📚 A-Level Mathematics: Strategies for Solving Absolute Value Equations and Inequalities | A-Level 数学：绝对值方程与不等式的求解策略
Absolute value equations and inequalities are a cornerstone of A-Level Mathematics. They require not only fluent algebraic manipulation but also a clear understanding of distance, intervals, and piecewise functions. Mastering these strategies will help you tackle both pure mathematics and applied problems with confidence.
绝对值方程与不等式是 A-Level 数学的基石。它们不仅要求熟练的代数运算，还要求对距离、区间和分段函数有清晰的理解。掌握这些求解策略，将帮助你在纯数学和应用题中自信应对。
1. Definition and Basic Properties | 定义与基本性质
The absolute value of a real number x, written |x|, is defined piecewise:
实数 x 的绝对值，记作 |x|，按分段方式定义：
|x| = x 当 x ≥ 0; |x| = -x 当 x < 0
This means |x| represents the distance from x to 0 on the number line, so it is always non-negative.
这意味着 |x| 表示数轴上 x 到 0 的距离，因此它永远是非负的。
Two fundamental properties are the multiplicative property and the triangle inequality.
两个基本性质是乘法性质和三角不等式。
|ab| = |a| |b|, |a/b| = |a| / |b| (b ≠ 0)
|a + b| ≤ |a| + |b|
These properties are frequently used to simplify expressions before solving. For example, |−3x| = 3|x|.
这些性质常用来在求解前简化表达式。例如，|−3x| = 3|x|。
2. Geometric Interpretation | 几何意义
On the number line, |x − a| is the distance between x and a. This simple observation turns many equations and inequalities into interval statements.
在数轴上，|x − a| 表示 x 与 a 之间的距离。这个简单的观察将许多方程和不等式转化为区间描述。
For example, the equation |x − 3| = 5 means x is exactly 5 units away from 3, so x = 8 or x = −2.
例如，方程 |x − 3| = 5 表示 x 距 3 恰好 5 个单位，因此 x = 8 或 x = −2。
For inequalities, |x − a| < d means the distance is less than d, which gives the open interval (a − d, a + d). Similarly, |x − a| > d gives the union of two rays: x < a − d or x > a + d.
对于不等式，|x − a| < d 表示距离小于 d，即开区间 (a − d, a + d)。类似地，|x − a| > d 给出两个射线之并：x < a − d 或 x > a + d。
Always draw a number line for these problems; the visual intuition prevents sign errors.
解决这类问题时务必画一条数轴；直观图形可以防止符号错误。
3. Solving Absolute Value Equations by Casework | 分段讨论法
The formal definition of |x| naturally leads to the casework method: split the domain where the expression inside the absolute value changes sign.
绝对值的正式定义自然引出分段讨论法：在绝对值内部表达式改变符号的地方划分定义域。
To solve |f(x)| = g(x), follow these steps:
求解 |f(x)| = g(x) 的步骤如下：
Find the critical points where f(x) = 0. These divide the real line into intervals.
求出 f(x) = 0 的临界点，它们将实轴分成若干区间。
On each interval, replace |f(x)| by either f(x) or −f(x), then solve the resulting linear or quadratic equation.
在每个区间上，将 |f(x)| 替换为 f(x) 或 −f(x)，然后求解所得的一次或二次方程。
Check that each candidate solution lies in the assumed interval; discard any that do not.
检查每个候选解是否落在假设的区间内；不满足的应舍去。
Example: Solve |2x − 1| 1) = x + 3 gives −2x + 1 = x + 3, so x = −2/3, also valid. The solution set is {4, −2/3}.
例：解 |2x − 1| = x + 3。临界点为 x = 1/2。当 x ≥ 1/2 时，2x − 1 = x + 3，得 x = 4，有效。当 x < 1/2 时，−(2x − 1) = x + 3，即 −2x + 1 = x + 3，得 x = −2/3，同样有效。解集为 {4, −2/3}。
4. Solving by Squaring | 平方法
Since |u| ue signs. However, squaring an equation may introduce extraneous roots, so verification is essential.
因为 |u|² = u²，平方可以去掉绝对值符号。但平方可能引入增根，因此必须验根。
For two absolute values, |f(x)| = |g(x)| hich simplifies to (f(x) − g(x))(f(x) + g(x)) = 0. This gives f(x) = g(x) or f(x) = −g(x).
对于两个绝对值，|f(x)| = |g(x)| 等价于 f(x)² = g(x)²，即 (f(x) − g(x))(f(x) + g(x)) = 0。由此得 f(x) = g(x) 或 f(x) = −g(x)。
For |f(x)| = g(x), squaring gives f(x)² = g(x)², which is equivalent to |f(x)| = |g(x)| iginal equation requires g(x) ≥ 0, any resulting solution with g(x) < 0 must be discarded.
