First-Degree Inequalities: Worked Exercises and Basic Rules. A practical guide to solving linear inequalities step by step. Learn how to handle the reversal of the inequality sign, apply the equivalence principles, and write solutions correctly in interval notation.
Exercise of 26/03/2026 - 09:00 — level ★★☆☆☆
\[ 2x + 3 > 7 \]
Answer
\[ x > 2 \]
Working
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 of 26/03/2026 - 09:10 — level ★★☆☆☆
\[ 3x - 5 \leq 4 \]
Answer
\[ x \leq 3 \]
Working
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 of 26/03/2026 - 09:20 — level ★★☆☆☆
\[ -2x + 1 < 5 \]
Answer
\[ x > -2 \]
Working
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 of 26/03/2026 - 09:30 — level ★★☆☆☆
\[ 4x - 8 \geq 0 \]
Answer
\[ x \geq 2 \]
Working
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 of 26/03/2026 - 09:45 — level ★★★☆☆
\[ 3x + 2 > x + 8 \]
Answer
\[ x > 3 \]
Working
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 of 26/03/2026 - 10:00 — level ★★★☆☆
\[ 5x - 3 \leq 2x + 9 \]
Answer
\[ x \leq 4 \]
Working
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 of 26/03/2026 - 10:15 — level ★★★☆☆
\[ 2(x + 1) < 3(x - 1) \]
Answer
\[ x > 5 \]
Working
Expanding the brackets
\[ 2x+2 < 3x-3 \]
Collecting terms
\[ 2-3x < -3-2x \implies \text{or: } 2+3 < 3x-2x \implies 5 < x \]
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 of 26/03/2026 - 10:30 — level ★★★☆☆
\[ \frac{x}{2} + 1 > \frac{x}{3} \]
Answer
\[ x > -6 \]
Working
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 of 26/03/2026 - 10:45 — level ★★★☆☆
\[ \frac{x - 1}{2} \leq \frac{x + 3}{4} \]
Answer
\[ x \leq 5 \]
Working
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 of 26/03/2026 - 11:00 — level ★★★☆☆
\[ 3(2x - 1) \geq 2(x + 5) \]
Answer
\[ x \geq \dfrac{13}{4} \]
Working
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 of 26/03/2026 - 11:20 — level ★★★★☆
\[ \begin{cases} x + 1 > 0 \\ 2x - 3 < 5 \end{cases} \]
Answer
\[ -1 < x < 4 \]
Working
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 of 26/03/2026 - 11:35 — level ★★★★☆
\[ \begin{cases} 3x - 2 \geq 1 \\ x + 5 > 2x \end{cases} \]
Answer
\[ 1 \leq x < 5 \]
Working
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 of 26/03/2026 - 11:50 — level ★★★★☆
\[ -1 < 2x + 3 < 7 \]
Answer
\[ -2 < x < 2 \]
Working
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 of 26/03/2026 - 12:05 — level ★★★★☆
\[ \begin{cases} 2x - 1 > 3 \\ 3x + 2 < 14 \end{cases} \]
Answer
\[ 2 < x < 4 \]
Working
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 of 26/03/2026 - 12:20 — level ★★★★☆
\[ \begin{cases} \dfrac{x}{2} - 1 \geq 0 \\[6pt] \dfrac{x + 3}{3} < 2 \end{cases} \]
Answer
\[ 2 \leq x < 3 \]
Working
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 of 26/03/2026 - 12:35 — level ★★★★☆
\[ \begin{cases} x > 5 \\ x < 3 \end{cases} \]
Answer
No solution
Working
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 of 26/03/2026 - 12:50 — level ★★★★★
\[ \frac{2x - 3}{4} - \frac{x + 1}{3} > \frac{1}{6} \]
Answer
\[ x > \dfrac{15}{2} \]
Working
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 of 26/03/2026 - 13:05 — level ★★★★★
\[ 3(x - 2) - 2(2x + 1) \geq x - 5 \]
Answer
\[ x \leq -\dfrac{3}{2} \]
Working
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 of 26/03/2026 - 13:20 — level ★★★★★
\[ \begin{cases} \dfrac{x-1}{2} < \dfrac{x}{3} + 1 \\[8pt] 2x - 3 > x - 7 \end{cases} \]
Answer
\[ -4 < x < 9 \]
Working
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 of 26/03/2026 - 13:35 — 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
\[ x \geq -4 \]
Working
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
\[ \boxed{x \geq -4} \]