A practical guide to solving linear inequalities step by step. Learn how to reverse the inequality sign correctly, apply equivalence principles, and write solutions in proper interval notation.
Exercise 1 — level ★★☆☆☆
\[ 2x + 3 > 7 \]
Answer
\[ x > 2 \]
Solution
Key idea
We isolate \(x\) on the left-hand side using the same operations as for an equation. Since we divide by a positive number, the direction of the inequality does not change.
Isolating the unknown
Subtract \(3\) from both sides:
\[ 2x > 7-3 \implies 2x > 4 \]
Divide by \(2\) (positive, so the direction remains unchanged):
\[ x > 2 \]
Solution set
\[ S = \{x \in \mathbb{R} \mid x > 2\} = (2,\,+\infty) \]
Answer
\[ \boxed{x > 2} \]
Exercise 2 — level ★★☆☆☆
\[ 3x - 5 \leq 4 \]
Answer
\[ x \leq 3 \]
Solution
Isolating the unknown
Add \(5\) to both sides:
\[ 3x \leq 9 \]
Divide by \(3\) (positive, direction unchanged):
\[ x \leq 3 \]
Solution set
\[ S = (-\infty,\,3] \]
Answer
\[ \boxed{x \leq 3} \]
Exercise 3 — level ★★☆☆☆
\[ -2x + 1 < 5 \]
Answer
\[ x > -2 \]
Solution
Sign caution
When dividing or multiplying by a negative number, the direction of the inequality reverses.
Isolating the unknown
Subtract \(1\) from both sides:
\[ -2x < 4 \]
Divide by \(-2\) (negative): the direction reverses from \(<\) to \(>\):
\[ x > -2 \]
Solution set
\[ S = (-2,\,+\infty) \]
Answer
\[ \boxed{x > -2} \]
Exercise 4 — level ★★☆☆☆
\[ 4x - 8 \geq 0 \]
Answer
\[ x \geq 2 \]
Solution
Isolating the unknown
Add \(8\) to both sides:
\[ 4x \geq 8 \]
Divide by \(4\) (positive, direction unchanged):
\[ x \geq 2 \]
Solution set
\[ S = [2,\,+\infty) \]
Answer
\[ \boxed{x \geq 2} \]
Exercise 5 — level ★★★☆☆
\[ 3x + 2 > x + 8 \]
Answer
\[ x > 3 \]
Solution
Collecting the \(x\) terms
Collect the \(x\) terms on the left-hand side and the constants on the right:
\[ 3x-x > 8-2 \implies 2x > 6 \implies x > 3 \]
Solution set
\[ S = (3,\,+\infty) \]
Answer
\[ \boxed{x > 3} \]
Exercise 6 — level ★★★☆☆
\[ 5x - 3 \leq 2x + 9 \]
Answer
\[ x \leq 4 \]
Solution
Collecting terms
\[ 5x-2x \leq 9+3 \implies 3x \leq 12 \implies x \leq 4 \]
Solution set
\[ S = (-\infty,\,4] \]
Answer
\[ \boxed{x \leq 4} \]
Exercise 7 — level ★★★☆☆
\[ 2(x + 1) < 3(x - 1) \]
Answer
\[ x > 5 \]
Solution
Expanding the brackets
\[ 2x+2 < 3x-3 \]
Collecting terms
\[ 2x-3x < -3-2 \implies -x < -5 \implies x > 5 \]
More precisely: \(2x-3x < -3-2 \implies -x < -5 \implies x > 5\) (the direction reverses when dividing by \(-1\)).