对于 |f(x)| = g(x)，平方得 f(x)² = g(x)²，这等价于 |f(x)| = |g(x)|。由于原方程要求 g(x) ≥ 0，因此任何使 g(x) < 0 的结果都必须舍去。
Example: Solve |x − 2| = −3. Substitute x = −3 into 2x + 1 = −5 < 0, invalid. Thus only x = 1/3 is the solution.
例：解 |x − 2| = 2x + 1。平方得 (x − 2)² = (2x + 1)²。展开得 x² − 4x + 4 = 4x² + 4x + 1，即 3x² + 8x − 3 = 0，因此 x = 1/3 或 x = −3。将 x = −3 代入 2x + 1 = −5 < 0，无效。所以只有 x = 1/3 是解。
5. Basic Absolute Value Inequalities | 基本绝对值不等式
For a positive constant a, the simplest absolute value inequalities have two standard forms.
对于正常数 a，最简单的绝对值不等式有两种标准形式。
|x| < a ⇔ −a < x < a
|x| > a ⇔ x < −a 或 x > a
These are the building blocks for all more complex problems. For example, |2x − 3| < 5 becomes −5 < 2x − 3 < 5, so −2 < 2x < 8, giving −1 < x < 4.
这些是解决所有更复杂问题的基础。例如，|2x − 3| < 5 变为 −5 < 2x − 3 < 5，于是 −2 < 2x < 8，得到 −1 < x < 4。
If a ≤ 0, the solution sets are different: |x| < 0 has no solution, |x| ≤ 0 has solution x = 0, and |x| > a (with a < 0) is true for all real x.
如果 a ≤ 0，解集不同：|x| < 0 无解，|x| ≤ 0 的解为 x = 0，而 |x| > a（a < 0）对所有实数 x 都成立。
Always check the sign of the right-hand side before applying the standard pattern.
在套用标准模式之前，务必检查右边的符号。
6. Interval Method for Inequalities | 区间检验法
For inequalities with more than one absolute value term, such as |x − 1| + |x + 2| > 3, the interval method is reliable.
对于含有多个绝对值项的不等式，如 |x − 1| + |x + 2| > 3，区间检验法非常可靠。
1 and x = −2, dividing the number line into three regions: x < −2, −2 ≤ x < 1, and x ≥ 1.
首先找出所有临界点：即使任意绝对值表达式为零的值。这里临界点为 x = 1 和 x = −2，将数轴分为三个区域：x < −2，−2 ≤ x < 1，x ≥ 1。
For x < −2: |x − 1| = 1 − x and |x + 2| − x − x − 2 > 3, i.e. −2x − 1 > 3, hence x < −2. Combined with x < −2, this gives x < −2.
当 x < −2 时：|x − 1| = 1 − x，|x + 2| = −x − 2，不等式化为 1 − x − x − 2 > 3，即 −2x − 1 > 3，得 x < −2。与 x < −2 合并，结果为 x < −2。
For −2 ≤ x < 1: |x − 1| = 1 − x and |x + 2| + 2, giving 1 − x + x + 2 > 3, i.e. 3 > 3, which is false. No solutions in this interval.
当 −2 ≤ x < 1 时：|x − 1| = 1 − x，|x + 2| = x + 2，得 1 − x + x + 2 > 3，即 3 > 3，不成立。该区间内无解。
For x ≥ 1: |x − 1| = x − 1 and |x + 2| so x − 1 + x + 2 > 3, i.e. 2x + 1 > 3, hence x > 1. Combined with x ≥ 1, this gives x > 1.
当 x ≥ 1 时：|x − 1| = x − 1，|x + 2| = x + 2，于是 x − 1 + x + 2 > 3，即 2x + 1 > 3，得 x > 1。与 x ≥ 1 合并，结果为 x > 1。
The final solution is x < −2 or x > 1.
最终解为 x < −2 或 x > 1。
7. Graphical Method | 图像法
Graphing both sides of an equation or inequality provides a powerful visual check. For inequalities, the solution set is the x-values where the graph of the left-hand side lies above or below the graph of the right-hand side.
将方程或不等式的两边同时画出图像，是一种有力的直观检验。对于不等式，解集是左边图像位于右边图像上方或下方所对应的 x 值。
For example, to solve |x − 1| < 3, draw y = |x − 1| he inequality holds when the V-shaped graph lies below the horizontal line, so −2 < x < 4.
例如，解 |x − 1| < 3，画 y = |x − 1| 和 y = 3。交点位于 x = −2 和 x = 4；当 V 形图像位于水平线下方时不等式成立，所以 −2 < x < 4。
When dealing with multiple absolute values, graph the piecewise linear sum to identify intervals at a glance. Always mark the intersection points precisely, as they become the boundaries of the solution intervals.
