Inequalities — linear, solving, number line representation: Exercises

This page is your self-testing gym for linear inequalities. Each problem is graded by difficulty. Try it first with the solution collapsed, then open the > [!recall]- box to check every step.

Before we start, one shared picture. Every solution to a one-variable inequality is a ray (a half-line) or a segment on the number line: a starting point (the boundary) and a direction. Keep this image in mind.

Figure — Inequalities — linear, solving, number line representation

Level 1 — Recognition

Problem 1.1

Which of these is a linear inequality in one variable? $(a) 3 x - 4 < 0 (b) x^{2} > 9 (c) \frac{1}{x} \leq 2 (d) 5 \geq 2 x + 1$

Recall Solution 1.1

A linear inequality has the variable to the first power only — no $x^{2}$ , no $\frac{1}{x}$ , no roots.

(a) $3 x - 4 < 0$ → $x$ to the first power ✓ linear
(b) $x^{2} > 9$ → has $x^{2}$ ✗ (this is quadratic)
(c) $\frac{1}{x} \leq 2$ → $x$ in a denominator ✗
(d) $5 \geq 2 x + 1$ → $x$ to the first power ✓ linear

Answer: (a) and (d).

Problem 1.2

For the solution $x < 2$ , is the circle at $2$ open or closed, and does the arrow point left or right?

Recall Solution 1.2

The symbol is $<$ (strict) — it does not include the boundary, so the circle is open (hollow). "Less than" means "smaller numbers", which live to the left on the number line, so the arrow points left.

Answer: open circle at $2$ , arrow pointing left. In interval notation, $(- \infty, 2)$ .

Level 2 — Application

Problem 2.1

Solve $5 x - 3 \leq 12$ and write the answer in interval notation.

Recall Solution 2.1

$5 x - 3 \leq 12$ Add 3 to both sides (adding preserves order): $5 x \leq 15$ Divide by 5 — a positive number, so no flip: $x \leq 3$ Boundary included ( $\leq$ ) → closed circle at $3$ , arrow left. Answer: $x \in (- \infty, 3]$ .

Problem 2.2

Solve $- 4 x + 5 > 17$ .

Recall Solution 2.2

$- 4 x + 5 > 17$ Subtract 5 (order preserved): $- 4 x > 12$ Divide by $- 4$ — a negative number, so flip $>$ to $<$ : $x < - 3$ Answer: $x \in (- \infty, - 3)$ . Quick check with $x = - 4$ (should be inside): $- 4 (- 4) + 5 = 21 > 17$ ✓. With $x = 0$ (outside): $0 + 5 = 5 \neq > 17$ ✓ correctly excluded.

Problem 2.3

Solve $\frac{2 - 3 x}{5} \geq 4$ .

Recall Solution 2.3

$\frac{2 - 3 x}{5} \geq 4$ Multiply by 5 (positive, no flip — this clears the denominator): $2 - 3 x \geq 20$ Subtract 2: $- 3 x \geq 18$ Divide by $- 3$ (negative → flip $\geq$ to $\leq$ ): $x \leq - 6$ Answer: $x \in (- \infty, - 6]$ .

Level 3 — Analysis

Problem 3.1

Solve $3 (x - 2) < 5 x + 4$ .

Recall Solution 3.1

Expand the bracket first (distribute the 3): $3 x - 6 < 5 x + 4$ Gather $x$ -terms on one side. Subtract $5 x$ from both sides: $3 x - 5 x - 6 < 4 ⟹ - 2 x - 6 < 4$ Add 6: $- 2 x < 10$ Divide by $- 2$ (negative → flip): $x > - 5$ Answer: $x \in (- 5, \infty)$ . Check $x = 0$ : LHS $3 (0 - 2) = - 6$ , RHS $5 (0) + 4 = 4$ , and $- 6 < 4$ ✓.

Problem 3.2

Find all integers $x$ satisfying $- 7 < 2 x - 1 \leq 5$ (a compound inequality — two conditions at once).

Recall Solution 3.2

This says $2 x - 1$ is trapped between $- 7$ and $5$ . Do the same operation to all three parts. Add 1 to every part: $- 7 + 1 < 2 x \leq 5 + 1 ⟹ - 6 < 2 x \leq 6$ Divide every part by 2 (positive, no flip): $- 3 < x \leq 3$ The real solution is $(- 3, 3]$ . The integers inside: $- 3$ is excluded (strict), $3$ is included. Answer (integers): $- 2, - 1, 0, 1, 2, 3$ .

