Grade 9 Unit 4: Equations and Inequalities | 九年级第四单元：方程与不等式

📚 Grade 9 Unit 4: Equations and Inequalities | 九年级第四单元：方程与不等式

In Grade 9 Unit 4, students consolidate their understanding of algebraic equations and inequalities. The unit focuses on solving linear equations with brackets, unknown terms on both sides, and fractional coefficients. It also introduces inequality notation, compound inequalities, and rearrangement of formulae. These skills are fundamental for IGCSE and A Level mathematics, so a secure grasp of every method is essential.

在九年级第四单元，学生将巩固代数方程与不等式的理解。本单元重点包括解含括号的线性方程、未知数在两边、分数系数方程，并引入不等式符号、复合不等式及公式变形。这些技能是 IGCSE 和 A Level 数学的基础，因此牢固掌握每一种方法至关重要。

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1. What Is an Equation? | 什么是方程？

An equation states that two algebraic expressions are equal. For example, 3x + 5 = 20 means ‘three times x, plus five, equals twenty’. A solution is a value of the variable that makes the statement true. Here x = 5 because 3(5) + 5 = 20.

方程表示两个代数表达式相等。例如 3x + 5 = 20 表示 ‘3 乘以 x，再加 5，等于 20’。解就是使等式成立的变量值。这里 x = 5，因为 3(5) + 5 = 20。

Every equation must stay balanced. Whatever operation you perform on one side, such as adding 4 or dividing by 2, you must also perform on the other side. This balance idea is the foundation for all equation solving in Unit 4.

每个方程都必须保持平衡。无论你对等号一边进行什么运算，例如加 4 或除以 2，另一边也必须进行相同的运算。这种平衡思想是第四单元所有方程求解的基础。

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2. Solving Linear Equations with Brackets | 解含括号的线性方程

Always expand brackets first using the distributive law: a(b + c) = ab + ac. Solve 2(x + 3) = 14. Expand: 2x + 6 = 14. Subtract 6: 2x = 8. Divide by 2: x = 4. Check: 2(4 + 3) = 14.

解题时先用分配律展开括号：a(b + c) = ab + ac。解方程 2(x + 3) = 14。展开得 2x + 6 = 14。两边减 6 得 2x = 8。两边除以 2 得 x = 4。检验：2(4 + 3) = 14。

Be careful with negative signs: solve 10 − 3(x − 2) = 4. Expand −3(x − 2) as −3x + 6, so the equation becomes 10 − 3x + 6 = 4, then 16 − 3x = 4, so −3x = −12 and x = 4.

处理负号时要小心：解方程 10 − 3(x − 2) = 4。将 −3(x − 2) 展开为 −3x + 6，于是方程变为 10 − 3x + 6 = 4，即 16 − 3x = 4，所以 −3x = −12，x = 4。

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3. Unknowns on Both Sides | 未知数在两边

When variables appear on both sides, collect like terms by adding or subtracting the smaller variable term. Solve 5x − 3 = 2x + 9. Subtract 2x: 3x − 3 = 9. Add 3: 3x = 12. Divide by 3: x = 4. Always keep the variable coefficient positive if possible.

当未知数出现在等号两边时，通过加减较小未知数项来合并同类项。解方程 5x − 3 = 2x + 9。两边减 2x 得 3x − 3 = 9。加 3 得 3x = 12。除以 3 得 x = 4。尽量保持未知数系数为正。

If the variable term on the left is smaller, move it to the right. For 6 − x = 2x + 3, add x to both sides: 6 = 3x + 3. Subtract 3: 3 = 3x. Divide by 3: x = 1. This avoids negative coefficients.

如果左边的未知数项较小，就把它移到右边。例如 6 − x = 2x + 3，两边加 x 得 6 = 3x + 3。减 3 得 3 = 3x。除以 3 得 x = 1。这样可以避免负系数。

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4. Equations with Fractional Coefficients | 分数系数方程

Multiply every term by the lowest common denominator to clear fractions. Solve x/2 + x/3 = 5. The LCD of 2 and 3 is 6, so multiply: 3x + 2x = 30. Therefore 5x = 30, and x = 6. This method avoids fraction arithmetic mistakes.

将每一项乘以最小公分母来去分母。解方程 x/2 + x/3 = 5。2 和 3 的最小公分母是 6，所以乘以 6 得 3x + 2x = 30。因此 5x = 30，x = 6。这种方法可以避免分数运算错误。

For a fraction that includes a bracket, treat it as one expression. Solve (x + 2)/3 = 4. Multiply both sides by 3: x + 2 = 12. Subtract 2: x = 10. Never multiply only part of the numerator.

