📚 Unit 2 Higher Homework: Linear Equations and Inequalities | 第2单元 高等作业：一次方程与不等式
This revision article covers the key algebra skills tested in Unit 2 Higher Homework, pages 27 to 40. It focuses on linear equations, inequalities, rearranging formulae, and the graphical interpretation of solutions. The material is aimed at higher-tier students targeting grades 7 to 9.
本文复习第2单元高等作业第27至40页考查的核心代数技能，重点是一次方程、不等式、公式变形以及解的图像意义。内容面向目标7至9级的高层级学生。
1. Understanding Linear Equations | 理解一次方程
A linear equation is an equation of the first degree, meaning the variable has an exponent of 1. It can usually be written in the form ax + b = c, where a, b, and c are constants and a ≠ 0. Solving a linear equation means finding the value of the unknown that makes the statement true.
一次方程是一次式方程，即未知数的指数为1。它通常可以写成 ax + b = c 的形式，其中 a、b、c 为常数且 a ≠ 0。解一次方程就是求出使等式成立的未知数的值。
For example, in the equation 3x + 5 = 20, the aim is to isolate x by performing inverse operations. Subtract 5 from both sides to get 3x = 15, then divide both sides by 3 to obtain x = 5.
例如，在方程 3x + 5 = 20 中，目标是通过逆运算将 x 分离出来。两边同时减去5得到 3x = 15，然后两边同时除以3得到 x = 5。
3x + 5 = 20 ⇒ x = 5
2. Solving Equations with Brackets | 解含括号的方程
When an equation contains brackets, expand them first. Use the distributive law: a(b + c) = ab + ac. Then collect like terms and solve using inverse operations. Always check your answer by substituting it back into the original equation.
当方程含有括号时，首先展开括号。使用分配律：a(b + c) = ab + ac。然后合并同类项，并用逆运算求解。一定要将答案代回原方程进行检验。
Consider the equation 2(x + 3) − 4 = 10. The steps are shown below.
考虑方程 2(x + 3) − 4 = 10。步骤如下。
- Expand: 2(x + 3) − 4 = 10
- Simplify: 2x + 6 − 4 = 10
- Combine: 2x + 2 = 10
- Subtract 2: 2x = 8
- Divide by 2: x = 4
Always expand brackets before adding or subtracting terms. This prevents sign errors and keeps the working clear.
一定要先展开括号，再进行加减项。这样可以避免符号错误，并使解题过程保持清晰。
3. Equations with Fractions | 含分数的方程
Equations with fractions are easier to handle if you multiply every term by the lowest common denominator (LCD). This clears the fractions and produces an integer equation. Be careful to multiply both sides and every term, including whole numbers.
含分数的方程，最好先找出最小公分母（LCD），并将其乘以每一项。这样可以消去分母，得到整数方程。注意两边以及每一项（包括整数项）都要乘以最小公分母。
For example, solve x/3 + 1 = x/2. The LCD of 3 and 2 is 6, so multiply every term by 6.
例如，解方程 x/3 + 1 = x/2。3和2的最小公分母是6，所以将每一项都乘以6。
x/3 + 1 = x/2 ⇒ 2x + 6 = 3x ⇒ x = 6
This method removes fractions in one step, making the equation much simpler to solve.
这种方法能一步消去分数，使方程更容易求解。
4. Forming Equations from Word Problems | 由文字题列方程
Many higher-tier questions ask you to form an equation from a word problem. Identify the unknown quantity, represent it with a letter, and translate words into algebraic expressions. Key words: “more than” means +, “less than” means −, “times” means ×, and “is” means =.
许多高层级题目要求根据文字题列出方程。确定未知量，用字母表示，并将文字转化为代数表达式。关键词：“多于”表示 +，“少于”表示 −，“倍”表示 ×，“是”表示 =。
Example: Three more than twice a number is 19.
例如：一个数的两倍再加3等于19。
2n + 3 = 19 ⇒ n = 8
When forming equations from real-life problems, always define the variable first, then write the equation, then solve it, and finally answer in context.
从实际问题中列方程时，一定要先定义变量，再写出方程，然后求解，最后根据题意作答。
5. Inequalities on a Number Line | 数轴上的不等式
An inequality compares two expressions using <, >, ≤, or ≥. On a number line, an open circle represents < or >, while a closed circle represents ≤ or ≥. The direction of the arrow shows all possible values of the variable.
不等式用 <、>、≤ 或 ≥ 来比较两个表达式。在数轴上，空心圆表示 < 或 >，实心圆表示 ≤ 或 ≥。箭头的方向表示未知数的所有可能取值。
For example, x > 2 is shown with an open circle at 2 and an arrow pointing to the right. This means all values greater than 2, but not 2 itself.
例如，x > 2 在数轴上用2处的空心圆和指向右边的箭头表示。这意味着所有大于2的值，但不包括2本身。
It is essential to use the correct circle type in diagrams, as examiners often award marks for accurate notation.
在图中使用正确的圆圈类型至关重要，因为考官通常会对准确的表示方法给分。
6. Solving Linear Inequalities | 解一次不等式
Solving a linear inequality is similar to solving an equation, but there is one critical difference: if you multiply or divide both sides by a negative number, you must reverse the inequality sign. For example, −2x < 8 becomes x > −4 after dividing by −2.
