Algebra 9: Equations, Inequalities and Functions | 代数9：方程、不等式与函数

📚 Algebra 9: Equations, Inequalities and Functions | 代数9：方程、不等式与函数

Algebra 9 is a core revision topic for Edexcel IGCSE Mathematics. It brings together manipulation, equation solving, graphing and problem-solving skills that are examined across both Paper 1 and Paper 2.

代数9是Edexcel IGCSE数学的核心复习专题。它将代数变形、方程求解、图像绘制和问题解决技能整合在一起，这些内容在Paper 1和Paper 2中都会考到。

1. Expanding and Simplifying | 展开与化简

Expanding means removing brackets by multiplying every term inside the bracket by the term outside. Simplify by collecting like terms afterwards.

展开是指将括号外的每一项乘以括号内的每一项，从而去掉括号。之后合并同类项进行化简。

Example: 3(2x + 5) = 6x + 15.

例如：3(2x + 5) = 6x + 15。

For double brackets, use FOIL or the grid method:

双括号展开可使用FOIL法或网格法：

(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15

Multiply each term from the first bracket by each term from the second.
将第一个括号中的每一项与第二个括号中的每一项相乘。
Always collect like terms after expanding.
展开后务必合并同类项。

2. Factorising | 因式分解

Factorising is the reverse of expanding. Take out the highest common factor first, then look for double brackets.

因式分解是展开的逆运算。先提取最大公因式，再看是否能写成双括号形式。

Example of taking out a common factor:

提取公因式的例子：

6x² + 12x = 6x(x + 2)

For a quadratic expression x² + bx + c, find two numbers that multiply to c and add to b.

对于二次表达式x² + bx + c，找到两个数，它们的乘积是c，和是b。

x² + 7x + 12 = (x + 3)(x + 4)

Check your factorisation by expanding the brackets again.

通过再次展开括号来检查因式分解是否正确。

3. Solving Linear Equations | 解一次方程

A linear equation has the unknown raised to the power 1. The key is to keep the equation balanced while performing the same operation on both sides.

一次方程中未知数的指数为1。关键在于对等号两边进行相同运算，以保持方程平衡。

Steps for solving:

解题步骤：

Remove fractions by multiplying both sides by the denominator.
通过两边乘以分母来去掉分数。
Expand any brackets.
展开所有括号。
Collect variable terms on one side and constants on the other.
将含未知数的项移到一边，常数项移到另一边。
Divide both sides by the coefficient of the variable.
两边同时除以未知数的系数。

2x + 5 = 17 → 2x = 12 → x = 6

Always substitute your answer back into the original equation to verify.

一定要将答案代回原方程检验。

4. Simultaneous Equations | 联立方程

Simultaneous equations are solved by finding values of unknowns that satisfy all equations at once. For two linear equations, use elimination or substitution.

联立方程是寻找同时满足所有方程的未知数值。对于两个线性方程，可使用消元法或代入法。

Elimination method example:

消元法示例：

2x + y = 9 and x − y = 3

Add the equations to eliminate y:

将两个方程相加以消去y：

3x = 12 → x = 4

Then substitute x = 4 into the first equation:

然后将x = 4代入第一个方程：

8 + y = 9 → y = 1

A quadratic and a linear equation can be solved using substitution, then solving the resulting quadratic.

二次方程与一次方程联立时，通常用代入法，然后解所得二次方程。

5. Quadratic Equations | 二次方程

A quadratic equation has the form ax² + bx + c = 0. Solve it by factorising, completing the square, or using the quadratic formula.

二次方程具有ax² + bx + c = 0的形式。可通过因式分解、配方法或二次公式求解。

If the quadratic can be factorised, set each bracket equal to zero:

如果二次式可以因式分解，则令每个括号等于零：

x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3

The quadratic formula also works for any quadratic:

二次公式适用于任何二次方程：

x = (−b ± √(b² − 4ac)) ÷ 2a

The discriminant b² − 4ac tells you the number of solutions:

判别式b² − 4ac告诉你解的数量：

If b² − 4ac > 0, there are two distinct roots.
如果b² − 4ac > 0，有两个不同的根。
If b² − 4ac = 0, there is one repeated root.
如果b² − 4ac = 0，有一个重根。
If b² − 4ac < 0, there are no real roots.
如果b² − 4ac < 0，没有实数根。

6. Inequalities | 不等式

Inequalities are solved using the same rules as equations, except that multiplying or dividing by a negative number reverses the inequality sign.

