📚 A-Level Maths: Solving and Representing Linear Inequalities | A-Level 数学：线性不等式的求解与表示
Linear inequalities are a fundamental topic in A-Level mathematics, bridging algebra and graphical reasoning. This article covers the key techniques for solving single and double linear inequalities, representing solution sets on number lines, and interpreting regions on graphs.
线性不等式是 A-Level 数学中的基础内容，连接了代数运算与图形推理。本文将系统讲解单重与双重线性不等式的求解技巧、在数轴上表示解集，以及如何在平面图形中解读可行区域。
1. Basic Rules of Solving Linear Inequalities | 线性不等式求解的基本规则
Solving a linear inequality follows the same steps as solving a linear equation: simplify, collect like terms, and isolate the variable. However, one crucial difference exists: when multiplying or dividing both sides by a negative number, the inequality sign must be reversed.
求解线性不等式的步骤与解线性方程相同：化简、合并同类项并分离变量。但有一个关键区别：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。
For example, if we have −2x > 6, dividing both sides by −2 gives x < −3. This reversal is essential because the order on the number line is reflected when multiplied by a negative value.
例如，对于 −2x > 6，两边同时除以 −2 后得到 x < −3。这种反转至关重要，因为乘以负数会导致数轴上的顺序发生镜像翻转。
Another important rule is that you may add or subtract any real number from both sides without changing the direction of the inequality. This also applies to adding or subtracting the same expression involving the variable.
另一个重要规则是：可以对两边同时加或减任意实数，而不改变不等号方向。这也适用于同时加减含有变量的相同表达式。
2. Solving Single Linear Inequalities | 解单一线性不等式
h sides to obtain x ≤ −12. The solution set is all real numbers less than or equal to −12.
考虑不等式 3x + 7 ≤ 2x − 5。求解时，先在两边减去 2x，得到 x + 7 ≤ −5，再两边减去 7，得到 x ≤ −12。解集为所有小于或等于 −12 的实数。
When the variable appears on both sides, always collect the terms so that the coefficient of the variable is positive if possible. This reduces the chance of sign errors. For example, 5 − 4x > x + 2 can be rearranged to 5 − 2 > x + 4x, giving 3 > 5x, or x < 3/5.
当变量同时出现在两边时，尽量整理使变量的系数为正，这样能减少符号错误。例如，5 − 4x > x + 2 可整理为 5 − 2 > x + 4x，即 3 > 5x，也即 x < 3/5。
Always check your final answer by substituting a boundary value and a value from the solution interval into the original inequality. This verification step is highly recommended in exams.
务必通过将边界值和解区间内的一个值代入原不等式来检查最终答案。在考试中，强烈建议进行这一步验证。
3. Solving Double Inequalities | 解双重不等式
7 means that 2x + 1 is strictly greater than −3 and less than or equal to 7. The variable is in the middle, so we apply operations to all three parts simultaneously.
双重不等式如 −3 < 2x + 1 ≤ 7 表示 2x + 1 严格大于 −3 且小于或等于 7。变量位于中间，因此我们需要同时对三个部分进行相同的运算。
is the final solution, meaning x can take any value greater than −2 and up to 3 inclusive.
三个部分同时减去 1：−4 < 2x ≤ 6。然后三个部分同时除以 2：−2 < x ≤ 3。这就是最终解，意味着 x 可以取大于 −2 且不超过 3（含 3）的任意值。
2x − 3 ≤ x + 5. Solve each separately and then take the intersection of the solution sets.
对于变量出现在外侧的双重不等式，如 x + 1 < 2x − 3 ≤ x + 5，可将其拆分为两个独立不等式：x + 1 < 2x − 3 和 2x − 3 ≤ x + 5。分别求解后取两个解集的交集。
4. Representing Solutions on a Number Line | 在数轴上表示解集
On a number line, an open circle (○ indicates that the endpoint is not included (strict inequality < or >). A closed circle (● d (≤ or ≥). A shaded line or ray extends from the circle in the direction of the solution.
