Solved Examples

How to Use This Section

Thirty fully worked examples sweep the whole chapter — solving over naturals, integers and reals; sign-flip situations; fractional and bracketed inequalities; number-line graphs; double inequalities; systems; and every standard class of word problem — averages, consecutive integers, geometry, formula ranges and mixtures — solved in full.

[Board Tip] Work each example yourself before reading the solution. In this chapter the errors are stereotyped — a missed flip, a wrong bracket, an ignored domain — so comparing your attempt against the worked steps is the fastest diagnostic there is.

Solving over Different Number Sets

Example 1: One inequality, two domains

Solve $24x < 100$ when (i) $x$ is a natural number (ii) $x$ is an integer.

Solution:

Step 1 — Divide by the positive 24. $x < \frac{100}{24} = \frac{25}{6} \approx 4.17$.

Step 2 — (i) Filter by naturals. $\{1, 2, 3, 4\}$.

Step 3 — (ii) Filter by integers. $\{\ldots, -2, -1, 0, 1, 2, 3, 4\}$.

Takeaway: Solve once, filter per domain.

Example 2: Strict boundary check

Solve $5x - 3 < 7$ when (i) $x$ is an integer (ii) $x$ is a real number.

Solution:

Step 1 — Collect. Add 3: $5x < 10$; divide by 5: $x < 2$.

Step 2 — (i) Integers. $\{\ldots, -1, 0, 1\}$ — 2 itself is excluded by the strict sign.

Step 3 — (ii) Reals. $(-\infty, 2)$.

Takeaway: The strict sign evicts the boundary in every domain.

Example 3: A right-facing ray

Solve $3x + 8 > 2$ when (i) $x$ is an integer (ii) $x$ is a real number.

Solution:

Step 1 — Collect. Subtract 8: $3x > -6$, so $x > -2$.

Step 2 — (i) Integers. $\{-1, 0, 1, 2, \ldots\}$.

Step 3 — (ii) Reals. $(-2, \infty)$.

Takeaway: Rays open away from the boundary; list integers from the first admitted value.

Example 4: The flip via −x

Solve $4x + 3 < 5x + 7$ for real $x$.

Solution:

Step 1 — Collect. Subtract $5x$ and 3: $-x < 4$.

Step 2 — Multiply by $-1$ and FLIP. $x > -4$.

Step 3 — Interval. $(-4, \infty)$; check x = 0: $3 < 7$ ✓.

Takeaway: $-x < 4$ is NOT $x < -4$ — the flip is compulsory.

Example 5: Flip with a coefficient

Solve $3x - 7 > 5x - 1$ for real $x$.

Solution:

Step 1 — Collect. $3x - 5x > -1 + 7$ gives $-2x > 6$.

Step 2 — Divide by $-2$ and FLIP. $x < -3$.

Step 3 — Interval. $(-\infty, -3)$; check x = $-4$: $-19 > -21$ ✓.

Takeaway: Same flip, negative coefficient this time — the rule does not care how the negative arose.

Brackets and Fractions

Example 6: Brackets both sides

Solve $3(2 - x) \ge 2(1 - x)$.

Solution:

Step 1 — Expand. $6 - 3x \ge 2 - 2x$.

Step 2 — Collect. Add 3x, subtract 2: $4 \ge x$, i.e. $x \le 4$.

Step 3 — Interval. $(-\infty, 4]$ — slack sign, closed end.

Takeaway: Reading "$4 \ge x$" as "$x \le 4$" is a free rewrite — no flip involved.

Example 7: Fractions of x on one side

Solve $x + \frac{x}{2} + \frac{x}{3} < 11$.

Solution:

Step 1 — Combine coefficients. $1 + \frac{1}{2} + \frac{1}{3} = \frac{6 + 3 + 2}{6} = \frac{11}{6}$, so the LHS is $\frac{11x}{6}$.

Step 2 — Solve. $\frac{11x}{6} < 11$ gives $x < 6$.

Step 3 — Interval. $(-\infty, 6)$.

Takeaway: Combine like terms BEFORE clearing denominators — here the 11's cancel beautifully.