对于带括号的分数，把它看作一个整体。解方程 (x + 2)/3 = 4。两边乘以 3 得 x + 2 = 12。减 2 得 x = 10。绝不要只乘以分子的一部分。

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5. Checking Your Solution | 检验解

Substitute your answer back into the original equation to verify it. For example, if you solved 4(x − 2) = 5x − 13, test x = 5: LHS = 4(5 − 2) = 12, RHS = 5(5) − 13 = 12. Since LHS = RHS, the solution is correct.

把你的答案代回原方程进行验证。例如，如果你解出 4(x − 2) = 5x − 13 的解是 x = 5，检验：左边 = 4(5 − 2) = 12，右边 = 5(5) − 13 = 12。左边等于右边，说明解正确。

In an exam, checking takes only a few seconds and catches sign errors, distribution errors, and arithmetic slips. Write LHS and RHS separately, then show that they are equal.

考试中，检验只需要几秒钟，就能发现符号错误、分配律错误和计算失误。分别写出左边和右边，然后证明它们相等。

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6. Forming Equations from Word Problems | 应用题列方程

Translate written statements into algebra. Define the unknown as x. Example: ‘The sum of three consecutive integers is 72.’ Let the integers be x, x + 1, x + 2. Then x + (x + 1) + (x + 2) = 72. Solve: 3x + 3 = 72, so 3x = 69 and x = 23. The integers are 23, 24, 25.

将文字表述转化为代数式。先设未知数为 x。例题：’三个连续整数的和是 72。’ 设这三个整数为 x、x + 1、x + 2。则 x + (x + 1) + (x + 2) = 72。解方程：3x + 3 = 72，所以 3x = 69，x = 23。三个整数为 23、24、25。

Look for key words: ‘sum’ means add, ‘difference’ means subtract, ‘product’ means multiply, and ‘more than’ often reverses the order. Always define your variable clearly before building the equation.

注意关键词：’和’ 表示相加，’差’ 表示相减，’积’ 表示相乘，’比…多’ 常常需要调整顺序。列方程前一定要清楚地定义变量。

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7. What Is an Inequality? | 什么是不等式？

An inequality compares two expressions using <, ≤, >, or ≥. For example, x > 3 means all real numbers greater than 3. Use an open circle for < or > on a number line, and a closed circle for ≤ or ≥.

不等式使用 <、≤、> 或 ≥ 来比较两个表达式。例如 x > 3 表示所有大于 3 的实数。在数轴上表示时，< 或 > 用空心圆圈，≤ 或 ≥ 用实心圆圈。

The table below summarises the symbols and their number-line representation.

下表总结了这些符号及其在数轴上的表示方法。

Symbol | 符号	Meaning | 含义	Circle | 圆圈
<	less than | 小于	open | 空心
≤	less than or equal to | 小于等于	closed | 实心
>	greater than | 大于	open | 空心
≥	greater than or equal to | 大于等于	closed | 实心

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8. Solving Linear Inequalities | 解一元一次不等式

Solve inequalities like equations, but reverse the inequality sign when multiplying or dividing by a negative number. Solve 4 − 3x ≥ 10. Subtract 4: −3x ≥ 6. Divide by −3 and reverse: x ≤ −2. The solution is all values less than or equal to −2.

解不等式与解方程类似，但当乘以或除以负数时，必须反转不等号方向。解不等式 4 − 3x ≥ 10。两边减 4 得 −3x ≥ 6。两边除以 −3 并反转不等号：x ≤ −2。解集为所有小于或等于 −2 的值。

Do not reverse the sign when adding, subtracting, multiplying by a positive number, or dividing by a positive number. The reversal only happens with a negative multiplier or divisor.

加、减、乘以正数或除以正数时，不要反转不等号。只有在乘以或除以负数时，才需要反转不等号方向。

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9. Compound Inequalities | 复合不等式

A compound inequality joins two inequalities with ‘and’. Solve 3 < 2x + 1 ≤ 7. Subtract 1 throughout: 2 < 2x ≤ 6. Divide by 2: 1 < x ≤ 3. This means x is greater than 1 and at most 3.

复合不等式用 ‘且’ 连接两个不等式。解 3 < 2x + 1 ≤ 7。整体减 1 得 2 < 2

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