📚 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
x + 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
ps 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
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。
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