The absolute value of a number is a fundamental concept in mathematics that frequently appears in equations, inequalities, and in the study of functions. On the YÖS exam, problems involving absolute value may be standalone or form part of more complex problems. A deep understanding of the definition of absolute value and of the methods for working with it is a necessary condition for success.
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Definition: The absolute value (modulus) of a number $a$ is the number defined by the following rule:
$$|a| = \begin{cases} a, & \text{if } a \geq 0 \\-a, & \text{if } a < 0 \end{cases}$$
Examples:- $|5| = 5$ (since $5 \geq 0$)- $|-3| = -(-3) = 3$ (since $-3 < 0$)- $|0| = 0$- $|-7| = 7$
Property: The absolute value of any number is always nonnegative:$$|a| \geq 0 \text{ for all } a \in \mathbb{R}$$
Moreover, $|a| = 0$ only when $a = 0$.
For any positive number $a > 0$:$$|x| = a \Leftrightarrow x = a \text{ or } x = -a$$
This means that the equation $|x| = a$ has two solutions (if $a > 0$), one solution $x = 0$ (if $a = 0$), and no solutions (if $a < 0$).
Definition: The absolute value of a number $a$ represents the distance from the point $a$ on the number line to the origin (the point 0).
$$|a| = \text{distance from } a \text{ to } 0$$
Example:```Number line:←────●────●────●────→ -3 0 3
|-3| = 3 (the distance from -3 to 0 is 3)|3| = 3 (the distance from 3 to 0 is 3)```
Definition: The distance between the points $a$ and $b$ on the number line is equal to $|a - b|$.
$$\text{distance}(a, b) = |a - b|$$
Examples:- Distance between 2 and 7: $|2 - 7| = |-5| = 5$- Distance between -3 and 4: $|-3 - 4| = |-7| = 7$- Distance between -5 and -2: $|-5 - (-2)| = |-3| = 3$
Property 1 (Absolute value of a product):$$|ab| = |a| \cdot |b|$$
Proof: Consider the four cases depending on the signs of $a$ and $b$. In each case we obtain a correct equality.
Property 2 (Absolute value of a quotient):$$\left|\frac{a}{b}\right| = \frac{|a|}{|b|}, \quad b \neq 0$$
Property 3 (Absolute value of a power):$$|a^n| = |a|^n \text{ for natural } n$$
For even $n$ this is simply $|a^n| = a^n$.
Theorem (Triangle inequality):$$|a + b| \leq |a| + |b|$$
Equality holds only when $a$ and $b$ have the same sign (both nonnegative or both nonpositive).
Proof: It is known that $-|a| \leq a \leq |a|$ and $-|b| \leq b \leq |b|$.
Adding these inequalities:$$-(|a| + |b|) \leq a + b \leq |a| + |b|$$
By the definition of absolute value:$$|a + b| \leq |a| + |b|$$
Example:- $|3 + 4| = 7 = |3| + |4|$ (equality, since both are positive)- $|3 - 4| = 1 < 7 = |3| + |-4|$ (strict inequality)
Corollary (Reverse triangle inequality):$$||a| - |b|| \leq |a - b|$$
Rule: $|f(x)| = a \Leftrightarrow f(x) = a$ or $f(x) = -a$
Example 1: Solve $|x - 3| = 5$
Solution:$$x - 3 = 5 \quad \text{or} \quad x - 3 = -5$$$$x = 8 \quad \text{or} \quad x = -2$$
Check: $|8 - 3| = 5$ ✓, $|-2 - 3| = 5$ ✓
Rule: $|f(x)| = 0 \Leftrightarrow f(x) = 0$
Example 2: Solve $|2x + 1| = 0$
Solution:$$2x + 1 = 0$$$$x = -\frac{1}{2}$$
Rule: The equation $|f(x)| = a$ has no solutions when $a < 0$.
