📚 Core Knowledge Review for Year 8 SQA Advanced Mathematics | Year 8 SQA 进阶数学：核心知识点梳理
This comprehensive guide summarises the essential topics in Year 8 Advanced Mathematics under the Scottish SQA curriculum. Mastering these fundamentals will build a strong foundation for National 5 and beyond.
本指南总结了苏格兰 SQA 课程下 Year 8 进阶数学的核心主题。掌握这些基础知识将为 National 5 及更高级别的学习奠定坚实基础。
1. Integers, Powers and Roots | 整数、幂与根
tive, add its positive value. For example, 5 + (-3) = 5 - 3 = 2, and 5 - (-3) = 5 + 3 = 8.
负数表示零以下的量。加上一个负数相当于减去它的绝对值；减去一个负数相当于加上它的绝对值。例如，5 + (-3) = 5 - 3 = 2，5 - (-3) = 5 + 3 = 8。
The order of operations (BIDMAS/BODMAS: Brackets, Indices, Division and Multiplication, Addition and Subtraction) must be followed to avoid mistakes. Evaluate exponents before multiplication, and perform left‑to‑right for division and multiplication.
计算时必须遵守运算顺序（括号、指数、乘除、加减），先算括号和指数，再算乘除（从左到右），最后算加减。这样可以避免错误。
Powers express repeated multiplication: 43 = 4 × 4 × 4 = 64. The square root √ reverses squaring: √81 = 9 because 92 = 81. Cube roots follow the same principle. Any non‑zero number raised to 0 equals 1.
幂表示重复相乘：43 = 4 × 4 × 4 = 64。平方根 √ 是平方的逆运算：√81 = 9，因为 92 = 81。立方根同理。任何非零数的 0 次幂都等于 1。
2. Fractions, Decimals and Percentages | 分数、小数与百分比
Fractions, decimals and percentages are interchangeable. The fraction ½ equals 0.5 and 50%. To convert a fraction to a decimal, divide the numerator by the denominator; to turn a decimal to a percentage, multiply by 100.
分数、小数和百分比可以互化。½ = 0.5 = 50%。分数化小数用分子除以分母；小数化百分数乘以 100。
ogether. To divide by a fraction, multiply by its reciprocal. Example: ¾ ÷ ½ = ¾ × 2 = 1½.
加减分数先通分，分母相同后再加减分子。分数相乘时分子乘分子、分母乘分母。除以一个分数等于乘以它的倒数，如 ¾ ÷ ½ = ¾ × 2 = 1½。
Finding a percentage of a quantity: multiply by the decimal form. To increase 80 by 15%, multiply 80 by 1.15. Percentage change = (change ÷ original) × 100%.
求一个数的百分比：用该数乘以对应的小数。80 增加 15% 就是 80 × 1.15。百分比变化 = (变化量 ÷ 原量) × 100%。
3. Algebraic Expressions and Simplification | 代数表达式与化简
Simplify by collecting like terms – part. For instance, 5a + 3b - 2a + b = 3a + 4b. Constants are like terms with each other.
通过合并同类项化简——变量部分完全相同的项。例如 5a + 3b - 2a + b = 3a + 4b。常数之间也是同类项。
4) = 3x + 12, and -2(y - 5) = -2y + 10. Double brackets such as (x + 2)(x + 3) expand to x2 + 5x + 6.
展开括号时，用外面的项乘以里面的每一项：3(x + 4) = 3x + 12；-2(y - 5) = -2y + 10。双重括号如 (x + 2)(x + 3) 展开得 x2 + 5x + 6。
e: take out the highest common factor. 6y + 9 = 3(2y + 3). Always check by expanding back.
因式分解是展开的逆运算：提取最大公因子。6y + 9 = 3(2y + 3)。展开检验即可确认正确。
4. Solving Linear Equations and Inequalities | 解一元一次方程与不等式
t 3x = 9, then divide by 3 to find x = 3. Always perform the same operation on both sides.
利用逆运算使变量单独出现。解 3x + 2 = 11 时，两边同减 2 得 3x = 9，再除以 3 得 x = 3。每一步都要在等式两边做相同运算。
imilarly, but if you multiply or divide by a negative number, reverse the inequality sign. E.g., -2x < 6 becomes x > -3.
解不等式（<, >, ≤, ≥）方法类似，但若乘或除以负数，必须调转不等号。例如 -2x < 6 变为 x > -3。
can be shown on a number line: open circles for strict inequalities, closed circles for ≤
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