📚 Solving Equations: From Rearrangement to Final Answer | 解方程：从移项到最终答案
ough the key techniques and common pitfalls so you can confidently reach the final answer.
在 Edexcel A-Level 数学中，要求解方程的题目几乎都需要清晰的步骤：移项、分离变量、因式分解或使用逆运算，最后写出 x = ... 。本文将梳理关键技巧与常见陷阱，帮助你自信地得出最终答案。
1. What Does Solving an Equation Mean? | 解方程的含义
should be written explicitly as x = ... or x = a, x = b. Simply writing the number without 'x =' can lose the answer mark in an exam.
解方程意味着找出使等式成立的所有未知数（通常是 x）的值。最终答案应明确写成 x = ... 或 x = a, x = b。如果只写出数字而没有 “x =”，在考试中可能会丢失答案分。
a balanced statement, so whatever operation you apply to one side must also be applied to the other side.
例如，解方程 2x + 3 = 7 得到 x = 2，因为 2(2) + 3 = 7。方程是一个平衡的等式，因此对一边进行的任何运算都必须同时对另一边进行。
In Edexcel papers, marks are often awarded for method as well as accuracy. Showing the rearrangement step by step, even if the final answer is wrong, can still earn method marks.
在 Edexcel 试卷中，分数通常同时给方法和答案。即使最终答案错误，逐步展示移项过程仍然可以获得方法分。
2. Linear Equations and Basic Rearrangement | 线性方程与基本移项
. To solve, perform inverse operations in reverse order: first subtract b, then divide by a. This is the most fundamental skill in algebra and appears in almost every A-Level topic.
线性方程形式为 ax + b = c。求解时按逆序使用逆运算：先减去 b，再除以 a。这是代数中最基本的技能，几乎出现在 A-Level 的每个主题中。
ax + b = c ⇒ x = (c - b) / a
Example: Solve 5x - 12 = 3.
例题：解方程 5x - 12 = 3。
- Add 12 to both sides: 5x = 15.
- 两边加 12：5x = 15。
- Divide both sides by 5: x = 3.
- 两边除以 5：x = 3。
3, not just 3. If the coefficient of x is negative, divide by the negative coefficient carefully and keep the sign correct.
最终答案要写成 x = 3，而不仅仅是 3。如果 x 的系数为负，除以负数系数时要小心，确保符号正确。
sides, collect like terms first. For example, 7x + 4 = 3x - 8 becomes 4x = -12, so x = -3.
当未知数出现在等号两边时，先合并同类项。例如，7x + 4 = 3x - 8 可化为 4x = -12，所以 x = -3。
3. Quadratic Equations: Factorising, Completing the Square, and the Formula | 二次方程：因式分解、配方法与公式法
mon: factorising, completing the square, and the quadratic formula. The best method depends on the coefficients and the required form of the answer.
二次方程通常形式为 ax² + bx + c = 0。常用三种方法：因式分解、配方法和二次公式。最佳方法取决于系数以及题目要求的答案形式。
be written as (px + q)(rx + s) = 0. Then set each bracket equal to zero to find the roots.
当二次式可写成 (px + q)(rx + s) = 0 时，可用因式分解。然后令每个括号等于零来求根。
Example: Solve x² - 5x + 6 = 0.
例题：解方程 x² - 5x + 6 = 0。
Factorise: (x - 2)(x - 3) = 0, so x = 2 or x = 3.
因式分解：(x - 2)(x - 3) = 0，所以 x = 2 或 x = 3。
If factorising is difficult, use the quadratic formula. This is given in the Edexcel formula booklet, but you must know how to use it correctly.
如果因式分解困难，可使用二次公式。Edexcel 公式手册中提供该公式，但你必须知道如何正确使用。
x = (-b ± √b² - 4ac) / 2a
Always compute the discriminant b² - 4ac first. If it is positive, there are two distinct real roots; if zero, one repeated root; if negative, no real roots.
务必先计算判别式 b² - 4ac。如果为正，则有两个不相等的实根；如果为零，则有一个重根；如果为负，则没有实根。
required. For example, x² + 6x + 2 = 0 can be written as (x + 3)² - 7 = 0, giving x = -3 ± √7.
当题目要求求顶点或需要精确答案时，配方法很有用。例如，x² + 6x + 2 = 0 可写成 (x + 3)² - 7 = 0，从而 x = -3 ± √7。
4. Simultaneous Equations: Substitution and Elimination | 联立方程：代入法与消元法
When two equations involve two unknowns, we solve them simultaneously. Substitution is usually best when one equation is already solved for y or x. Elimination is often quicker for two linear equations.
