📚 7S Answers Page 1: Core Maths Worked Solutions | 7S 答案第1页：核心数学分步解析
Welcome to this 7S Answers page, designed to provide clear, step-by-step solutions to core mathematics topics typically covered in international middle school and IGCSE foundation courses. This revision sheet focuses on worked examples, common misconceptions and efficient methods to help you check your understanding and improve accuracy.
欢迎使用本页 7S 答案与解析，旨在针对国际初中和 IGCSE 基础课程中的核心数学主题提供清晰、分步的解答。本复习页侧重于例题详解、常见误区与高效方法，帮助你检验理解并提高准确性。
1. Simplifying Algebraic Expressions | 化简代数表达式
Example: Simplify 3x + 5x - 2x.
例题：化简 3x + 5x - 2x。
ne like terms by adding or subtracting the coefficients: 3x + 5x - 2x = (3 + 5 - 2)x = 6x.
解答：合并同类项，将系数相加或相减：3x + 5x - 2x = (3 + 5 - 2)x = 6x。
Always check that the variables and their powers are identical before combining; x and x² are not like terms.
合并前务必确认变量及其指数完全相同；x 与 x² 不是同类项。
Answer: 6x.
答案：6x。
2. Solving Linear Equations | 解一元一次方程
Example: Solve 2(x + 3) = 14.
例题：解方程 2(x + 3) = 14。
+ 6 = 14. Then subtract 6 from both sides: 2x = 8. Finally divide both sides by 2: x = 4.
解答：先去括号：2x + 6 = 14。然后两边同时减去 6：2x = 8。最后两边同时除以 2：x = 4。
u can verify by substituting x = 4 back into the original equation: 2(4 + 3) = 2 × 7 = 14.
可将 x = 4 代回原方程检验：2(4 + 3) = 2 × 7 = 14。
Answer: x = 4.
答案：x = 4。
3. Working with Fractions | 分数运算
Example: Evaluate 3/4 + 2/5.
例题：计算 3/4 + 2/5。
on denominator of 4 and 5, which is 20. Convert each fraction: 3/4 = 15/20 and 2/5 = 8/20.
解答：求出 4 和 5 的最小公分母为 20。将各分数转换：3/4 = 15/20，2/5 = 8/20。
Add the numerators: 15/20 + 8/20 = 23/20 = 1 3/20.
分子相加：15/20 + 8/20 = 23/20 = 1 3/20。
Answer: 23/20 or 1 3/20.
答案：23/20 或 1 3/20。
4. Ratio and Proportion | 比例与比率
Example: Divide £120 in the ratio 2 : 3.
例题：将 120 英镑按 2 : 3 的比例分配。
Solution: Add the parts of the ratio: 2 + 3 = 5 parts. One part is £120 ÷ 5 = £24.
解答：将比例的份数相加：2 + 3 = 5 份。一份为 120 ÷ 5 = 24 英镑。
The first share is 2 × £24 = £48; the second share is 3 × £24 = £72.
第一份为 2 × 24 = 48 英镑；第二份为 3 × 24 = 72 英镑。
Check: £48 + £72 = £120.
检验：48 + 72 = 120 英镑。
Answer: £48 and £72.
答案：48 英镑和 72 英镑。
5. Percentage Increase and Decrease | 百分比增减
Example: A jacket costing $80 is reduced by 15%. Find the sale price.
例题：一件夹克原价 80 美元，降价 15%。求售价。
Solution: Calculate 15% of $80: 0.15 × 80 = $12. Subtract the discount from the original price: $80 - $12 = $68.
解答：计算 80 美元的 15%：0.15 × 80 = 12 美元。从原价中减去折扣：80 - 12 = 68 美元。
Alternatively, multiply by 85% (100% - 15%) directly: 0.85 × $80 = $68.
也可直接乘以 85%（100% - 15%）：0.85 × 80 = 68 美元。
Answer: $68.
答案：68 美元。
6. Angle Properties in Triangles | 三角形角度性质
Example: In triangle ABC, angle A = 48° and angle B = 63°. Find angle C.
