This set of intermediate algebra problems focuses on solving inequalities involving absolute value. Each question is followed by a detailed solution provided below. These exercises can be used alongside the tutorial on How to Solve Inequalities with Absolute Value to reinforce understanding and practice key skills.
Problem: Solve the absolute value inequality \( |x| \leq 3 \).
Solution:
\[ x \in [-3, 3] \]Because the absolute value of \( x \) is at most 3, \( x \) lies between \(-3\) and \(3\).
Problem: Solve the absolute value inequality \( |x| \geq 5 \).
Solution:
\[ x \in (-\infty, -5] \cup [5, +\infty) \]Values of \( x \) are those whose distance from zero is at least 5, so either less than or equal to \(-5\) or greater than or equal to \(5\).
Problem: Solve the absolute value inequality \( |x| \leq 0 \).
Solution:
\[ x = \{0\} \]The absolute value is zero only at \( x = 0 \).
Problem: Solve the absolute value inequality \( |x| \lt 0 \).
Solution:
No solutions
Absolute value is never negative, so no real number \( x \) satisfies this inequality.
Problem: Solve the absolute value inequality \( |x| \geq 0 \).
Solution:
\[ x \in (-\infty, +\infty) \]Absolute value is always non-negative, so all real numbers satisfy this inequality.
Problem: Solve the absolute value inequality \( |x| > 0 \).
Solution:
\[ x \in (-\infty, 0) \cup (0, +\infty) \]All real numbers except zero have a positive absolute value.
Problem: Solve the absolute value inequality \( |2x + 2| \leq 10 \).
Solution:
The expression inside the absolute value lies between \(-10\) and \(10\):
\[ -10 \leq 2x + 2 \leq 10 \]Subtract 2 from all parts:
\[ -12 \leq 2x \leq 8 \]Divide by 2:
\[ -6 \leq x \leq 4 \]Final Answer: \[ x \in [-6, 4] \]
Problem: Solve the absolute value inequality \( |3x - 7| \leq 0 \).
Solution:
An absolute value can never be strictly less than zero. Therefore, the inequality holds only when the absolute value equals zero:
\[ 3x - 7 = 0 \implies 3x = 7 \implies x = \dfrac{7}{3} \]Final Answer: \[ x = \left\{\dfrac{7}{3}\right\} \]
Problem: Solve the absolute value inequality \( |-x + 3| \geq 7 \).
Solution:
Split the inequality into two cases based on the definition of absolute value:
\[ -x + 3 \leq -7 \quad \text{or} \quad -x + 3 \geq 7 \]Solve the first inequality:
\[ -x \leq -10 \implies x \geq 10 \]Solve the second inequality:
\[ -x \geq 4 \implies x \leq -4 \]Final Answer: \[ x \in (-\infty, -4] \cup [10, +\infty) \]
Problem: Solve the absolute value inequality \( |3x + 9| > 0 \).
Solution:
The absolute value expression is strictly positive for all real numbers except where the inside expression equals zero:
\[ 3x + 9 = 0 \implies 3x = -9 \implies x = -3 \]Thus, all real numbers except \( x = -3 \) satisfy the inequality.
Final Answer: \[ x \in (-\infty, -3) \cup (-3, +\infty) \]
Problem: Solve the absolute value inequality \( 2|4x - 10| + 3 \geq 17 \).
Solution:
First, isolate the absolute value term:
\[ 2|4x - 10| \geq 17 - 3 \] \[ 2|4x - 10| \geq 14 \] \[ |4x - 10| \geq 7 \]Split into two inequalities:
\[ 4x - 10 \leq -7 \quad \text{or} \quad 4x - 10 \geq 7 \]Solve the first:
\[ 4x \leq 3 \implies x \leq \dfrac{3}{4} \]Solve the second:
\[ 4x \geq 17 \implies x \geq \dfrac{17}{4} \]Final Answer: \[ x \in (-\infty, \dfrac{3}{4}] \cup \left[\dfrac{17}{4}, +\infty\right) \]
Problem: Solve the absolute value inequality \( -3|-5x + 2| + 5 \lt -10 \).
Solution:
Isolate the absolute value term:
\[ -3|-5x + 2| \lt -10 - 5 \] \[ -3|-5x + 2| \lt -15 \]Divide both sides by \(-3\) (remember to reverse the inequality sign):
\[ |-5x + 2| > 5 \]Split into two inequalities:
\[ -5x + 2 \lt -5 \quad \text{or} \quad -5x + 2 > 5 \]Solve the first:
\[ -5x \lt -7 \implies x > \dfrac{7}{5} \]Solve the second:
\[ -5x > 3 \implies x \lt -\dfrac{3}{5} \]Final Answer: \[ x \in (-\infty, -\dfrac{3}{5}) \cup \left(\dfrac{7}{5}, +\infty\right) \]
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