在数轴上，空心圆圈（○）表示端点不包含在内（严格不等式 < 或 >）。实心圆圈（●）表示端点包含在内（≤ 或 ≥）。从圆圈向解的方向画阴影线或射线。
≤ 3, use an open circle at −2 and a closed circle at 3, with a solid line connecting them.
例如，x > 2 在数轴上表示为在 2 处画空心圆圈，并向右侧画射线。对于 x ≤ −1，在 −1 处画实心圆圈并向左侧阴影。对于双重不等式 −2 < x ≤ 3，在 −2 处用空心圆圈、在 3 处用实心圆圈，并用实线连接两者。
In A-Level exams, always label the key numbers on the line and use a ruler to draw the shading. A clear diagram helps examiners award full marks for the representation.
在 A-Level 考试中，务必在数轴上标注关键数字，并用直尺绘制阴影线。清晰的图示有助于阅卷人给出满分。
5. Solving Inequalities with Fractions | 含分数的不等式求解
When an inequality involves fractions, multiply both sides by the lowest common denominator (LCD) to clear the denominators. However, since the denominator is usually positive in linear contexts, the direction of the inequality remains unchanged.
当不等式中含有分数时，两边同时乘以最小公分母（LCD）以消去分母。由于在线性情境中分母通常为正，不等号方向保持不变。
can be verified by substituting x = 30, which gives 10 − 2 < 7.5 + 1, i.e. 8 < 8.5, true.
例如，解 (x/3) − 2 < (x/4) + 1。每项乘以 12：4x − 24 < 3x + 12。然后减去 3x 并加上 24：x < 36。代入 x = 30 验证：10 − 2 < 7.5 + 1，即 8 < 8.5，成立。
If a denominator is negative or contains a variable, extra care is needed. Since multiplying by a negative quantity reverses the inequality, such cases are usually handled by considering separate intervals or by bringing all terms to one side with a common denominator.
如果分母为负数或含变量，则需要格外小心。由于乘以负数会反转不等号方向，此类问题通常通过分区间讨论或将所有项移到一侧并化为公分母来处理。
6. Graphical Representation of Linear Inequalities | 线性不等式的图形表示
ght line. Use a solid line if the inequality is ≤ or ≥, and a dashed line if it is < or >.
两个变量的线性不等式，如 y > 2x − 1 或 3x + 2y ≤ 6，在坐标平面中定义了一个半平面。先画出对应的边界直线。若不等式为 ≤ 或 ≥，则画实线；若为 < 或 >，则画虚线。
To determine which side of the line to shade, choose a test point not on the line, usually (0,0) if it is not on the line. Substitute it into the inequality. If the result is true, shade the side containing the test point; otherwise shade the other side.
要确定应对方程的哪一侧进行阴影，选择一个不在直线上的测试点，通常选 (0,0)（只要它不在直线上）。将其代入不等式。若结果成立，则阴影包含该测试点的一侧；否则阴影另一侧。
2,0). Testing (0,0) gives 0 ≤ 4, which is true. Therefore shade the region below the line.
例如，对于 y ≤ −2x + 4，画一条经过 (0,4) 和 (2,0) 的实线。测试 (0,0) 得到 0 ≤ 4，成立。因此阴影直线下方的区域。
7. Finding the Feasible Region for a System | 求不等式组的可行区域
When several linear inequalities are given together, the feasible region is the intersection of all corresponding half-planes. This region is often a polygon, and its vertices are found by solving pairs of boundary line equations simultaneously.
当多个线性不等式同时给出时，可行区域是所有对应半平面的交集。该区域通常是一个多边形，其顶点通过联立求解边界直线方程得到。
and shade the region that satisfies every inequality. The vertices of the feasible region are (0,0), (4,0), (2,4), and (0,6).
