OCR Mathematics, appearing across both Foundation and Higher tiers. They build on equation-solving skills but introduce a new layer of reasoning about ranges of values rather than exact solutions. In this guide, you will learn how to solve linear inequalities, represent them on number lines, tackle double inequalities, shade graphical regions, and interpret real-life problems. Every explanation is paired with Chinese translation to support bilingual learners. Let’s dive in and turn inequalities into one of your strongest topics.
不等式是 GCSE OCR 数学中的基础考点，在基础卷和高阶卷中都会出现。它在等式求解的基础上引入了关于取值范围的新逻辑。本指南将系统讲解如何解线性不等式、在数轴上表示、处理双重不等式、图形区域着色以及解读应用题。每个知识点都配有中英双语解释，帮助你彻底掌握不等式，让它成为你的强项。
1. Understanding Inequalities: Symbols and Meanings | 理解不等式：符号与意义
The four inequality symbols are strict and non-strict. '>' means ‘greater than’, '<' means ‘less than’, '≥' means ‘greater than or equal to’, and '≤' means ‘ , x > 5 means x can be 5.1, 6, 10, but not 5. x ≤ 3 means x can be 3, 2, 0, -5, and so on.
四个不等式符号分为严格和非严格两类。'>' 表示“大于”，'<' 表示“小于”，'≥' 表示“大于等于”，'≤' 表示“小于等于”。严格不等式不包含边界值，非严格不等式则包含。例如 x > 5 意味着 x 可以是 5.1、6、10，但不能是 5；而 x ≤ 3 则允许 x 取 3、2、0、-5 等。
row points towards the values that make the inequality true. This visual representation is a common exam question.
在数轴上，空心圆圈表示严格不等式（如 x < 4 或 x > 4），实心圆圈表示非严格不等式（x ≤ 4 或 x ≥ 4）。箭头指向使不等式成立的值方向。这种可视化表示是热门考题。
2. Solving Linear Inequalities | 解一元一次不等式
Solving a linear inequality is much like solving an equation: you can add, subtract, multiply, or divide both sides by the same positive number without changing the inequality sign. For example, solve 3x + 7 < 19. Subtract 7 from both sides to get 3x < 12, then divide by 3 to obtain x < 4.
解一元一次不等式与解方程非常相似：你可以对两边进行加减运算，或同时乘以、除以同一个正数，不等号方向不变。例如解 3x + 7 < 19，两边减 7 得 3x < 12，再两边除以 3 得到 x < 4。
The critical rule you must remember is: when you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign. For instance, -2x > 6. Dividing by -2 gives x < -3. The '>' flips to '<' because of the negative divisor.
必须牢记的关键法则是：当两边同时乘以或除以一个负数时，必须反转不等号的方向。例如 -2x > 6，两边除以 -2 得 x < -3。由于除数是负数，'>' 变成了 '<'。
t x = 3: 3(3)+7 = 16, and 16 < 19 holds true. This habit helps avoid sign errors in exams.
始终通过将解集中的某个值代回原不等式来验证答案。例如 x < 4，试 x = 3：3(3)+7 = 16，16 < 19 成立。养成这个习惯有助于在考试中避免符号错误。
3. Representing Inequalities on a Number Line | 在数轴上表示不等式
w a number line, locate 2, place a closed circle at 2, and draw an arrow extending to the left to show all numbers less than or equal to 2. The direction of the arrow matches the number line orientation: left for smaller numbers, right for larger numbers.
对于 x ≤ 2 这样的单一不等式，画出数轴，标记点 2，用实心圆圈表示 2 包含在内，然后向左画箭头表示所有小于等于 2 的数值。箭头的方向与数轴方向一致：向左表示较小数，向右表示较大数。
all integers x: -2, -1, 0, 1, 2, 3, 4, 5. Notice -3 is not included because of the strict '<' sign, but 5 is included due to '≤'. Always read the circle type carefully.
常考题要求列出满足不等式的所有整数值。例如 -3 < x ≤ 5，列举所有整数 x：-2, -1, 0, 1, 2, 3, 4, 5。注意 -3 由于是严格 '<' 号不包含，而 5 因 '≤' 包含在内。务必仔细辨别圆圈类型。
the left often helps you visualise the direction correctly before drawing the number line.