Example 8: Cross-shaped fractions

Solve $\frac{3(x - 2)}{5} \le \frac{5(2 - x)}{3}$.

Solution:

Step 1 — Multiply by the positive 15. $9(x - 2) \le 25(2 - x)$.

Step 2 — Expand. $9x - 18 \le 50 - 25x$.

Step 3 — Collect. $34x \le 68$, so $x \le 2$: $(-\infty, 2]$.

Takeaway: LCM-multiply once; the rest is bracket algebra.

Example 9: Three fractions

Solve $\frac{x}{2} \ge \frac{5x - 2}{3} - \frac{7x - 3}{5}$.

Solution:

Step 1 — Multiply by 30 (LCM of 2, 3, 5). $15x \ge 10(5x - 2) - 6(7x - 3)$.

Step 2 — Simplify the right side. $50x - 20 - 42x + 18 = 8x - 2$.

Step 3 — Collect. $15x \ge 8x - 2$ gives $7x \ge -2$, so $x \ge -\frac{2}{7}$: $\left[-\frac{2}{7}, \infty\right)$.

Takeaway: Distribute the minus over BOTH terms of the second fraction — the $+18$ is where errors live.

Example 10: Brackets with a constant

Solve $2(2x + 3) - 10 < 6(x - 2)$.

Solution:

Step 1 — Expand. $4x + 6 - 10 < 6x - 12$, i.e. $4x - 4 < 6x - 12$.

Step 2 — Collect. $8 < 2x$, so $x > 4$.

Step 3 — Interval. $(4, \infty)$.

Takeaway: Keep x's coefficient positive by collecting onto the larger side — no flip needed then.

Example 11: Three fractions, strict

Solve $\frac{x}{4} < \frac{5x - 2}{3} - \frac{7x - 3}{5}$.

Solution:

Step 1 — Multiply by 60. $15x < 20(5x - 2) - 12(7x - 3)$.

Step 2 — Simplify the right. $100x - 40 - 84x + 36 = 16x - 4$.

Step 3 — Collect. $15x < 16x - 4$ gives $4 < x$: $(4, \infty)$.

Takeaway: Moving 15x right instead of 16x left keeps everything positive — choose the collection direction wisely.

Example 12: A heavier three-fraction chain

Solve $\frac{2x - 1}{3} \ge \frac{3x - 2}{4} - \frac{2 - x}{5}$.

Solution:

Step 1 — Multiply by 60. $20(2x - 1) \ge 15(3x - 2) - 12(2 - x)$.

Step 2 — Simplify each side. LHS $= 40x - 20$; RHS $= 45x - 30 - 24 + 12x = 57x - 54$.

Step 3 — Collect. $-20 + 54 \ge 57x - 40x$ gives $34 \ge 17x$, so $x \le 2$: $(-\infty, 2]$.

Takeaway: $-12(2 - x) = -24 + 12x$ — sign care inside the subtracted bracket decides the whole problem.

Number-Line Graphs

Example 13: Solve and graph

Solve $3x - 2 < 2x + 1$ and show the graph of the solutions.

Solution:

Step 1 — Collect. Subtract 2x, add 2: $x < 3$.

Step 2 — Graph. Strict sign → OPEN circle at 3; solutions below → dark line to the left: $(-\infty, 3)$.

Takeaway: Circle from the sign, direction from the inequality.

Example 14: Slack version

Solve $5x - 3 \ge 3x - 5$ and graph the solutions.

Solution:

Step 1 — Collect. $2x \ge -2$, so $x \ge -1$.

Step 2 — Graph. Slack sign → FILLED circle at $-1$; dark line rightwards: $[-1, \infty)$.

Takeaway: The filled circle earns the "boundary included" mark.

Example 15: Brackets then graph

Solve $3(1 - x) < 2(x + 4)$ and graph the solutions.

Solution:

Step 1 — Expand. $3 - 3x < 2x + 8$.

Step 2 — Collect. $-5 < 5x$, so $x > -1$.

Step 3 — Graph. Open circle at $-1$, dark line rightwards: $(-1, \infty)$.

Takeaway: Collecting to keep the coefficient positive avoided a flip entirely.