Level 4 — Synthesis

Problem 4.1

A phone plan charges a fixed ₹200 plus ₹2 per minute. You want your monthly bill to stay strictly under ₹500. For how many whole minutes $m$ can you talk?

Recall Solution 4.1

Translate to an inequality. Bill $= 200 + 2 m$ , and we want it strictly under 500: $200 + 2 m < 500$ Subtract 200: $2 m < 300$ Divide by 2 (positive): $m < 150$ Minutes are whole numbers and $m$ cannot be negative, so $m \in {0, 1, 2, \dots, 149}$ . Answer: at most $149$ whole minutes (150 would hit exactly ₹500, which is not under 500).

Problem 4.2

For which values of $k$ does the inequality $k x > 6$ have solution $x < \frac{6}{k}$ (i.e. the sign flips)? Explain using cases.

Recall Solution 4.2

To solve $k x > 6$ we divide by $k$ . The direction depends entirely on the sign of $k$ — this is a full case analysis.

Case $k > 0$ : dividing by a positive keeps $>$ . Solution $x > \frac{6}{k}$ . No flip.
Case $k < 0$ : dividing by a negative flips to $x < \frac{6}{k}$ . Flip happens here.
Case $k = 0$ : the inequality becomes $0 > 6$ , which is false for every $x$ — no solution (you can't divide by 0 at all).

Answer: the flip (giving $x < \frac{6}{k}$ ) occurs exactly when $k < 0$ . When $k = 0$ there is no solution; when $k > 0$ the sign stays.

Level 5 — Mastery

Problem 5.1

Solve $\frac{x - 1}{3} - \frac{2 x + 1}{4} \geq \frac{1}{2}$ and draw the solution on a number line.

Recall Solution 5.1

Clear all denominators at once. The least common multiple of $3, 4, 2$ is $12$ . Multiply every term by $12$ (positive → no flip): $12 \cdot \frac{x - 1}{3} - 12 \cdot \frac{2 x + 1}{4} \geq 12 \cdot \frac{1}{2}$ $4 (x - 1) - 3 (2 x + 1) \geq 6$ Expand both brackets: $4 x - 4 - 6 x - 3 \geq 6$ Combine like terms on the left ( $4 x - 6 x = - 2 x$ , and $- 4 - 3 = - 7$ ): $- 2 x - 7 \geq 6$ Add 7: $- 2 x \geq 13$ Divide by $- 2$ (negative → flip $\geq$ to $\leq$ ): $x \leq - \frac{13}{2} = - 6.5$ Boundary included ( $\leq$ ) → closed circle at $- 6.5$ , arrow left. Answer: $x \in (- \infty, - \frac{13}{2}]$ . Check $x = - 7$ : $\frac{- 8}{3} - \frac{- 13}{4} = - 2.667 + 3.25 = 0.583 \geq 0.5$ ✓.

Problem 5.2

Find every value of $x$ satisfying both $2 x + 1 > - 3$ and $3 - x \geq 1$ . Give the answer as a single interval.

Recall Solution 5.2

Solve each inequality separately, then take the overlap (both must hold — this is a mini system). First: $2 x + 1 > - 3 \Rightarrow 2 x > - 4 \Rightarrow x > - 2$ . Set: $(- 2, \infty)$ . Second: $3 - x \geq 1 \Rightarrow - x \geq - 2 \Rightarrow$ divide by $- 1$ (flip) $\Rightarrow x \leq 2$ . Set: $(- \infty, 2]$ . Overlap (intersection): numbers greater than $- 2$ and at most $2$ . Answer: $x \in (- 2, 2]$ . Left end open (from strict $>$ ), right end closed (from $\leq$ ).

Recall Master Recall

The one flip rule ::: Multiplying or dividing both sides by a negative reverses the inequality direction. Boundary rule ::: $<, >$ → open circle (excluded); $\leq, \geq$ → closed circle (included). Compound "and" ::: Take the intersection (overlap) of the two solution sets. Dividing by an unknown ::: Split into cases $> 0$ , $< 0$ , $= 0$ before dividing.

Connections

Linear Equations — solve these first, then just watch the direction.
Absolute Value Equations and Inequalities — each $∣ x ∣$ inequality splits into a compound one like Problem 5.2.
Systems of Inequalities — Problem 5.2 is the one-variable seed of this.
Quadratic Inequalities — the case-splitting of Problem 4.2 becomes essential.
Linear Programming — many inequalities at once, optimised.
Number Line and Real Numbers — the picture every solution lives on.

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