例题：在三角形 ABC 中，∠A = 48°，∠B = 63°。求 ∠C。
s 180°. Add the two known angles: 48° + 63° = 111°. Subtract from 180°: 180° - 111° = 69°.
解答：三角形内角和为 180°。将两个已知角相加：48° + 63° = 111°。从 180° 中减去：180° - 111° = 69°。
Answer: angle C = 69°.
答案：∠C = 69°。
7. Area of Composite Shapes | 组合图形面积
Example: Find the area of a shape made from a rectangle of length 8 cm and width 5 cm with a semicircle of radius 2 cm removed.
例题：一个组合图形由一个长 8 cm、宽 5 cm 的长方形挖去一个半径 2 cm 的半圆组成，求其面积。
Solution: Area of rectangle = length × width = 8 × 5 = 40 cm².
解答：长方形面积 = 长 × 宽 = 8 × 5 = 40 cm²。
Area of semicircle = (1/2) × π × r² = (1/2) × π × 2² = 2π cm².
半圆面积 = (1/2) × π × r² = (1/2) × π × 2² = 2π cm²。
micircle area: 40 - 2π cm². Using π ≈ 3.14, the approximate area is 40 - 6.28 = 33.72 cm².
减去半圆面积：40 - 2π cm²。取 π ≈ 3.14，面积约为 40 - 6.28 = 33.72 cm²。
Answer: (40 - 2π) cm² or approximately 33.7 cm².
答案：(40 - 2π) cm² 或约 33.7 cm²。
8. Coordinates and Straight Lines | 坐标与直线
Example: Find the midpoint of the line segment joining A(2, 5) and B(8, -1).
例题：求连接 A(2, 5) 和 B(8, -1) 两点的线段中点。
y₁ + y₂)/2). Substitute the coordinates: ((2 + 8)/2, (5 + (-1))/2) = (10/2, 4/2) = (5, 2).
解答：中点公式为 ((x₁ + x₂)/2, (y₁ + y₂)/2)。代入坐标：((2 + 8)/2, (5 + (-1))/2) = (10/2, 4/2) = (5, 2)。
Answer: (5, 2).
答案：(5, 2)。
9. Probability Basics | 基础概率
Example: A bag contains 3 red, 2 blue and 5 green counters. Find the probability of picking a blue counter at random.
例题：一个袋子中有 3 个红色、2 个蓝色和 5 个绿色筹码。求随机抽出一个蓝色筹码的概率。
lution: Total number of counters = 3 + 2 + 5 = 10. Number of favourable outcomes = 2 blue.
解答：筹码总数 = 3 + 2 + 5 = 10。有利结果数 = 2 个蓝色。
Probability = favourable outcomes / total outcomes = 2/10 = 1/5.
概率 = 有利结果数 / 总结果数 = 2/10 = 1/5。
Answer: 1/5 or 0.2.
答案：1/5 或 0.2。
10. Averages and Range | 平均数与极差
Example: Find the mean, median, mode and range of the data set: 4, 7, 7, 9, 12.
例题：求数据集 4、7、7、9、12 的平均数、中位数、众数和极差。
Solution: Mean = (4 + 7 + 7 + 9 + 12) ÷ 5 = 39 ÷ 5 = 7.8.
解答：平均数 = (4 + 7 + 7 + 9 + 12) ÷ 5 = 39 ÷ 5 = 7.8。
Median is the middle value when ordered: 4, 7, 7, 9, 12 → median = 7.
中位数是排序后中间的值：4、7、7、9、12 → 中位数 = 7。
Mode is the most frequent value: 7 appears twice, so mode = 7.
众数是出现最频繁的值：7 出现两次，因此众数 = 7。
Range = highest - lowest = 12 - 4 = 8.
极差 = 最大值 - 最小值 = 12 - 4 = 8。
Answer: mean = 7.8, median = 7, mode = 7, range = 8.
答案：平均数 = 7.8，中位数 = 7，众数 = 7，极差 = 8。
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