Example 16: A graph read backwards

A number line shows a filled circle at 2 with the dark line running left. Write the inequality and the interval.

Solution:

Step 1 — Decode the circle. Filled: 2 is INCLUDED — slack sign.

Step 2 — Decode the direction. Dark line left: solutions below 2.

Step 3 — Write both forms. $x \le 2$; interval $(-\infty, 2]$.

Takeaway: The graph-to-algebra dictionary runs both ways — exams test the reverse direction too.

Double Inequalities

Example 17: The basic closed chain

Solve $2 \le 3x - 4 \le 5$.

Solution:

Step 1 — Add 4 to all three parts. $6 \le 3x \le 9$.

Step 2 — Divide by the positive 3. $2 \le x \le 3$.

Step 3 — Interval. The closed interval $[2, 3]$ — both signs slack throughout.

Takeaway: Whatever you do to the middle, do to both ends.

Example 18: Chain with a negative multiplier

Solve $-3 \le 4 - \frac{7x}{2} \le 18$.

Solution:

Step 1 — Subtract 4 throughout. $-7 \le -\frac{7x}{2} \le 14$.

Step 2 — Multiply by $-\frac{2}{7}$ — negative, BOTH signs flip. $2 \ge x \ge -4$.

Step 3 — Reorder. $-4 \le x \le 2$, i.e. $[-4, 2]$.

Takeaway: Flip both signs, then rewrite smallest-to-largest — two moves, one deliberate step.

Example 19: Mixed strict-slack chain

Solve $-15 < \frac{3(x - 2)}{5} \le 0$.

Solution:

Step 1 — Multiply by the positive $\frac{5}{3}$. $-25 < x - 2 \le 0$.

Step 2 — Add 2 throughout. $-23 < x \le 2$.

Step 3 — Interval. $(-23, 2]$ — the strict and slack ends keep their own brackets.

Takeaway: Each end of a chain carries its own sign type all the way to the final bracket.

Example 20: Fraction in the middle

Solve $7 \le \frac{3x + 11}{2} \le 11$.

Solution:

Step 1 — Multiply all parts by 2. $14 \le 3x + 11 \le 22$.

Step 2 — Subtract 11. $3 \le 3x \le 11$.

Step 3 — Divide by 3. $1 \le x \le \frac{11}{3}$: interval $\left[1, \frac{11}{3}\right]$.

Takeaway: Peel the middle expression layer by layer — multiply, subtract, divide.

Example 21: A double negative to simplify first

Solve $-12 < 4 - \frac{3x}{-5} \le 2$.

Solution:

Step 1 — Simplify the double negative. $-\frac{3x}{-5} = +\frac{3x}{5}$, so the chain is $-12 < 4 + \frac{3x}{5} \le 2$.

Step 2 — Subtract 4 throughout. $-16 < \frac{3x}{5} \le -2$.

Step 3 — Multiply by the positive $\frac{5}{3}$. $-\frac{80}{3} < x \le -\frac{10}{3}$: interval $\left(-\frac{80}{3}, -\frac{10}{3}\right]$.

Takeaway: Tidy the expression BEFORE operating on the chain — the buried double negative changes everything.

Systems of Inequalities

Example 22: A two-sided trap

Solve the system $2(x - 1) < x + 5$ and $3(x + 2) > 2 - x$.

Solution:

Step 1 — Solve the first. $2x - 2 < x + 5$ gives $x < 7$.

Step 2 — Solve the second. $3x + 6 > 2 - x$ gives $4x > -4$, so $x > -1$.

Step 3 — Intersect. $(-1, 7)$ — the overlap of the two rays.

Takeaway: Opposite-facing conditions trap x in an interval; same-facing ones reduce to the tighter ray.

Example 23: A system with no solution

Solve $x - 4 > 0$ and $5 - x > 2$.

Solution:

Step 1 — Solve each. First: $x > 4$. Second: $x < 3$.

Step 2 — Attempt the intersection. The rays point away from each other with a gap between 3 and 4 — no overlap.

Step 3 — Conclude. Solution set $\varnothing$.

Takeaway: State the empty set explicitly — it is a complete, correct answer.

