Chapter 5 · Linear Inequalities

5.2 Solving Linear Inequalities in One Variable

Solve a linear inequality the way you solve an equation — clear fractions, expand, collect $x$ on one side — but turn the sign round whenever you divide by a negative number. The answer is an interval, drawn on the number line.

A mobile plan costs $\text{₹}\,199$ a month plus $\text{₹}\,1.50$ per minute of calls; another costs a flat $\text{₹}\,399$. For which call times is the first plan cheaper? Writing $m$ for the minutes, the question is $199+1.5m\lt399$. You already know how to solve the equation $199+1.5m=399$: subtract $199$, divide by $1.5$. Section 5.1 showed that the same two moves are allowed for inequalities, because $1.5$ is positive — so $m\lt\tfrac{200}{1.5}\approx133.3$, and the first plan wins for any call time up to $133$ whole minutes.

In this section we turn that idea into a reliable method. We solve linear inequalities with brackets, fractions and $x$ on both sides; meet the surprising cases where the $x$ disappears and the answer is "every real number" or "no number at all"; solve systems of inequalities and double inequalities such as $-5\le3-2x\lt9$; and finally translate word problems about marks, budgets, ages and mixtures into inequalities. Every answer will be written in the interval notation of Section 1.2 and drawn on the number line.

Solving a Linear Inequality

📐 Theorem 5.3 — Solution of $ax+b\lt0$

Let $a$ and $b$ be real numbers with $a\ne0$. Then

if $a\gt0$: $\ ax+b\lt0\iff x\lt-\dfrac ba$, with solution set $\left(-\infty,-\dfrac ba\right)$;
if $a\lt0$: $\ ax+b\lt0\iff x\gt-\dfrac ba$, with solution set $\left(-\dfrac ba,\infty\right)$.

For $ax+b\le0$ the same holds with the end point $-\dfrac ba$ included (a square bracket); for $ax+b\gt0$ and $ax+b\ge0$ the directions are reversed.

Proof. Subtracting $b$ (Rule 1) gives the equivalent inequality $ax\lt-b$. If $a\gt0$, dividing by $a$ keeps the sign (Rule 2): $x\lt-\dfrac ba$. If $a\lt0$, dividing by $a$ reverses it (Rule 3): $x\gt-\dfrac ba$. Every step can be undone, so the final inequality has exactly the same solutions as the first. $\blacksquare$

So the solution set of a linear inequality in one variable is always a ray of the number line, starting at the single point $x=-\tfrac ba$ where the two sides would be equal. That boundary point is a useful check: the solution of the equation $ax+b=0$ tells you where the ray starts, and one test value tells you which way it points.

🧭 Problem-solving strategy — Solving a linear inequality

Clear fractions by multiplying both sides by the LCM of the denominators — a positive number, so no flip.
Expand brackets and simplify each side.
Collect the $x$-terms on one side and the numbers on the other (Rule 1). Collecting $x$ on the side with the larger coefficient keeps that coefficient positive and avoids a flip.
Divide by the coefficient of $x$. If it is negative, reverse the sign.
Write and draw the solution set: a hollow end point ○ and a round bracket for $\lt$ or $\gt$; a filled end point ● and a square bracket for $\le$ or $\ge$. Next to $\pm\infty$ the bracket is always round.
Check with one value inside the solution set and one outside.

✏️ Example 1 — Three basic inequalities

Solve, and write the solution set as an interval: (a) $3x-5\lt7$ (b) $5-2x\ge11$ (c) $4x+3\lt6x+7$.

💡 Show solution▶

(a) Add $5$: $3x\lt12$. Divide by $3$: $x\lt4$. Solution set $(-\infty,4)$, a ray to the left of a hollow dot at $4$. Check: $x=0$: $-5\lt7$ ✓; $x=5$: $10\lt7$ ✗.

(b) Subtract $5$: $-2x\ge6$. Divide by $-2$ and reverse: $x\le-3$. Solution set $(-\infty,-3]$. Check: $x=-4$: $13\ge11$ ✓; $x=0$: $5\ge11$ ✗.

(c) Collect The larger coefficient is on the right, so move $4x$ there and $7$ to the left: $3-7\lt6x-4x$, i.e. $-4\lt2x$.