Rule: $|f(x)| = |g(x)| \Leftrightarrow f(x) = g(x)$ or $f(x) = -g(x)$
Example 3: Solve $|x + 2| = |2x - 1|$
Solution:$$x + 2 = 2x - 1 \quad \text{or} \quad x + 2 = -(2x - 1)$$
First equation:$$x + 2 = 2x - 1$$$$3 = x$$$$x = 3$$
Second equation:$$x + 2 = -2x + 1$$$$3x = -1$$$$x = -\frac{1}{3}$$
Check: - For $x = 3$: $|3 + 2| = 5$, $|2(3) - 1| = 5$ ✓- For $x = -\frac{1}{3}$: $|-\frac{1}{3} + 2| = \frac{5}{3}$, $|2(-\frac{1}{3}) - 1| = |-\frac{5}{3}| = \frac{5}{3}$ ✓
When solving complex equations involving absolute value one should:1. Find the critical points where the expressions under the absolute value are equal to zero2. Determine the sign of the expressions on each interval3. Expand the absolute values according to the definition on each interval4. Solve the equations on each interval
Example: Solve $|x - 2| = 3x + 1$
Critical point: $x - 2 = 0 \Rightarrow x = 2$
On the interval $x < 2$: $|x - 2| = -(x - 2) = -x + 2$
Equation:$$-x + 2 = 3x + 1$$$$1 = 4x$$$$x = \frac{1}{4}$$
Check: $\frac{1}{4} < 2$ ✓. Substituting: $|\frac{1}{4} - 2| = \frac{7}{4}$, $3(\frac{1}{4}) + 1 = \frac{7}{4}$ ✓
On the interval $x \geq 2$:$|x - 2| = x - 2$
Equation:$$x - 2 = 3x + 1$$$$-3 = 2x$$$$x = -\frac{3}{2}$$
Check: $-\frac{3}{2} < 2$ ✗. This value does not belong to the interval $x \geq 2$, so it is not a solution.
Answer: $x = \frac{1}{4}$
Example: Solve $|x + 1| + |x - 2| = 5$
Critical points: $x = -1$ and $x = 2$
On the interval $x < -1$:$|x + 1| = -(x + 1)$, $|x - 2| = -(x - 2)$
Equation:$$-(x + 1) - (x - 2) = 5$$$$-x - 1 - x + 2 = 5$$$$-2x + 1 = 5$$$$x = -2$$
Check: $-2 < -1$ ✓. Substituting: $|-2 + 1| + |-2 - 2| = 1 + 4 = 5$ ✓
On the interval $-1 \leq x < 2$:$|x + 1| = x + 1$, $|x - 2| = -(x - 2)$
Equation:$$(x + 1) - (x - 2) = 5$$$$x + 1 - x + 2 = 5$$$$3 = 5$$
Contradiction, no solutions on this interval.
On the interval $x \geq 2$:$|x + 1| = x + 1$, $|x - 2| = x - 2$
Equation:$$(x + 1) + (x - 2) = 5$$$$2x - 1 = 5$$$$x = 3$$
Check: $3 \geq 2$ ✓. Substituting: $|3 + 1| + |3 - 2| = 4 + 1 = 5$ ✓
Answer: $x = -2$ or $x = 3$
Nested absolute values are absolute values inside absolute values, for example, $||a| - |b||$ or $||x| - 3|$.
Example: Solve $||x| - 2| = 1$
Solution (in two stages):
Stage 1: Let $y = |x|$. Then we solve $|y - 2| = 1$:$$y - 2 = 1 \quad \text{or} \quad y - 2 = -1$$$$y = 3 \quad \text{or} \quad y = 1$$
Stage 2: We return to $x$:- From $|x| = 3$: $x = 3$ or $x = -3$- From $|x| = 1$: $x = 1$ or $x = -1$
Check:- $||3| - 2| = |3 - 2| = 1$ ✓- $||-3| - 2| = |3 - 2| = 1$ ✓- $||1| - 2| = |1 - 2| = 1$ ✓- $||-1| - 2| = |1 - 2| = 1$ ✓
Answer: $x \in \{-3, -1, 1, 3\}$
Graph of $y = |x|$:- Vertex at the point $(0, 0)$- On the interval $x < 0$: $y = -x$ (a line with slope $-1$)- On the interval $x \geq 0$: $y = x$ (a line with slope $1$)- Shape: the letter V
Graph of $y = |x - a|$:- Vertex at the point $(a, 0)$- This is a shift of the graph of $y = |x|$ by $a$ units to the right
Graph of $y = |x| + b$:- Vertex at the point $(0, b)$- This is a shift of the graph of $y = |x|$ by $b$ units upward
This is a V-shaped graph with vertex at the point $(2, 1)$.
