GCSE AQA Maths Unit Test: Algebraic Skills | GCSE AQA 数学单元测试卷：代数技能

📚 GCSE AQA Maths Unit Test: Algebraic Skills | GCSE AQA 数学单元测试卷：代数技能

Unit tests are a valuable way to check your understanding of a specific topic area within the GCSE AQA Mathematics syllabus. This resource provides a focused practice test on algebraic skills, ranging from simplifying expressions to solving quadratic equations and simultaneous equations. Use these questions to assess your strengths and identify areas that need further revision before sitting your Paper 1, 2 or 3 examinations.

单元测试是检查你对 GCSSE AQA 数学大纲中特定主题理解程度的有效方式。本资源提供了一套针对代数技能的专项练习，涵盖从化简表达式到解二次方程和联立方程等内容。利用这些问题来评估你的强项，并找出在参加试卷一、二或三考试之前需要进一步复习的地方。

1. Simplifying Algebraic Products | 化简代数乘积

Simplify 3a × 4b. Multiply the coefficients: 3 × 4 = 12. Multiply the variables: a × b = ab. The simplified expression is 12ab.

化简 3a × 4b。系数相乘得：3 × 4 = 12。变量相乘得：a × b = ab。最终化简结果为 12ab。

2. Expanding Brackets | 展开括号

Expand and simplify 2(x + 3) + 3(x – 1). First, expand each bracket: 2 × x + 2 × 3 = 2x + 6, and 3 × x + 3 × (-1) = 3x – 3. Then collect like terms: 2x + 3x = 5x and 6 – 3 = 3. The result is 5x + 3.

展开并化简 2(x + 3) + 3(x – 1)。先展开每个括号：2×x + 2×3 = 2x + 6，以及 3×x + 3×(-1) = 3x – 3。然后合并同类项：2x + 3x = 5x，6 – 3 = 3。最终结果为 5x + 3。

3. Factorising a Difference of Squares | 因式分解平方差

Factorise x² – 9. Recognise this as a difference of two squares, where x² = (x)² and 9 = 3². Apply the formula a² – b² = (a + b)(a – b). Hence, x² – 9 = (x + 3)(x – 3).

对 x² – 9 进行因式分解。识别出这是平方差的形式，其中 x² = (x)²，9 = 3²。运用公式 a² – b² = (a + b)(a – b)。因此，x² – 9 = (x + 3)(x – 3)。

4. Solving a Linear Equation | 解一元一次方程

Solve 2x + 5 = 13. Subtract 5 from both sides: 2x = 8. Then divide both sides by 2: x = 4. Check: 2(4) + 5 = 8 + 5 = 13.

解方程 2x + 5 = 13。两边同时减去 5：2x = 8。然后两边除以 2：x = 4。检验：2(4) + 5 = 8 + 5 = 13。

5. Solving a Quadratic Equation by Factorising | 因式分解法解二次方程

Solve x² – 5x + 6 = 0. Factorise the quadratic: we need two numbers that multiply to +6 and add to -5, which are -2 and -3. So, (x – 2)(x – 3) = 0. Set each factor equal to zero: x – 2 = 0 gives x = 2; x – 3 = 0 gives x = 3.

解 x² – 5x + 6 = 0。对二次式进行因式分解：需要找两个数，乘积为 +6，和为 -5，即 -2 和 -3。因此 (x – 2)(x – 3) = 0。令每个因式为零：x – 2 = 0 得 x = 2；x – 3 = 0 得 x = 3。

6. Solving a Linear Inequality | 解一元一次不等式

Solve 3x – 4 ≤ 5. Add 4 to both sides: 3x ≤ 9. Divide both sides by 3: x ≤ 3. The solution set includes all numbers less than or equal to 3.

解不等式 3x – 4 ≤ 5。两边加 4：3x ≤ 9。两边除以 3：x ≤ 3。解集包含所有小于或等于 3 的数。

7. Changing the Subject of a Formula | 变换公式的主项

Make y the subject of x = (3y + 2) / 5. Multiply both sides by 5: 5x = 3y + 2. Subtract 2: 5x – 2 = 3y. Divide by 3: y = (5x – 2) / 3.

将方程 x = (3y + 2) / 5 的主项变为 y。两边乘以 5：5x = 3y + 2。减去 2：5x – 2 = 3y。除以 3：y = (5x – 2) / 3。

8. Finding the Gradient of a Straight Line | 求直线的梯度

Find the gradient of the line with equation 2y + 3x = 6. Rearrange into the form y = mx + c. Subtract 3x from both sides: 2y = -3x + 6. Divide by 2: y = (-3/2)x + 3. The gradient m is -3/2.

求直线 2y + 3x = 6 的梯度。将其整理为 y = mx + c 的形式。两边减去 3x：2y = -3x + 6。除以 2：y = (-3/2)x + 3。梯度 m 为 -3/2。

9. Using the Quadratic Formula | 使用求根公式

Solve 2x² – 4x – 1 = 0 using the quadratic formula, leaving answers in surd form. Here a = 2, b = -4, c = -1. The formula is x = [ -b ± √(b² – 4ac) ] / (2a). Substitute: x = [ -(-4) ± √((-4)² – 4×2×(-1)) ] / (2×2) = [ 4 ± √(16 + 8) ] / 4 = [ 4 ± √24 ] / 4. Simplify √24 = 2√6, so x = [ 4 ± 2√6 ] / 4 = [ 2 ± √6 ] / 2.

使用求根公式解 2x² – 4x – 1 = 0，答案保留根式形式。这里 a = 2，b = -4，c = -1。公式为 x = [ -b ± √(b² – 4ac) ] / (2a)。代入：x = [ -(-4) ± √((-4)² – 4×2×(-1)) ] / (2×2) = [ 4 ± √(16 + 8) ] / 4 = [ 4 ± √24 ] / 4。化简 √24 = 2√6，因此 x = [ 4 ± 2√6 ] / 4 = [ 2 ± √6 ] / 2。

10. Solving Simultaneous Equations | 解联立方程组

Solve the simultaneous equations: 2x + y = 7, x – y = -1. From the second equation, x = y – 1. Substitute into the first: 2(y – 1) + y = 7. Expand: 2y – 2 + y = 7 → 3y – 2 = 7. Add 2: 3y = 9 → y = 3. Then x = 3 – 1 = 2. The solution is x = 2, y = 3. Check in the first equation: 2(2) + 3 = 7.

解联立方程组：2x + y = 7，x – y = -1。由第二个方程得 x = y – 1。代入第一个方程：2(y – 1) + y = 7。展开得 2y – 2 + y = 7，即 3y – 2 = 7。两边加 2：3y = 9，得 y = 3。然后 x = 3 – 1 = 2。解为 x = 2，y = 3。代入第一个方程检验：2(2) + 3 = 7。

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GCSE AQA Maths Unit Test: Algebraic Skills | GCSE AQA 数学单元测试卷：代数技能