📚 Solving Equations Leading to x = … | 解方程得出 x = …
In A-Level Edexcel Mathematics, a large part of Pure algebra involves changing equations and formulas until the unknown is written in the form x = something. This skill is tested directly in solving linear, quadratic and simultaneous equations, and it also underpins almost every applied topic from mechanics to statistics.
在 A-Level Edexcel 数学中，纯代数很大一部分内容都是对等式或公式进行变形，直到把未知数写成 x = 某个表达式。这一技能不仅直接考查解线性方程、二次方程和联立方程，而且也是从力学到统计几乎所有应用主题的基础。
1. The Goal of Rearranging | 移项整理的目标
When a question says “find x” or “solve for x”, it is asking you to rewrite the relationship so that x appears by itself on one side of the equals sign. The other side may be a number, an expression, or a formula.
当题目说 “find x” 或 “solve for x” 时，它要求你重新整理关系，使得 x 单独出现在等号的一侧。另一侧可以是数字、表达式或公式。
The equals sign acts as a balance: whatever operation you apply to one side, you must apply to the other side. This balance is the only rule needed for all legal rearrangements.
等号相当于一个天平：你对一边进行任何运算，另一边也必须进行相同运算。这个平衡原则是所有合法变形唯一需要的规则。
2. Linear Equations | 线性方程
A linear equation can always be reduced to the form ax + b = c, where a, b and c are constants. To solve it, undo the operations around x in reverse order.
线性方程总能化为 ax + b = c 的形式，其中 a、b 和 c 是常数。解方程时要按照与运算顺序相反的顺序逐步解除 x 周围的运算。
2x + 3 = 11 → 2x = 8 → x = 4
Check by substituting x = 4 back into the left-hand side: 2(4) + 3 = 11, so the solution is valid.
将 x = 4 代回左边检验：2(4) + 3 = 11，因此解成立。
3. Inverse Operations | 逆运算
The inverse of addition is subtraction, the inverse of multiplication is division, and the inverse of squaring is taking the square root. Applying inverse operations is how you isolate x.
加法的逆运算是减法，乘法的逆运算是除法，平方的逆运算是开平方。应用逆运算就是把 x 单独分离出来的方法。
For example, if x² = 25, taking the square root of both sides gives x = ±√25 = ±5. In A-Level work, always consider both positive and negative roots unless the context restricts the domain.
例如，若 x² = 25，两边开平方得到 x = ±√25 = ±5。在 A-Level 学习中，除非题目语境限制了取值范围，否则必须同时考虑正负两个根。
4. Brackets and Expanding | 括号与展开
If an equation contains brackets, it is usually easier to expand them first. Then collect like terms before isolating x.
如果方程中含有括号，通常先展开括号。然后合并同类项，再分离 x。
3(x − 2) = 2x + 1 → 3x − 6 = 2x + 1 → x = 7
Expanding gives 3x − 6 = 2x + 1. Subtract 2x from both sides, then add 6 to both sides, leading to x = 7.
展开得到 3x − 6 = 2x + 1。两边同时减去 2x，再同时加上 6，得出 x = 7。
5. Fractions in Equations | 含分式的方程
When x appears in a denominator, multiply every term by the lowest common denominator to clear fractions. However, check that no solution makes an original denominator zero.
当 x 出现在分母中时，每一项都乘以最小公分母以消去分式。但要检查所得解不会使原分母为零。
2/x = 3 → 2 = 3x → x = 2/3
After multiplying by x, the equation becomes 2 = 3x. Dividing both sides by 3 gives x = 2/3, which is valid because x ≠ 0.
乘以 x 后，方程变为 2 = 3x。两边除以 3 得到 x = 2/3，这个解有效，因为 x ≠ 0。
6. Factorising Quadratics | 因式分解二次方程
For a quadratic equation such as x² + bx + c = 0, factorising is often the quickest method. If the product of two brackets is zero, at least one bracket must be zero.
对于 x² + bx + c = 0 这样的二次方程，因式分解通常是最快的方法。如果两个括号的乘积为零，那么至少有一个括号为零。
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3
Set each factor equal to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. Both are solutions.
令每个因式等于零：x − 2 = 0 得 x = 2，x − 3 = 0 得 x = 3。两者都是解。
7. The Quadratic Formula | 二次公式
When factorising is difficult, use the quadratic formula for ax² + bx + c = 0:
当因式分解较困难时，对于 ax² + bx + c = 0 可使用二次公式：
x = (−b ± √(b² − 4ac)) ÷ 2a
The expression b² − 4ac is called the discriminant. It tells you how many real solutions exist: positive gives two, zero gives one repeated root, and negative gives no real roots.
表达式 b² − 4ac 称为判别式。它告诉你实根的数量：大于零有两个实根，等于零有一个重根，小于零没有实根。
2x² + 3x − 2 = 0 → x = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4 → x = 1/2 or x = −2
This gives x = 1/2 or x = −2 after simplifying the two cases.
化简两种情况后得到 x = 1/2 或 x = −2。
8. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q = 0. It is especially useful for finding the vertex of a graph and deriving the quadratic formula.
配方法将二次式改写为 a(x + p)² + q = 0 的形式。它对于求图像的顶点以及推导二次公式特别有用。
x² + 6x + 5 = 0 → (x + 3)² − 4 = 0 → x + 3 = ±2 → x = −1 or x = −5
After taking the square root, remember the ± sign, then subtract 3 from both sides to isolate x.
开平方后要记得 ± 号，然后两边同时减去 3 以求出 x。
9. Equations with x on Both Sides | 两边含 x 的方程
If x appears on both sides of an equation, collect the x terms on one side and the constant terms on the other. It is often helpful to subtract the smaller coefficient of x first.
如果 x 同时出现在等号两边，就把含 x 的项移到一边，常数项移到另一边。通常先减去较小的 x 系数会比较容易。
5x − 3 = 2x + 9 → 3x = 12 → x = 4
Subtracting 2x from both sides gives 3x − 3 = 9. Adding 3 gives 3x = 12, so x = 4.
两边减去 2x 得到 3x − 3 = 9。加上 3 得到 3x = 12，所以 x = 4。
10. Simultaneous Equations | 联立方程
For two equations in two unknowns, eliminate one variable by substitution or by adding/subtracting equations. The result is a single equation leading to x = a value, which can then be used to find y.
对于含有两个未知数的两个方程，可以通过代入法或加减消元法消去一个变量。结果是一个单一方程，从而得出 x = 某个值，再代入求出 y。
2x + y = 10 and x − y = 2 → adding gives 3x = 12 → x = 4
Substitute x = 4 into either original equation to obtain y = 2, so the solution is (4, 2).
将 x = 4 代入任意一个原方程可得 y = 2，因此解为 (4, 2)。
11. Rearranging Formulas with x Appearing Twice | 公式中 x 出现两次的整理
A-Level exam questions often ask you to make x the subject when x occurs in several places, such as ax + b = cx + d or fractional formulas. The first step is to gather all terms containing x on one side.
A-Level 考试题经常要求在 x 出现多次的式子中把 x 作为主项，例如 ax + b = cx + d 或分式公式。第一步是把所有含 x 的项集中到一边。
ax + b = cx + d → ax − cx = d − b → x(a − c) = d − b → x = (d − b) ÷ (a − c)
This factorising step is essential because x cannot be divided out until it is written as a single factor.
这一步因式分解非常关键，因为只有把 x 写成单个因式之后才能将其除
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