当不等式中的 x 写在右边时，如 8 ≥ x，等价于 x ≤ 8。在画数轴前将不等式改写成 x 在左边的形式，有助于你正确把握方向。
4. Solving Inequalities with Brackets | 解含括号的不等式
Divide by 6: x ≤ 4. The only special caution is that if a negative multiplier appears outside brackets, pay attention when dividing later.
首先展开括号，再化简两边，然后分离变量。例如解 2(3x - 5) ≤ 14。展开得 6x - 10 ≤ 14。两边加 10：6x ≤ 24。除以 6：x ≤ 4。唯一需要特别注意的是，如果括号外有负号相乘，后续除法时留心符号方向。
Consider -4(2x + 3) > 20. Expand to -8x - 12 > 20. Add 12 to both sides: -8x > 32. Now divide by -8 and reverse the sign: x < -4. The double negative effect can be confusing; take your time to avoid mistakes.
考虑 -4(2x + 3) > 20。展开得 -8x - 12 > 20。两边加 12：-8x > 32。然后除以 -8 并反转符号：x < -4。双重负号容易混淆，务必步步谨慎，避免出错。
5. Inequalities with Unknowns on Both Sides | 含两边未知数的不等式
When the variable appears on both sides, collect all variable terms on one side and constants on the other. Solve 5x - 3 > 2x + 9. Subtract 2x from both sides: 3x - 3 > 9. Add 3: 3x > 12. Divide by 3: x > 4.
当未知数出现在两边时，将所有含变量的项移到一边，常数项移到另一边。解 5x - 3 > 2x + 9。两边减 2x：3x - 3 > 9。加 3：3x > 12。除以 3：x > 4。
If the variable terms cancel out, you get a statement that is either always true or never true. For example, 2x + 5 > 2x + 1 simplifies to 5 > 1, which is always true, so the solution is all real numbers. Conversely, 2x + 5 < 2x + 1 gives 5 < 1, which is false, so there is no solution.
如果变量的项抵消了，你会得到一个永远成立或永不成立的陈述。例如 2x + 5 > 2x + 1 化简为 5 > 1，恒成立，因此解为所有实数。反之，2x + 5 < 2x + 1 得到 5 < 1，是假命题，所以无解。
6. Double Inequalities | 双重不等式
eans that the expression 3x + 1 is simultaneously greater than -5 and less than or equal to 7. You can solve it by breaking it into two separate inequalities, or by performing the same operation on all three parts at once.
双重不等式如 -5 < 3x + 1 ≤ 7 表示表达式 3x + 1 同时大于 -5 且小于等于 7。你可以将其拆成两个单独的不等式来求解，也可以同时对三部分进行相同运算。
all x greater than -2 and up to 2. This method is faster but remember that any multiplication or division by a negative number would reverse both inequality signs – so handle with care.
使用三部分同步法：三部分同时减 1：-6 < 3x ≤ 6。三部分同时除以 3：-2 < x ≤ 2。解为所有大于 -2 且不超过 2 的 x。这种方法更快捷，但要记住，乘以或除以负数时需要同时反转两个不等号，务必小心处理。
7. Graphical Inequalities: Shading Regions | 图形不等式：区域着色
d line for strict inequalities (< or >) and a solid line for non-strict ones (≤ or ≥). Then test a point (x=0, y=0 is easiest) to decide which side of the line to shade.
含两个变量的不等式在坐标平面上定义区域。首先画出相应等式的边界线。严格不等式（< 或 >）用虚线，非严格不等式（≤ 或 ≥）用实线。然后选取测试点（(0,0) 最简单）来判断着色哪一侧。
a dashed line. Test (0,0): 0 > 2(0)+1 gives 0 > 1, which is false, so shade the side that does not contain (0,0). The shaded region represents all points that satisfy the inequality.
对于 y > 2x + 1，用虚线画出 y = 2x + 1。测试 (0,0)：0 > 2(0)+1 即 0 > 1，不成立，因此着色不含 (0,0) 的一侧。阴影区域表示所有满足不等式的点。
When two or more inequalities are given, the required region is where all individual shaded regions overlap. In the exam, you may be asked to clearly label this region, often labelled R. Make sure your shading is neat and boundaries are correctly dashed or solid according to the inequality signs.