(c) Divide $x\gt-2$ (dividing by positive $2$ — no flip needed). Solution set $(-2,\infty)$. Check: $x=0$: $3\lt7$ ✓; $x=-3$: $-9\lt-11$ ✗.

✏️ Example 2 — Brackets, fractions, and the cases where $x$ disappears

Solve: (a) $3(x-2)-2(1-x)\le4(x+1)$ (b) $\dfrac{3x-1}{4}-\dfrac{x+2}{3}\le\dfrac{x-5}{6}$ (c) $2(x+3)\gt2x+1$ (d) $3x-4\le3(x-2)$

💡 Show solution▶

(a) Expand: $3x-6-2+2x\le4x+4$, i.e. $5x-8\le4x+4$. Collect: $x\le12$. Solution set $(-\infty,12]$. Check at the boundary $x=12$: $30-2(-11)=52$ and $4(13)=52$, equal ✓.

(b) Clear fractions The LCM of $4$, $3$, $6$ is $12$. Multiplying by $12$: $3(3x-1)-4(x+2)\le2(x-5)$.

(b) Solve $9x-3-4x-8\le2x-10$, so $5x-11\le2x-10$, $3x\le1$, $x\le\tfrac13$. Solution set $\left(-\infty,\tfrac13\right]$. Check: $x=0$: $-\tfrac14-\tfrac23=-\tfrac{11}{12}\le-\tfrac56$ ✓ (since $-\tfrac{11}{12}\lt-\tfrac{10}{12}$).

(c) $2x+6\gt2x+1$. Subtracting $2x$ leaves $6\gt1$ — a true statement with no $x$ in it. It is true whatever $x$ is, so the solution set is $\mathbb{R}=(-\infty,\infty)$.

(d) $3x-4\le3x-6$. Subtracting $3x$ leaves $-4\le-6$ — false for every $x$. The solution set is the empty set $\phi$.

Systems of Inequalities and Double Inequalities

Often a quantity must satisfy several conditions at once: a vaccine fridge must be at least $2\,^\circ\text{C}$ and at most $8\,^\circ\text{C}$; a number of guests must be within the budget and above the minimum booking. The values that pass every test are the ones common to all the solution sets — their intersection, from Section 1.3.

📘 Definition 5.4 — System of linear inequalities in one variable

Two or more linear inequalities in the same variable that must hold simultaneously form a system of linear inequalities. Its solution set is the intersection of the solution sets of the individual inequalities; if the intersection is empty, the system has no solution.

A double inequality $p\lt ax+b\lt q$ is a system of two inequalities, $p\lt ax+b$ and $ax+b\lt q$, written in one line (similarly with $\le$).

📐 Theorem 5.4 — Working with a double inequality

Let $p\lt E\lt q$, where $E$ is an expression in $x$. Then, for every real $c$ and every real $k\ne0$:

$p+c\lt E+c\lt q+c$;
if $k\gt0$: $\ kp\lt kE\lt kq$;
if $k\lt0$: $\ kp\gt kE\gt kq$ — both signs reverse, which we then rewrite as $kq\lt kE\lt kp$.

The same holds with $\le$ in either or both places. Proof. Apply the corresponding rule of Theorem 5.1 to each of the two inequalities $p\lt E$ and $E\lt q$ separately. $\blacksquare$

✏️ Example 3 — Solving a system

Solve each system and show the solution set on the number line.
(a) $2x+1\ge x-3$ and $5-3x\gt-7$ (b) $x-2\gt3$ and $2x+1\lt7$

💡 Show solution▶

(a) First $2x+1\ge x-3\iff x\ge-4$: the ray $[-4,\infty)$.

(a) Second $5-3x\gt-7\iff-3x\gt-12\iff x\lt4$ (flip): the ray $(-\infty,4)$.

(a) Intersect Both hold for $-4\le x\lt4$: solution set $[-4,4)$ — filled dot at $-4$, hollow dot at $4$, shaded between. Check $x=0$: $1\ge-3$ ✓ and $5\gt-7$ ✓; $x=4$: $5-12=-7\gt-7$ ✗.

(b) The first gives $x\gt5$, the second $2x\lt6$, i.e. $x\lt3$. No number is both greater than $5$ and less than $3$: the rays do not overlap and the solution set is $\phi$.