解一次不等式与解方程类似，但有一个关键区别：如果将两边同时乘以或除以一个负数，必须改变不等号的方向。例如，−2x < 8 两边除以 −2 后变为 x > −4。
Here is a typical solution for the inequality 3x − 5 ≤ 7.
下面是不等式 3x − 5 ≤ 7 的典型解法。
- 3x − 5 ≤ 7
- Add 5: 3x ≤ 12
- Divide by 3: x ≤ 4
Remember to reverse the sign only when multiplying or dividing by a negative number. Adding or subtracting a negative number does not require reversing the sign.
请记住，只有在乘以或除以负数时才需要改变不等号方向。加减负数并不需要改变不等号。
7. Compound Inequalities | 复合不等式
A compound inequality combines two inequalities, often written as a < x < b. Solve each part separately, or keep the variable in the middle while performing inverse operations. The solution is the intersection of the two inequalities.
复合不等式将两个不等式组合在一起，通常写作 a < x < b。可以分别解每个部分，或者在保持未知数在中间的同时进行逆运算。解是两个不等式的交集。
For example, solve 4 < 2x + 2 ≤ 10.
例如，解 4 < 2x + 2 ≤ 10。
4 < 2x + 2 ≤ 10 ⇒ 1 < x ≤ 4
This means x is greater than 1 and less than or equal to 4. When writing the final answer, keep the variable in the middle and the smaller value on the left.
这意味着 x 大于1且小于等于4。书写最终答案时，将变量放在中间，较小的值放在左边。
8. Rearranging Formulae | 公式变形
Rearranging a formula means making a different variable the subject. Treat the variable you want as the unknown and apply inverse operations, just as in solving equations. If the variable appears in two terms, factor it out first.
公式变形是指将另一个变量作为主项。把要求的变量当作未知数，像解方程一样使用逆运算。如果该变量出现在两个项中，先将其因式分解出来。
For example, make x the subject of y = mx + c.
例如，将 x 作为 y = mx + c 的主项。
y = mx + c ⇒ x = (y − c) / m
This skill is very common in science and higher mathematics, where a formula may need to be rearranged before substituting values.
这项技能在科学和高等数学中非常常见，通常需要先对公式进行变形，再代入数值。
9. Simultaneous Equations Basics | 联立方程基础
Simultaneous equations are two equations with two unknowns. You can solve them by elimination or substitution. The solution is the pair of values that satisfies both equations. Graphically, it is the point where the two lines intersect.
联立方程是含有两个未知数的两个方程。可以通过消元法或代入法求解。解是同时满足两个方程的一对值。从图像上看，它是两条直线相交的点。
For example, solve the system x + y = 10 and x − y = 4.
例如，解方程组 x + y = 10 和 x − y = 4。
x + y = 10, x − y = 4 ⇒ x = 7, y = 3
In higher-tier exams, simultaneous equations may be linked to real-life contexts such as cost, age, or geometry. Always check both original equations.
在高层级考试中，联立方程可能结合实际情境，如费用、年龄或几何问题。一定要代回两个原方程进行检验。
10. Using Graphs to Solve Equations | 利用图像解方程
The solution of f(x) = g(x) is the x-coordinate of the point where the graphs of y = f(x) and y = g(x) intersect. For linear equations, this can be read directly from the graph. This method is especially useful for checking algebraic solutions.
方程 f(x) = g(x) 的解是函数 y = f(x) 与 y = g(x) 图像交点的 x 坐标。对于一次方程，可以直接从图像上读出解。这种方法对于检验代数解特别有用。
When drawing linear graphs, use a table of values and plot at least three points to ensure the line is accurate. The intersection point must be clearly marked.
画一次函数图像时，使用数值表并至少描出三个点，以确保直线准确。交点必须清晰标出。
11. Common Errors and Exam Tips | 常见错误与考试技巧
Many marks are lost through slips with signs, brackets, and inequality reversal. Always show your working, write one operation per line, and double-check your final answer by substitution. In higher-tier papers, questions often combine algebra with geometry or ratio, so practise mixed problems.
许多失分是由于正负号、括号和不等式变号方面的疏忽。一定要写出步骤，每行只写一个运算，并通过代入检验最终答案。在高层级试卷中，题目常将代数与几何或比例结合，因此要练习混合题型。
| Common error | Correct approach |
| Forgetting to reverse the inequality sign when dividing by a negative number. | Always reverse the sign for multiply or divide by a negative value. |
| Expanding brackets incorrectly. | Use the distributive law carefully and check each term. |
| Not multiplying every term by the LCD in fraction equations. | Multiply both sides and every term, including whole numbers. |
Careful presentation is one of the easiest ways to secure marks in algebra questions.
在代数题中，清晰的表达是获得分数最简单的方法之一。
12. Practice Check | 练习自测
Try these quick questions to test yourself. Answers are given after the list.
试试以下快速自测题。答案列在后面。
- Solve 5x − 4 = 21.
- Solve 2(x − 3) + 4 = 16.
- Solve x/2 + 3 = x/3 + 5.
- Solve −3x + 7 > 1.
- Make r the subject of C = 2πr.
Answers: x = 5; x = 9; x = 12; x < 2; r = C / (2π).
答案：x = 5；x = 9；x = 12；x < 2；r = C / (2π)。
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