考虑不等式组：x ≥ 0，y ≥ 0，x + y ≤ 6，以及 2x + y ≤ 8。画出所有边界线，并对满足所有不等式的区域进行阴影。可行区域的顶点为 (0,0)、(4,0)、(2,4) 和 (0,6)。
In optimization problems, the objective function is evaluated at each vertex to find its maximum or minimum value. This is the essence of linear programming, which relies on the graphical representation of linear inequalities.
在优化问题中，目标函数会在每个顶点处求值，以找到最大值或最小值。这正是线性规划的核心，它依赖于线性不等式的图形表示。
8. Converting Word Problems into Linear Inequalities | 将应用题转化为线性不等式
Many exam questions describe a real-world constraint using words like "at least", "no more than", or "cannot exceed". Translate these phrases into inequality symbols: "at least" means ≥, "no more than" means ≤, "more than" means >, and "fewer than" means <.
许多考题用“至少”“不超过”“不能超过”等文字描述现实约束。将这些短语转换为不等号：“至少”表示 ≥，“不超过”表示 ≤，“大于”表示 >，“小于”表示 <。
t 200 people and 60 are already inside, then the additional people y must satisfy y ≤ 140.
例如，若学生至少需要 40 分才能及格，且已得 22 分，则剩余分数应满足 x ≥ 18。若大厅最多能容纳 200 人，已有 60 人在内，则额外人数 y 必须满足 y ≤ 140。
tems purchased, also include x ≥ 0 and y ≥ 0 as non-negativity constraints in your system.
当变量受自然条件约束时，例如购买物品的数量，还应在不等式组中加入非负约束 x ≥ 0 和 y ≥ 0。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
heck the sign change explicitly. For example, −3x ≤ 9 becomes x ≥ −3 after dividing by −3.
常见错误之一是在乘以或除以负数时忘记反转不等号。请务必写出步骤并明确检查符号变化。例如，−3x ≤ 9 在除以 −3 后变为 x ≥ −3。
pen circle, while ≤ or ≥ gives a closed circle. This distinction is often tested directly.
另一个错误是在绘制数轴时混淆空心与实心圆圈。记住：严格不等式（< 或 >）用空心圆圈，而 ≤ 或 ≥ 用实心圆圈。这一区别经常被直接考查。
When dealing with double inequalities, do not perform different operations on different parts. All three parts must receive the same operation at each step. Otherwise, the relationship between the left and right bounds is destroyed.
处理双重不等式时，不能对不同部分进行不同运算。每一步必须对三个部分施加相同运算；否则会破坏左、右边界之间的关系。
10. Exam-Style Practice Questions | 考试风格练习
Here are three short questions that reflect typical A-Level items:
以下是三个能反映典型 A-Level 题型的练习：
Solve the inequality 4(x − 3) > 2x + 5 and represent the solution on a number line.
求解不等式 4(x − 3) > 2x + 5，并在数轴上表示解集。
ind the range of x such that −5 ≤ 2x − 1 < 7. Express your answer using interval notation.
求满足 −5 ≤ 2x − 1 < 7 的 x 的取值范围，并用区间记号表示答案。
Sketch the feasible region defined by y ≥ x − 2, y ≤ 3, and x + y ≥ 1. Label all vertices.
画出由 y ≥ x − 2、y ≤ 3 和 x + y ≥ 1 定义的可行区域，并标注所有顶点。
tion 2, add 1 to all parts: −4 ≤ 2x < 8, then divide by 2: −2 ≤ x < 4, written as [−2, 4).
对于第 1 题，展开得 4x − 12 > 2x + 5，然后减去 2x 并加 12：2x > 17，故 x > 8.5。对于第 2 题，三部分加 1：−4 ≤ 2x < 8，再除以 2：−2 ≤ x < 4，写作 [−2, 4)。
tices are (−1, 2), (5, 3), and (−2, 3). The feasible region is the triangular region bounded by these points.
对于第 3 题，三条边界直线分别为 y = x − 2、y = 3 和 x + y = 1。顶点为 (−1, 2)、(5, 3) 和 (−2, 3)。可行区域是以这三个点为界的三角形区域。
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