✏️ Example 4 — Double inequalities

Solve: (a) $-5\le3-2x\lt9$ (b) $-3\le\dfrac{5x-2}{4}\le2$

💡 Show solution▶

(a) Subtract 3 From all three parts: $-8\le-2x\lt6$.

(a) Divide by −2 Both signs reverse: $4\ge x\gt-3$. Rewrite with the smaller number on the left: $-3\lt x\le4$. Solution set $(-3,4]$.

(a) Check $x=4$: $3-8=-5$, and $-5\le-5\lt9$ ✓ (included). $x=-3$: $3+6=9$, and $9\lt9$ ✗ (excluded) — matching the brackets.

(b) Multiply by $4$: $-12\le5x-2\le8$. Add $2$: $-10\le5x\le10$. Divide by $5$: $-2\le x\le2$. Solution set $[-2,2]$.

Word Problems

The mathematics in a word problem is usually easy; the skill is the translation. Most of it comes down to a few phrases, and to remembering at the end what kind of number the answer must be.

🧭 Problem-solving strategy — Inequality word problems

Name the unknown, with its unit: "let $x$ be the number of guests", "let $w$ be the volume of water in mL".
Translate the phrases. "at least $a$", "not less than $a$", "minimum $a$": $\ \ge a$. "at most $a$", "not more than $a$", "maximum $a$", "does not exceed $a$": $\ \le a$. "more than", "exceeds": $\gt$. "less than", "below": $\lt$. "between $a$ and $b$": $a\lt x\lt b$, unless the problem says the ends are included.
Add hidden conditions: lengths and quantities are positive, marks are at most the maximum, numbers of people or objects are whole numbers.
Solve, then interpret: round in the safe direction ($n\le16.7$ allows at most $16$ cartons; $w\ge10.9$ needs at least $11$ weeks), and answer in words with units.

✏️ Example 5 — Consecutive odd numbers

Find all pairs of consecutive odd natural numbers, both greater than $10$, whose sum is less than $40$.

💡 Show solution▶

Set up Let the smaller number be $x$ (odd); the next odd number is $x+2$. Conditions: $x\gt10$ and $x+(x+2)\lt40$.

Solve $2x+2\lt40\iff2x\lt38\iff x\lt19$. Together: $10\lt x\lt19$.

Interpret The odd numbers strictly between $10$ and $19$ are $11, 13, 15, 17$. The pairs are $(11,13)$, $(13,15)$, $(15,17)$, $(17,19)$.

Check $17+19=36\lt40$ ✓, but the next pair gives $19+21=40$, which is not less than $40$ ✓.

✏️ Example 6 — Diluting a solution

A lab has $300$ mL of a solution that is $30\%$ acid. Water is added so that the acid concentration ends up between $12\%$ and $20\%$ (end values excluded). How much water may be added?

💡 Show solution▶

Set up Let $w$ mL of water be added ($w\ge0$). The acid stays at $30\%$ of $300=90$ mL; the total becomes $300+w$ mL. Condition: $0.12\lt\dfrac{90}{300+w}\lt0.20$.

Clear the fraction $300+w$ is positive, so multiplying every part by it keeps the signs: $0.12(300+w)\lt90\lt0.20(300+w)$ — two linear inequalities.

Solve each Left: $36+0.12w\lt90\iff0.12w\lt54\iff w\lt450$. Right: $90\lt60+0.2w\iff30\lt0.2w\iff w\gt150$.

Answer $150\lt w\lt450$: more than $150$ mL but less than $450$ mL of water. Check $w=300$: $\tfrac{90}{600}=15\%$, inside ✓; $w=150$: $\tfrac{90}{450}=20\%$, exactly the excluded end ✓.

⚠️ Common mistakes when solving

1. Flipping only one sign of a double inequality. Dividing $-8\le-2x\lt6$ by $-2$ reverses both signs: $4\ge x\gt-3$. Then rewrite it as $-3\lt x\le4$ — and keep each end point with its own sign ($4$ was attached to $\le$, so it stays included).

2. Square brackets at infinity. $[2,\infty]$ is wrong: $\infty$ is not a number and cannot be included. Write $[2,\infty)$.