On $x < 2$: $y = -(x - 2) + 1 = -x + 3$On $x \geq 2$: $y = (x - 2) + 1 = x - 1$
Form 1: $|f(x)| < a$ (where $a > 0$)$$|f(x)| < a \Leftrightarrow -a < f(x) < a$$
Form 2: $|f(x)| > a$ (where $a > 0$)$$|f(x)| > a \Leftrightarrow f(x) < -a \text{ or } f(x) > a$$
Form 3: $|f(x)| \leq a$$$|f(x)| \leq a \Leftrightarrow -a \leq f(x) \leq a$$
Form 4: $|f(x)| \geq a$$$|f(x)| \geq a \Leftrightarrow f(x) \leq -a \text{ or } f(x) \geq a$$
Example 1: Solve $|x + 3| < 2$
Solution:$$-2 < x + 3 < 2$$$$-5 < x < -1$$
Answer: $x \in (-5, -1)$
Example 2: Solve $|2x - 5| > 3$
Solution:$$2x - 5 < -3 \quad \text{or} \quad 2x - 5 > 3$$$$2x < 2 \quad \text{or} \quad 2x > 8$$$$x < 1 \quad \text{or} \quad x > 4$$
Answer: $x \in (-\infty, 1) \cup (4, +\infty)$
Statement: Solve $|3x - 6| = 9$
Solution:$$3x - 6 = 9 \quad \text{or} \quad 3x - 6 = -9$$$$3x = 15 \quad \text{or} \quad 3x = -3$$$$x = 5 \quad \text{or} \quad x = -1$$
Check:- $|3(5) - 6| = |9| = 9$ ✓- $|3(-1) - 6| = |-9| = 9$ ✓
Answer: $x = -1$ or $x = 5$
Statement: Solve $|x - 1| = |x + 3|$
Solution:$$x - 1 = x + 3 \quad \text{or} \quad x - 1 = -(x + 3)$$
First equation:$$-1 = 3$$Contradiction, no solutions.
Second equation:$$x - 1 = -x - 3$$$$2x = -2$$$$x = -1$$
Check: $|-1 - 1| = 2$, $|-1 + 3| = 2$ ✓
Alternative solution (geometric): $|x - 1| = |x + 3|$ means that the distances from $x$ to the points 1 and $-3$ are equal. Such a point lies exactly midway between them: $x = \frac{1 + (-3)}{2} = -1$.
Answer: $x = -1$
Statement: Solve $|x + 2| < 4$
Solution:$$-4 < x + 2 < 4$$$$-6 < x < 2$$
Answer: $x \in (-6, 2)$
Statement: Solve $|x| + |x - 1| = 3$
Solution:Critical points: $x = 0$ and $x = 1$
On the interval $x < 0$:$$-x + (-(x - 1)) = 3$$$$-x - x + 1 = 3$$$$-2x = 2$$$$x = -1$$
Check: $-1 < 0$ ✓. $|-1| + |-1 - 1| = 1 + 2 = 3$ ✓
On the interval $0 \leq x < 1$:$$x + (-(x - 1)) = 3$$$$x - x + 1 = 3$$$$1 = 3$$
Contradiction, no solutions.
On the interval $x \geq 1$:$$x + (x - 1) = 3$$$$2x - 1 = 3$$$$x = 2$$
Check: $2 \geq 1$ ✓. $|2| + |2 - 1| = 2 + 1 = 3$ ✓
Answer: $x = -1$ or $x = 2$
Statement: The distance from a point on the number line to the numbers -2 and 4 is 3. Find this point.
Solution:Let the point be $x$.
Conditions: $|x - (-2)| = 3$ and $|x - 4| = 3$
From the first: $|x + 2| = 3$$$x + 2 = 3 \quad \text{or} \quad x + 2 = -3$$$$x = 1 \quad \text{or} \quad x = -5$$
We check against the second condition $|x - 4| = 3$:- For $x = 1$: $|1 - 4| = 3$ ✓- For $x = -5$: $|-5 - 4| = 9 \neq 3$ ✗
Answer: $x = 1$
Wrong: $|x + 1| = -5$ has a solution.
Correct: The absolute value is always nonnegative, therefore when $a < 0$ there are no solutions.
Wrong: $|x - 3| = 5$ solved only as $x - 3 = 5$.
Correct: $|x - 3| = 5 \Rightarrow x - 3 = 5$ or $x - 3 = -5$.
Wrong: $|a - b| = |a| - |b|$
Correct: $|a - b| \leq |a| + |b|$ (triangle inequality)
Wrong: $||x - 2|| = |x - 2|$ computed incorrectly.
Correct: First handle the inner absolute value, then the outer one.
Wrong: Solved on one interval but forgot to check that the solution actually belongs to that interval.
Correct: Always check that the found solution satisfies the conditions of the interval.
$|x - a|$ is the distance from $x$ to $a$. This helps intuitively understand what an equation or inequality means.
When there are several absolute values, draw a number line, mark the critical points and solve on each interval separately.
When using the interval method, always substitute the found solution back into the original equation.
For some problems (especially with a parameter) it is useful to draw graphs and find the points of intersection.
The equation $|f(x)| = a$ when $a < 0$ has no solutions. This is often tested on exams.
Methods for solving equations and inequalities with absolute value differ slightly. Make sure you are using the correct approach.
The absolute value of a number is an important and powerful tool used in many areas of mathematics. From simple definitions and properties to complex equations and inequalities, understanding all aspects of working with absolute value is necessary for successfully solving YÖS problems. Geometric intuition, knowledge of the basic forms and solution methods, and attention to detail are the keys to success in this topic. Constant practice and checking your answers will help you master this important tool.
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