Solution set
\[ S = (5,\,+\infty) \]
Answer
\[ \boxed{x > 5} \]
Exercise 8 — level ★★★☆☆
\[ \frac{x}{2} + 1 > \frac{x}{3} \]
Answer
\[ x > -6 \]
Solution
Clearing the fractions
The LCM of \(2\) and \(3\) is \(6\). Multiply everything by \(6\) (positive, direction unchanged):
\[ 3x + 6 > 2x \]
Collecting terms
\[ 3x-2x > -6 \implies x > -6 \]
Solution set
\[ S = (-6,\,+\infty) \]
Answer
\[ \boxed{x > -6} \]
Exercise 9 — level ★★★☆☆
\[ \frac{x - 1}{2} \leq \frac{x + 3}{4} \]
Answer
\[ x \leq 5 \]
Solution
Clearing the fractions
The LCM of \(2\) and \(4\) is \(4\). Multiply everything by \(4\):
\[ 2(x-1) \leq x+3 \implies 2x-2 \leq x+3 \]
Collecting terms
\[ 2x-x \leq 3+2 \implies x \leq 5 \]
Solution set
\[ S = (-\infty,\,5] \]
Answer
\[ \boxed{x \leq 5} \]
Exercise 10 — level ★★★☆☆
\[ 3(2x - 1) \geq 2(x + 5) \]
Answer
\[ x \geq \dfrac{13}{4} \]
Solution
Expanding the brackets
\[ 6x-3 \geq 2x+10 \]
Collecting terms
\[ 6x-2x \geq 10+3 \implies 4x \geq 13 \implies x \geq \frac{13}{4} \]
Solution set
\[ S = \left[\frac{13}{4},\,+\infty\right) \]
Answer
\[ \boxed{x \geq \dfrac{13}{4}} \]
Exercise 11 — level ★★★★☆
\[ \begin{cases} x + 1 > 0 \\ 2x - 3 < 5 \end{cases} \]
Answer
\[ -1 < x < 4 \]
Solution
Key idea
Each inequality is solved separately; then we take the intersection of the solution sets.
First inequality
\[ x+1>0 \implies x>-1 \]
Second inequality
\[ 2x-3<5 \implies 2x<8 \implies x<4 \]
Intersection
\[ x>-1 \;\text{ and }\; x<4 \implies -1<x<4 \]
Solution set
\[ S = (-1,\,4) \]
Answer
\[ \boxed{-1 < x < 4} \]
Exercise 12 — level ★★★★☆
\[ \begin{cases} 3x - 2 \geq 1 \\ x + 5 > 2x \end{cases} \]
Answer
\[ 1 \leq x < 5 \]
Solution
First inequality
\[ 3x-2\geq1 \implies 3x\geq3 \implies x\geq1 \]
Second inequality
\[ x+5>2x \implies 5>x \implies x<5 \]
Intersection
\[ x\geq1 \;\text{ and }\; x<5 \implies 1\leq x<5 \]
Solution set
\[ S = [1,\,5) \]
Answer
\[ \boxed{1 \leq x < 5} \]
Exercise 13 — level ★★★★☆
\[ -1 < 2x + 3 < 7 \]
Answer
\[ -2 < x < 2 \]
Solution
Key idea
This is a compound inequality. The same operations are applied to all three parts simultaneously.
Subtracting \(3\) throughout
\[ -1-3 < 2x+3-3 < 7-3 \implies -4 < 2x < 4 \]
Dividing throughout by \(2\)
The divisor is positive, so the directions remain unchanged:
\[ -2 < x < 2 \]
Solution set
\[ S = (-2,\,2) \]
Answer
\[ \boxed{-2 < x < 2} \]
Exercise 14 — level ★★★★☆
\[ \begin{cases} 2x - 1 > 3 \\ 3x + 2 < 14 \end{cases} \]
Answer
\[ 2 < x < 4 \]
Solution
First inequality
\[ 2x-1>3 \implies 2x>4 \implies x>2 \]
Second inequality
\[ 3x+2<14 \implies 3x<12 \implies x<4 \]
Intersection
\[ x>2 \;\text{ and }\; x<4 \implies 2<x<4 \]
Solution set
\[ S = (2,\,4) \]
Answer
\[ \boxed{2 < x < 4} \]
Exercise 15 — level ★★★★☆
\[ \begin{cases} \dfrac{x}{2} - 1 \geq 0 \\[6pt] \dfrac{x + 3}{3} < 2 \end{cases} \]
Answer
\[ 2 \leq x < 3 \]
Solution
First inequality
\[ \frac{x}{2}\geq1 \implies x\geq2 \]
Second inequality
Multiply by \(3\) (positive):
\[ x+3<6 \implies x<3 \]
Intersection
\[ x\geq2 \;\text{ and }\; x<3 \implies 2\leq x<3 \]
Solution set
\[ S = [2,\,3) \]
Answer
\[ \boxed{2 \leq x < 3} \]
Exercise 16 — level ★★★★☆
\[ \begin{cases} x > 5 \\ x < 3 \end{cases} \]
Answer
No solution
Solution
First inequality
\[ x>5 \implies S_1=(5,\,+\infty) \]
Second inequality
\[ x<3 \implies S_2=(-\infty,\,3) \]
Intersection
\[ S_1 \cap S_2 = (5,\,+\infty) \cap (-\infty,\,3) = \emptyset \]
There is no real number that is simultaneously greater than \(5\) and less than \(3\).
