Year 9 SQA Advanced Mathematics: Unit Test Mock Paper Walkthrough | 九年级SQA进阶数学：单元测试模拟卷解析

📚 Year 9 SQA Advanced Mathematics: Unit Test Mock Paper Walkthrough | 九年级SQA进阶数学：单元测试模拟卷解析

This walkthrough dissects a mock unit test tailored for Year 9 SQA Advanced Mathematics. It provides step-by-step solutions to essential problems, combining clear English explanations with their Chinese translations. Topics span linear equations, quadratic factorisation, simultaneous equations, inequalities, Pythagoras' theorem, basic trigonometry, and straight-line graphs. Use it to identify gaps and build exam confidence.

本解析针对九年级 SQA 进阶数学的单元模拟测试卷。它以中英双语逐步拆解核心题目，涵盖线性方程、二次因式分解、联立方程、不等式、勾股定理、基础三角学以及直线图像。通过逐题解析，帮助你查漏补缺，提升应考信心。

1. Solving Linear Equations | 解线性方程

Question: Solve 3x - 7 = 2x + 5 for x.

题目：解出方程 3x - 7 = 2x + 5 中的 x。

Step 1: Write down the equation clearly.

步骤1：清晰写下方程。

3x - 7 = 2x + 5

Step 2: Subtract 2x from both sides to begin isolating the variable. This eliminates x on the right-hand side.

步骤2：两边同时减去 2x，开始分离变量。这将消去右边的 x 项。

3x - 2x - 7 = 2x - 2x + 5 becomes x - 7 = 5.

得到 x - 7 = 5。

Step 3: Add 7 to both sides to solve for x.

步骤3：两边同时加 7，求出 x。

x - 7 + 7 = 5 + 7 gives x = 12.

得到 x = 12。

de: 3(12) - 7 = 29. Right side: 2(12) + 5 = 29. Both sides match, confirming the solution.

步骤4：将 x = 12 代入原方程进行验证。左边：3(12) - 7 = 29。右边：2(12) + 5 = 29。两边相等，确认解正确。

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2. Factorising Quadratic Expressions | 因式分解二次式

Question: Factorise x² + 5x + 6 completely.

题目：将 x² + 5x + 6 彻底因式分解。

1: Identify the coefficients in the standard form ax² + bx + c. Here a = 1, b = 5, c = 6.

步骤1：确定标准形式 ax² + bx + c 中的系数。这里 a = 1，b = 5，c = 6。

ultiply to c (6) and add to b (5). The pair 2 and 3 works because 2 x 3 = 6 and 2 + 3 = 5.

步骤2：找出两个数，它们的乘积为 c（6），和为 b（5）。数字 2 和 3 满足条件，因为 2 × 3 = 6，2 + 3 = 5。

Step 3: Write the quadratic as (x + 2)(x + 3). This is the factorised form.

步骤3：将二次式写成 (x + 2)(x + 3)。这就是因式分解的形式。

anding (x+2)(x+3) = x² + 3x + 2x + 6 = x² + 5x + 6, which matches the original expression.

步骤4：通过展开来检验：(x+2)(x+3) = x² + 3x + 2x + 6 = x² + 5x + 6，与原式一致。

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3. Solving Quadratic Equations by Factorising | 用因式分解法解二次方程

Question: Solve x² - 4x - 12 = 0.

题目：解方程 x² - 4x - 12 = 0。

oduct -12 and sum -4. The numbers -6 and 2 satisfy this: (-6) x 2 = -12 and (-6) + 2 = -4.

步骤1：需要对左边进行因式分解。寻找乘积为 -12、和为 -4 的两个数。-6 和 2 符合要求：(-6) × 2 = -12，(-6) + 2 = -4。

Step 2: Write the factorised equation as (x - 6)(x + 2) = 0.

步骤2：将方程写成因式形式 (x - 6)(x + 2) = 0。

: Apply the zero product property. Set each bracket equal to zero: x - 6 = 0 or x + 2 = 0.

步骤3：运用零乘积性质，令每个括号等于零：x - 6 = 0 或 x + 2 = 0。

Step 4: Solve each simple equation. x - 6 = 0 gives x = 6; x + 2 = 0 gives x = -2.

步骤4：解这两个简单方程。x - 6 = 0 得 x = 6；x + 2 = 0 得 x = -2。

Therefore the solutions are x = 6 and x = -2.

因此解为 x = 6 和 x = -2。

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4. Solving Simultaneous Equations | 解联立方程组

Question: Solve the system: 2x + y = 10, x - y = 2.

题目：解方程组：2x + y = 10，x - y = 2。

Step 1: Write both equations one above the other.

步骤1：将两个方程上下对齐写出。

(1) 2x + y = 10
(2) x - y = 2

Adding the equations will eliminate y. (2x + y) + (x - y) = 10 + 2 simplifies to 3x = 12.

步骤2：注意到 y 项符号相反，两式相加可以消去 y。(2x + y) + (x - y) = 10 + 2，化简得 3x = 12。

Step 3: Divide by 3 to find x. x = 4.

步骤3：除以 3，得到 x = 4。

e original equations, say equation (2): 4 - y = 2. Solve for y: -y = 2 - 4 = -2, so y = 2.

步骤4：将 x = 4 代入其中一个原方程，例如方程 (2)：4 - y = 2。解出 y：-y = 2 - 4 = -2，因此 y = 2。

Step 5: Check with equation (1): 2(4) + 2 = 8 + 2 = 10. Correct.

步骤5：用方程 (1) 检验：2(4) + 2 = 8 + 2 = 10，正确。

The solution is x = 4, y = 2.

解为 x = 4，y = 2。

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5. Inequalities and the Number Line | 不等式与数轴表示

Question: Solve the inequality 3x - 4 > 5x + 2 and represent the solution on a number line.

题目：解不等式 3x - 4 > 5x + 2，并在数轴上表示解集。

Step 1: Gather x terms on one side. Subtract 5x from both sides: 3x - 5x - 4 > 2, giving -2x - 4 > 2.

步骤1：将含 x 的项移到一边。两边减去 5x：3x - 5x - 4 > 2，得到 -2x - 4 > 2。

Step 2: Add 4 to both sides to isolate the x term: -2x > 6.

步骤2：两边加 4 以分离含 x 的项：-2x > 6。

Step 3: Divide by -2. Remember to reverse the inequality sign when dividing by a negative number. This yields x < -3.

步骤3：两边除以 -2。注意除以负数时要反转不等号方向，得到 x < -3。

Step 4: Represent on a number line: draw a horizontal line, mark -3 with an open circle (since -3 is not included), and shade the region to the left of -3, showing all numbers less than -3.

步骤4：在数轴上表示：画一条水平线，在 -3 处画空心圆（因为 -3 不包含在内），并向左画阴影线，表示所有小于 -3 的数。

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6. Pythagoras' Theorem | 勾股定理

Question: A right-angled triangle has legs of 5 cm and 12 cm. Find the length of the hypotenuse to 1 decimal place if necessary.

题目：一个直角三角形的两直角边长分别为 5 厘米和 12 厘米。求斜边的长度（如有需要保留一位小数）。

Step 1: Recall Pythagoras' theorem: c² = a² + b², where c is the hypotenuse and a, b are the legs.

步骤1：回顾勾股定理：c² = a² + b²，其中 c 为斜边，a 和 b 为直角边。

Step 2: Substitute the given values a = 5, b = 12: c² = 5² + 12² = 25 + 144 = 169.

步骤2：代入已知值 a = 5，b = 12：c² = 5² + 12² = 25 + 144 = 169。

Step 3:

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