当给出两个或更多不等式时，所求区域是所有单独阴影区域的重叠部分。考试中可能会要求清晰标出该区域，常记为 R。确保你的着色整洁，边界线根据不等号正确使用虚线或实线。
8. Forming Inequalities from Word Problems | 从应用题建立不等式
s require you to translate a verbal description into an algebraic inequality. Key phrases: 'at least' means ≥, 'no more than' or 'at most' means ≤, 'greater than' means >, 'less than' means <. For example, "A rectangle has length x and width 5. The perimeter is at least 30 cm" gives 2(x+5) ≥ 30.
应用题要求你将文字描述转化为代数不等式。关键短语：“至少”对应 ≥，“不超过”或“最多”对应 ≤，“大于”对应 >，“小于”对应 <。例如“一个矩形长为 x，宽为 5，周长至少 30 cm”转化为 2(x+5) ≥ 30。
Solve the inequality normally: 2x + 10 ≥ 30 ⇒ 2x ≥ 20 ⇒ Always give the final answer in the context of the question, and check that it makes sense – a negative length would be invalid even if mathematically a solution.
正常解不等式：2x + 10 ≥ 30 ⇒ 2x ≥ 20 ⇒ x ≥ 10。然后将结果带回语境：长度至少为 10 cm。始终根据题意给出最终答案，并检查其合理性——即使数学上得到负值解，在实际情况中负的长度也是无效的。
9. Common Mistakes to Avoid | 常见错误避坑
Mistake 1: Forgetting to flip the inequality sign when multiplying or dividing by a negative. This is the most frequent error. Always ask yourself: "Am I using a negative number to multiply/divide?" If yes, flip the sign immediately.
错误 1：乘除负数时忘记翻转不等号。这是最常见的错误。时刻自问：“我是否在乘/除一个负数？”如果是，立即翻转符号。
Mistake 2: Misreading 'at least' or 'at most' ing up ≥ and ≤ can lose easy marks. Underline these phrases in the question to stay alert.
错误 2：在应用题中误读“至少”或“最多”。混淆 ≥ 和 ≤ 会丢掉容易得的分。在题目中划出这些短语，保持警惕。
is part of the answer. Similarly, on a number line, an open circle for ≤ or ≥ will be marked wrong.
错误 3：严格不等式画成实线，或非严格不等式画成空心圆。在图形题中，线型是答案的一部分。同样，数轴上对 ≤ 或 ≥ 使用空心圆会被判错。
Mistake 4: Incorrectly solving double inequalities by only performing an operation on one or two parts instead of all three. Always apply the operation to the left part, the middle, and the right part simultaneously.
错误 4：解双重不等式时只对其中一两部分运算，而没对三部分同时运算。务必对左、中、右三部分同步进行操作。
10. Exam-Style Questions and Tips | 考试题型与技巧
Question 1 (Foundation): Solve 5x - 4 < 21, representing the solution on a number line.
Solution: 5x < 25 → x < 5. Draw an open circle at 5, arrow to the left.
题目 1（基础）：解 5x - 4 < 21，并在数轴上表示解集。
解：5x < 25 → x < 5。在 5 处画空心圆，箭头向左。
Question 2 (Higher): Solve the inequality -3(2x - 1) ≤ 9.
Solution: Expand to -6x + 3 ≤ 9 → -6x ≤ 6 → x ≥ -1 (sign reversed because dividing by -6).
题目 2（高阶）：解不等式 -3(2x - 1) ≤ 9。
解：展开得 -6x + 3 ≤ 9 → -6x ≤ 6 → x ≥ -1（除以 -6，不等号反转）。
Question 3 (Graphs): (>) and x = 3 (<). Test point (0,0) and shade the overlapping region. Label the region R.
题目 3（图形）：在网格上标出满足 y ≤ x + 2，y > -1 和 x < 3 的区域。y = x+2 画实线（≤），y = -1 和 x = 3 画虚线（> 和 <）。测试点 (0,0)，叠加着色区域，标记为 R。
Exam tip: Always check your final answer by picking a value inside your solution and testing it in the original inequality. If it fails, re-check your algebraic steps, especially the direction of the sign. Time invested in verification can prevent losing multiple marks.
考试技巧：始终从解集中挑一个值代回原不等式验证。如果不成立，重新检查代数步骤，尤其是不等号方向。花时间验证可以避免丢失多分。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图