解不等式的方法与方程类似，但乘以或除以负数时，不等号方向会改变。

Solve the linear inequality:

解一次不等式：

3x − 7 ≤ 8 → 3x ≤ 15 → x ≤ 5

You must also be able to show solutions on a number line:

你还必须能在数轴上表示解集：

An open circle means the value is not included; a closed circle means it is included.

空心圆表示该值不包含在解集中；实心圆表示包含。

7. Functions | 函数

A function maps each input value to exactly one output value. It is often written as f(x) = expression.

函数将每个输入值映射到唯一的输出值。通常写作f(x) = 表达式。

Example function:

函数示例：

f(x) = 3x + 2

To find f(4), substitute x = 4:

求f(4)，令x = 4代入：

f(4) = 3 × 4 + 2 = 14

Composite functions combine two functions, such as f(g(x)). Inverse functions reverse the action of a function.

复合函数是两个函数的组合，如f(g(x))。反函数则是将函数的操作反转。

For f(x) = 2x + 3, the inverse is:

对于f(x) = 2x + 3，其反函数为：

f⁻¹(x) = (x − 3) ÷ 2

8. Graphs of Algebraic Functions | 代数函数图像

Linear graphs are straight lines with equation y = mx + c, where m is the gradient and c is the y-intercept.

线性函数图像是一条直线，方程为y = mx + c，其中m是斜率，c是y轴截距。

Quadratic graphs form a parabola. The roots appear where the graph crosses the x-axis, and the turning point can be found by completing the square.

二次函数图像呈抛物线形。根出现在图像与x轴相交处，通过配方法可找到顶点坐标。

For y = (x − 3)² + 1, the turning point is (3, 1).

对于y = (x − 3)² + 1，顶点坐标为(3, 1)。

Solving equations graphically means finding the x-coordinates of intersection points of the graphs.

用图像法解方程，就是找到图像交点的x坐标。

9. Sequences | 数列

An arithmetic sequence has a common difference between consecutive terms. Its nth term is found using:

等差数列的相邻项之差为常数。其第n项通过以下公式求：

aₙ = a₁ + (n − 1)d

where a₁ is the first term and d is the common difference.

其中a₁是首项，d是公差。

A quadratic sequence has a constant second difference. Its nth term is written as an² + bn + c.

二次数列具有恒定的二阶差。它的第n项写作an² + bn + c。

Example: find the nth term of 2, 5, 10, 17, …

例如：求2, 5, 10, 17, … 的第n项。

First differences: 3, 5, 7. Second difference: 2. So a = 1.

一阶差：3, 5, 7。二阶差：2。所以a = 1。

nth term = n² + 1

10. Algebraic Proof | 代数证明

Algebraic proof uses symbols to represent numbers and prove that a statement is always true.

代数证明使用符号代表数，以证明某个陈述总是成立。

Key facts to remember:

需要牢记的关键事实：

Any even number can be written as 2n.
任何偶数可以写作2n。
Any odd number can be written as 2n + 1.
任何奇数可以写作2n + 1。
Consecutive integers are n, n + 1, n + 2, …
连续整数是n, n + 1, n + 2, …

Example: prove that the sum of two odd numbers is even.

例如：证明两个奇数之和是偶数。

Let the odd numbers be 2m + 1 and 2n + 1. Their sum is:

设两个奇数分别为2m + 1和2n + 1。它们的和为：

2m + 1 + 2n + 1 = 2m + 2n + 2 = 2(m + n + 1)

This is a multiple of 2, so it is even.

这是2的倍数，所以结果是偶数。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Algebra 9: Equations, Inequalities and Functions | 代数9：方程、不等式与函数