KS3 Maths: Essential Maths Book 8i Answers – Key Concepts Explained | KS3 数学：Essential Maths Book 8i 答案知识点精讲

📚 KS3 Maths: Essential Maths Book 8i Answers – Key Concepts Explained | KS3 数学：Essential Maths Book 8i 答案知识点精讲

Essential Maths Book 8i is designed for Year 8 students following the KS3 curriculum. This guide focuses on the key concepts behind the exercises and answers, providing step-by-step explanations to help you master the topics. You will learn how to approach problems in number, algebra, geometry, and data handling with confidence.

《Essential Maths Book 8i》专为 KS3 八年级学生设计。本指南聚焦于练习和答案背后的核心概念，提供逐步讲解，帮助你掌握各主题。你将学会自信地处理数字、代数、几何和数据处理问题。

1. Operations with Integers and Decimals | 整数与小数的四则运算

When adding or subtracting decimals, align the decimal points vertically. For multiplication, ignore decimal points initially and multiply as whole numbers; then place the decimal point in the product by counting the total number of decimal places in the factors.

进行小数加减时，需将小数点垂直对齐。乘法运算时，先忽略小数点按整数相乘；然后根据因数中小数位数的总和，在积中点上小数点。

Example: Calculate 4.25 + 3.7. Align: 4.25 and 3.70, sum = 7.95. For 2.4 × 0.3, do 24 × 3 = 72; factors have total 2 decimal places, so answer is 0.72.

示例：计算 4.25 + 3.7。对齐：4.25 和 3.70，和为 7.95。对于 2.4 × 0.3，先算 24 × 3 = 72；因数共有两位小数，因此答案为 0.72。

Division of decimals: If the divisor has a decimal, multiply both divisor and dividend by 10, 100, etc., until the divisor is a whole number. For example, 3.6 ÷ 0.4 becomes 36 ÷ 4 = 9.

小数除法：如果除数有小数，将除数和被除数同时乘以 10、100 等，直到除数变为整数。例如，3.6 ÷ 0.4 变为 36 ÷ 4 = 9。

Remember the order of operations (BIDMAS): Brackets, Indices, Division/Multiplication (left to right), Addition/Subtraction (left to right). So 3 + 5 × 2 = 3 + 10 = 13, not 16.

牢记运算顺序（BIDMAS）：先算括号，再指数，然后乘除（从左到右），最后加减（从左到右）。因此 3 + 5 × 2 = 3 + 10 = 13，而不是 16。

---

2. Fractions, Decimals and Percentages | 分数、小数与百分比

To find a fraction of a quantity, divide by the denominator and multiply by the numerator. For instance, 2/5 of 60 = (60 ÷ 5) × 2 = 24.

求一个数量的几分之几，用分母除，再乘分子。例如，60 的 2/5 = (60 ÷ 5) × 2 = 24。

Converting between fractions, decimals and percentages: a fraction like 3/4 equals 0.75 as a decimal and 75% as a percentage. To convert a percentage to a decimal, divide by 100; to express a decimal as a percentage, multiply by 100.

分数、小数和百分比之间的转换：分数如 3/4 等于小数 0.75，百分比 75%。将百分比转换为小数，除以 100；将小数表示为百分比，乘以 100。

When comparing or ordering mixed forms, it is often easiest to convert all numbers into the same representation (e.g., all decimals or all percentages). For example, to order 0.2, 1/4, and 30%, convert: 0.2, 0.25, 0.3 → order: 0.2, 0.25, 0.3.

当比较或排序混合形式的数时，通常最容易将所有数转换为同一种表示形式（例如都转为小数或都转为百分比）。例如，要对 0.2、1/4 和 30% 排序，转换：0.2、0.25、0.3 → 排序：0.2、0.25、0.3。

---

3. Simplifying Algebraic Expressions | 代数表达式的化简

Like terms can be combined by adding or subtracting their coefficients. For example, 5a + 3b – 2a + 7b = (5-2)a + (3+7)b = 3a + 10b.

同类项可以通过加减它们的系数来合并。例如，5a + 3b – 2a + 7b = (5-2)a + (3+7)b = 3a + 10b。

Remember that a term like a means 1a. When simplifying expressions with powers, such as a × a = a², only combine terms with identical variable parts. a and a² are not like terms.

记住像 a 这样的项表示 1a。当化简含幂的表达式时，如 a × a = a²，只有变量部分完全相同的项才能合并。a 和 a² 不是同类项。

Use the distributive law to expand brackets: 3(2x + 4) = 6x + 12. Then combine any like terms after expansion. For subtraction: 5 − 2(y + 3) = 5 − 2y − 6 = −1 − 2y.

运用分配律展开括号：3(2x + 4) = 6x + 12。展开后合并所有同类项。对于减号：5 − 2(y + 3) = 5 − 2y − 6 = −1 − 2y。

---

4. Solving Linear Equations | 解一元一次方程

To solve an equation, perform the same operation on both sides to isolate the variable. For x + 5 = 12, subtract 5 from both sides: x = 7.

解方程时，在等号两边进行相同的运算，以隔离变量。对于 x + 5 = 12，两边同时减 5：x = 7。

For two-step equations like 2x + 3 = 11, first subtract 3: 2x = 8, then divide by 2: x = 4. Always check your solution by substituting back into the original equation.

对于 2x + 3 = 11 这类两步方程，先减 3：2x = 8，然后除以 2：x = 4。务必通过代回原方程来检验解。

Equations with brackets: expand first, then simplify. Example: 3(x − 2) = 9 → 3x − 6 = 9 → 3x = 15 → x = 5.

带括号的方程：先展开，再化简。示例：3(x − 2) = 9 → 3x − 6 = 9 → 3x = 15 → x = 5。

If the variable appears on both sides, collect like terms onto one side. Example: 5x + 2 = 3x + 10 → 2x = 8 → x = 4.

如果方程两边都有未知数，将含未知数的项移到一边。示例：5x + 2 = 3x + 10 → 2x = 8 → x = 4。

---

5. Sequences and the nth Term | 数列与第 n 项

An arithmetic sequence has a constant difference between terms. For the sequence 3, 7, 11, 15, …, the common difference is 4. The nth term rule can be written as 4n − 1 (since 4×1 −1 = 3).

等差数列的相邻两项之差是一个常数。对于数列 3, 7, 11, 15, …，公差为 4。第 n 项的通项公式可写为 4n − 1（因为 4×1 −1 = 3）。

To find the nth term: the coefficient of n is the common difference; then work out the value needed to reach the first term. For a decreasing sequence like 10, 7, 4, 1, …, the difference is −3, so nth term = −3n + 13.

求第 n 项：n 的系数就是公差；然后求出使首项成立的常数值。对于递减数列 10, 7, 4, 1, …，公差为 −3，因此通项公式为 −3n + 13。

Once you have the nth term, you can find any term, such as the 20th term. For 4n − 1, the 20th term is 4×20 − 1 = 79.

得到通项公式后，你可以求出任意一项，例如第 20 项。对于 4n − 1，第 20 项就是 4×20 − 1 = 79。Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）