Answer
\[ \boxed{\text{No solution} \quad S = \emptyset} \]
Exercise 17 — level ★★★★★
\[ \frac{2x - 3}{4} - \frac{x + 1}{3} > \frac{1}{6} \]
Answer
\[ x > \dfrac{15}{2} \]
Solution
Clearing the fractions
The LCM of \(4\), \(3\) and \(6\) is \(12\). Multiply everything by \(12\) (positive):
\[ 3(2x-3) - 4(x+1) > 2 \]
Expanding
\[ 6x-9-4x-4 > 2 \implies 2x-13 > 2 \implies 2x > 15 \implies x > \frac{15}{2} \]
Check with \(x=8\)
\[ \frac{13}{4}-\frac{9}{3}=\frac{13}{4}-3=\frac{1}{4}>\frac{1}{6} \]
Solution set
\[ S = \left(\frac{15}{2},\,+\infty\right) \]
Answer
\[ \boxed{x > \dfrac{15}{2}} \]
Exercise 18 — level ★★★★★
\[ 3(x - 2) - 2(2x + 1) \geq x - 5 \]
Answer
\[ x \leq -\dfrac{3}{2} \]
Solution
Expanding the brackets
\[ 3x-6-4x-2 \geq x-5 \implies -x-8 \geq x-5 \]
Collecting terms
\[ -x-x \geq -5+8 \implies -2x \geq 3 \]
Divide by \(-2\) (negative): the direction reverses from \(\geq\) to \(\leq\):
\[ x \leq -\frac{3}{2} \]
Check with \(x=-2\)
\[ 3(-4)-2(-3)=-12+6=-6 \] and \[ -2-5=-7 \]. Since \(-6\geq-7\)
Solution set
\[ S = \left(-\infty,\,-\frac{3}{2}\right] \]
Answer
\[ \boxed{x \leq -\dfrac{3}{2}} \]
Exercise 19 — level ★★★★★
\[ \begin{cases} \dfrac{x-1}{2} < \dfrac{x}{3} + 1 \\[8pt] 2x - 3 > x - 7 \end{cases} \]
Answer
\[ -4 < x < 9 \]
Solution
First inequality
Multiply through by the LCM \(6\):
\[ 3(x-1)<2x+6 \implies 3x-3<2x+6 \implies x<9 \]
Second inequality
\[ 2x-x>-7+3 \implies x>-4 \]
Intersection
\[ x>-4 \;\text{ and }\; x<9 \implies -4<x<9 \]
Solution set
\[ S = (-4,\,9) \]
Answer
\[ \boxed{-4 < x < 9} \]
Exercise 20 — level ★★★★★
\[ \begin{cases} \dfrac{x}{3} - 1 \leq \dfrac{x}{2} + \dfrac{1}{6} \\[8pt] 2x + 3 \geq \dfrac{x}{2} - 3 \end{cases} \]
Answer
\[ x \geq -4 \]
Solution
First inequality
Multiply through by the LCM \(6\):
\[ 2x-6 \leq 3x+1 \implies -x\leq7 \implies x\geq-7 \]
Second inequality
Multiply by \(2\):
\[ 4x+6 \geq x-6 \implies 3x\geq-12 \implies x\geq-4 \]
Intersection
\[ x\geq-7 \;\text{ and }\; x\geq-4 \]
The more restrictive condition is \(x\geq-4\).
Solution set
\[ S = [-4,\,+\infty) \]
Answer
\[ \boxed{x \geq -4} \]