当两个方程包含两个未知数时，需要联立求解。当一个方程已经解出 y 或 x 时，通常用代入法；对于两个线性方程，消元法往往更快。
Example: Solve y = 2x + 1 and x + y = 10.
例题：解方程组 y = 2x + 1 和 x + y = 10。
Substitute y = 2x + 1 into x + y = 10:
将 y = 2x + 1 代入 x + y = 10：
x + (2x + 1) = 10 ⇒ 3x + 1 = 10 ⇒ x = 3
Then y = 2(3) + 1 = 7. The solution is x = 3, y = 7.
然后 y = 2(3) + 1 = 7。解为 x = 3, y = 7。
For elimination, multiply one or both equations so that the coefficients of one unknown match. Then add or subtract the equations to eliminate that unknown.
对于消元法，将一个或两个方程乘以适当的数，使某个未知数的系数相同。然后将方程相加或相减以消去该未知数。
One linear and one quadratic simultaneous equation may produce two pairs of solutions. After substituting the linear equation into the quadratic, solve the resulting quadratic and then find the corresponding y-values.
一个一次方程和一个二次方程联立可能产生两组解。将一次方程代入二次方程后，解所得的二次方程，再求出对应的 y 值。
5. Equations with Algebraic Fractions | 分式方程
For equations involving algebraic fractions, multiply through by the common denominator to clear fractions. Then solve the resulting equation. This technique is common in Edexcel Pure Mathematics.
对于含有代数分式的方程，两边乘以公分母以去掉分母，然后解所得方程。这种技巧在 Edexcel 纯数学中很常见。
Example: Solve (x + 1) / 3 = (x - 2) / 2.
例题：解方程 (x + 1) / 3 = (x - 2) / 2。
Multiply both sides by 6:
两边乘以 6：
2(x + 1) = 3(x - 2) ⇒ 2x + 2 = 3x - 6 ⇒ x = 8
For example, in 1/(x - 2) = 3, x cannot be 2 because that would make the denominator zero.
如果分母含有变量，必须检查排除值。例如，在 1/(x - 2) = 3 中，x 不能为 2，因为这会使分母为零。
After solving, substitute the answer back into the original equation to ensure no denominator becomes zero and the equality holds.
求解后，将答案代回原方程，确保分母不为零且等式成立。
6. Exponential and Logarithmic Equations | 指数与对数方程
Exponential equations such as 2ˣ both sides. The natural logarithm ln is often used in Edexcel papers because it simplifies expressions involving e.
指数方程如 2ˣ = 7 通常通过两边取对数求解。Edexcel 试卷中常用自然对数 ln，因为它能简化含有 e 的表达式。
2ˣ = 7 ⇒ ln(2ˣ) = ln 7 ⇒ x ln 2 = ln 7 ⇒ x = ln 7 / ln 2
n also use log base a, but the change-of-base formula is often required. The final answer can be left in exact logarithmic form unless a decimal approximation is requested.
对于 aᵡ = b 形式的方程，也可以使用以 a 为底的对数，但通常需要换底公式。除非要求近似值，最终答案可以保留精确的对数形式。
ample, logₐ x + logₐ(x - 1) = logₐ 6 gives x(x - 1) = 6, leading to x = 3 (reject x = -2).
对数方程通常需要先用对数法则合并项。例如 logₐ x + logₐ(x - 1) = logₐ 6 得到 x(x - 1) = 6，解得 x = 3（舍去 x = -2）。
Always check that the final x keeps the argument of any logarithm positive. A solution that makes the argument zero or negative must be rejected.
始终检查最终 x 是否使所有对数的真数为正。使真数为零或负数的解必须舍去。
7. Trigonometric Equations and CAST | 三角方程与 CAST 图
e the CAST diagram or graph symmetry to find additional solutions beyond the calculator value.
三角方程需要在给定区间内求出所有解，通常是 0 ≤ x < 360° 或 0 ≤ x < 2π。使用 CAST 图或图像对称性来找出计算器给出的值之外的其他解。
Example: Solve sin x = 0.5 for 0° ≤ x < 360°.
例题：在 0° ≤ x < 360° 内解 sin x = 0.5。
n is positive in the first and second quadrants, the second solution is 180° - 30° = 150°.
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