Example 24: Counting integer solutions of a system

How many integers satisfy both $x \ge -2$ and $x < 3$?

Solution:

Step 1 — Intersect. $[-2, 3)$.

Step 2 — List the integers. $-2, -1, 0, 1, 2$.

Step 3 — Count. Five — the slack end admits $-2$, the strict end excludes 3.

Takeaway: Bracket types decide the count at each end.

Example 25: Always true, never true

Solve (i) $4x + 7 < 2(2x + 3)$ (ii) $3(x + 1) \ge 3x + 2$.

Solution:

Step 1 — (i) Expand. $4x + 7 < 4x + 6$ — the x-terms cancel, leaving $7 < 6$: FALSE for every x.

Step 2 — Conclude (i). Solution $\varnothing$.

Step 3 — (ii) Expand. $3x + 3 \ge 3x + 2$ collapses to $3 \ge 2$: TRUE for every x.

Step 4 — Conclude (ii). Solution $\mathbb{R}$.

Takeaway: When the variable cancels, the leftover numerical statement decides: everything or nothing.

Word Problems

Example 26: Average with four tests

Scores so far: 72, 78, 69. What must the fourth score be for an average of at least 75?

Solution:

Step 1 — Translate. $\frac{72 + 78 + 69 + x}{4} \ge 75$, i.e. $219 + x \ge 300$.

Step 2 — Solve. $x \ge 81$.

Step 3 — Interpret. At least 81 marks (and at most 100, in context).

Takeaway: The four-step recipe scales to any number of tests.

Example 27: Consecutive evens, tighter bounds

Find all pairs of consecutive even positive integers, both smaller than 14, whose sum is more than 20.

Solution:

Step 1 — Conditions on the smaller even x. Both below 14: $x + 2 < 14$, so $x < 12$; sum: $2x + 2 > 20$, so $x > 9$.

Step 2 — Filter by parity. Even x in $(9, 12)$: $x = 10$ only.

Step 3 — List. Single pair: $(10, 12)$.

Takeaway: Tight bounds can shrink the answer to one pair — or none; count carefully.

Example 28: Boric acid dilution

640 litres of 8% boric acid solution is to be diluted with 2% solution so the mixture is more than 4% but less than 6% boric acid. How much 2% solution must be added?

Solution:

Step 1 — Track acid and volume. Adding x litres: acid $= 0.02x + 0.08(640) = 0.02x + 51.2$; volume $= 640 + x$.

Step 2 — More than 4%. $0.02x + 51.2 > 0.04(640 + x) = 25.6 + 0.04x$ gives $25.6 > 0.02x$, so $x < 1280$.

Step 3 — Less than 6%. $0.02x + 51.2 < 38.4 + 0.06x$ gives $12.8 < 0.04x$, so $x > 320$.

Step 4 — Combine. Between 320 and 1280 litres.

Takeaway: Here the ADDED solution is the weaker one — the same acid-tracking method handles dilution and enrichment alike.

Example 29: IQ range, younger group

With $IQ = \frac{MA}{CA} \times 100$ and $CA = 10$, find the mental-age range if $90 \le IQ \le 120$.

Solution:

Step 1 — Substitute CA = 10. $\frac{MA}{10} \times 100 = 10\,MA$, so $90 \le 10\,MA \le 120$.

Step 2 — Divide all parts by 10. $9 \le MA \le 12$ years.

Takeaway: A clean-coefficient formula turns the chain into one division.

Example 30: Interval plus integrality

A parking lot charges ₹40 for the first hour and ₹25 for each additional hour. With at most ₹200, for how many whole hours can a car be parked?

Solution:

Step 1 — Model the cost. For $h \ge 1$ hours: cost $= 40 + 25(h - 1)$.

Step 2 — Translate the budget. $40 + 25(h - 1) \le 200$ gives $25(h - 1) \le 160$, so $h - 1 \le 6.4$, i.e. $h \le 7.4$.

Step 3 — Interpret with integrality. Whole hours: $h \in \{1, 2, \ldots, 7\}$ — at most 7 hours.

Takeaway: Step 4 of the recipe: trim to whole, positive values — 7.4 hours is not a purchasable quantity.