处理多个绝对值时，画出分段线性求和图像可以一目了然地确定区间。务必精确标出交点，因为它们成为解区间的边界。
This method is especially useful for checking answers obtained algebraically.
此方法尤其适合检验代数求解得到的结果。
8. Common Pitfalls and Misconceptions | 常见陷阱与误区
Several common mistakes lead to lost marks in examinations.
几个常见错误会导致考试失分。
Forgetting to verify whether a candidate solution satisfies the original equation after squaring. Squaring can create extraneous roots, so always substitute back.
平方求根后忘记验证候选解是否满足原方程。平方可能产生增根，所以务必代回检验。
Applying |x| < a directly without ensuring a > 0. If a is negative or zero, the solution set differs.
未确保 a > 0 就套用 |x| < a 的模式。如果 a 为负数或零，解集会不同。
When using casework, forgetting to include the critical points themselves. Because |x| is continuous, the solution often contains the critical point; exclude it only if the inequality is strict and the point makes the expression equal to the other side.
使用分段讨论时，忘记包含临界点本身。因为 |x| 是连续的，解常常包含临界点；只有当不等式严格且该点使两边相等时才排除它。
Confusing the union and intersection of intervals. For |f(x)| > a, the solution is the union of two rays; for |f(x)| < a, it is the intersection of two inequalities.
混淆区间的并集与交集。对于 |f(x)| > a，解是两个射线的并集；对于 |f(x)| < a，解是两个不等式的交集。
9. Worked Example: Advanced Inequality | 综合例题
Solve the inequality |x − 2| − |x + 3| ≥ 1.
解不等式 |x − 2| − |x + 3| ≥ 1。
ritical points are x = 2 and x = −3, dividing the line into x < −3, −3 ≤ x < 2, and x ≥ 2.
临界点为 x = 2 和 x = −3，将数轴分为 x < −3，−3 ≤ x < 2，x ≥ 2。
For x < −3: |x − 2| = 2 − x, |x + 3| nequality becomes 2 − x + x + 3 ≥ 1, i.e. 5 ≥ 1, always true. So all x < −3 are solutions.
当 x < −3 时：|x − 2| = 2 − x，|x + 3| = −x − 3。不等式为 2 − x + x + 3 ≥ 1，即 5 ≥ 1，恒成立。因此所有 x < −3 都是解。
For −3 ≤ x < 2: |x − 2| = 2 − x, |x + 3| ≥ 1, so −2x − 1 ≥ 1, hence −2x ≥ 2, x ≤ −1. Combined with −3 ≤ x < 2, we get −3 ≤ x ≤ −1.
当 −3 ≤ x < 2 时：|x − 2| = 2 − x，|x + 3| = x + 3。不等式为 2 − x − (x + 3) ≥ 1，即 −2x − 1 ≥ 1，所以 −2x ≥ 2，x ≤ −1。与 −3 ≤ x < 2 合并，得到 −3 ≤ x ≤ −1。
For x ≥ 2: |x − 2| = x − 2, |x + 3| he inequality becomes x − 2 − (x + 3) ≥ 1, i.e. −5 ≥ 1, which is false. No solutions here.
当 x ≥ 2 时：|x − 2| = x − 2，|x + 3| = x + 3。不等式为 x − 2 − (x + 3) ≥ 1，即 −5 ≥ 1，不成立。此区间无解。
Therefore the solution set is x ≤ −1, written in interval notation as (−∞, −1].
因此解集为 x ≤ −1，用区间表示为 (−∞, −1]。
10. Practice Questions | 练习与提示
Apply these strategies to the following questions. For each problem, try both an algebraic method and a graphical check.
请运用以上策略解决下列问题。每个题目尝试用代数方法和图像检验两种方式。
Solve |3x + 2| = |x − 4|.
解 |3x + 2| = |x − 4|。
Solve |x + 1| > 2x − 3.
解 |x + 1| > 2x − 3。
Find all values of x satisfying |x − 1| + |x + 2| ≤ 5.
求满足 |x − 1| + |x + 2| ≤ 5 的所有 x 值。
A triangle has side lengths x, 4, and |x − 2|. Determine the range of x for which such a triangle exists. (Hint: Use the triangle inequality on the side lengths.)
一个三角形的三边长为 x、4、|x − 2|。确定这样的三角形存在时 x 的取值范围。（提示：对三边长使用三角形不等式。）
Remember to draw number lines, check extraneous roots, and state answers in interval or set notation as required by the question.
记住画数轴、检验增根，并按题目要求用区间或集合记号写出答案。
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