3. Mixing up "at least" and "more than". "At least $60$" includes $60$ ($\ge$); "more than $60$" does not ($\gt$). The difference decides whether the end point is filled or hollow.

4. Rounding the wrong way. If $n\le12.8$ cartons fit, the answer is $12$, not $13$. If $w\ge10.86$ weeks are needed, the answer is $11$, not $10$.

5. "Solving" an empty system. $x\gt5$ and $x\lt3$ do not give $3\lt x\lt5$; they have no common solution, $\phi$. Always draw both rays before you write the intersection.

📘 In the CBSE syllabus — school assessment only (not in the year-end board exam)

Graphical solution of linear inequalities in two variables is taught and tested in school, but not in the year-end board exam. Neither JEE Main nor JEE Advanced 2026 lists it; the shaded regions come back in Class 12 as the feasible regions of linear programming.

Half-planes. The line $ax+by=c$ cuts the plane into two half-planes. At every point on one side $ax+by\lt c$, and at every point on the other side $ax+by\gt c$ (moving across the line is the only way to change the sign of $ax+by-c$). So the solution set of $ax+by\lt c$ is one whole half-plane.

Draw the boundary $ax+by=c$ through two easy points (often its intercepts). Use a solid line for $\le$ or $\ge$ (points on the line are solutions) and a dashed line for $\lt$ or $\gt$ (they are not).
Test a point not on the line — $(0,0)$ when the line misses the origin. If the test point satisfies the inequality, shade its side; otherwise shade the other side.
For a system, shade each inequality and keep the region common to all of them. It may be bounded, unbounded or empty.

Worked example 1. Solve $3x+2y\le12$ graphically.
Solution. The boundary $3x+2y=12$ passes through $(4,0)$ and $(0,6)$; draw it solid, since "$=$" is allowed. Test $(0,0)$: $0\le12$ is true, so the solution set is the closed half-plane on the origin's side of the line, including the line itself. Check $(4,3)$, on the far side: $12+6=18\not\le12$ ✓.

Worked example 2. Solve the system $x+2y\le8$, $2x+y\le10$, $x\ge0$, $y\ge0$ graphically.
Solution. $x+2y=8$ meets the axes at $(8,0)$ and $(0,4)$; $2x+y=10$ at $(5,0)$ and $(0,10)$. $(0,0)$ satisfies both ($0\le8$, $0\le10$), so shade towards the origin for each; $x\ge0$, $y\ge0$ keep the first quadrant. The two lines cross where $x+2y=8$ and $2x+y=10$: subtracting twice the first from the second gives $-3y=-6$, so $y=2$, $x=4$. The solution region is the quadrilateral with corners $(0,0)$, $(5,0)$, $(4,2)$, $(0,4)$, boundary included. Check $(4,2)$: $4+4=8\le8$ and $8+2=10\le10$ ✓.

🚀 Beyond the board syllabus — a challenge: modulus inequalities

Not examined in the Board exam, and not a listed topic in JEE Main 2026 or JEE Advanced 2026 either — but combining linear inequalities with the modulus function $|x|$ of Section 2.3 is useful practice, and the idea turns up inside many entrance-exam problems (domains, distances, error bounds). Since $|x-a|$ is the distance from $x$ to $a$ on the number line, for $r\gt0$:

$|x-a|\lt r\iff a-r\lt x\lt a+r$ (within distance $r$ of $a$: one interval), and $|x-a|\gt r\iff x\lt a-r$ or $x\gt a+r$ (further than $r$ from $a$: two rays).

Worked example. Solve (i) $|2x-3|\le5$ (ii) $|x-2|\gt3$.
Solution. (i) $-5\le2x-3\le5$, so $-2\le2x\le8$ and $-1\le x\le4$: the interval $[-1,4]$. Check: $x=4$ gives $|5|=5\le5$ ✓. (ii) $x-2\lt-3$ or $x-2\gt3$, so $x\lt-1$ or $x\gt5$: the union $(-\infty,-1)\cup(5,\infty)$. Check: $x=0$ gives $|-2|=2$, not greater than $3$, and indeed $0$ is not in the set ✓.

Exercises

Solve each inequality for real $x$, write the solution set in interval notation and show it on the number line (except in part 12).

1 $4x+3\lt19$
2 $7-3x\ge-8$
3 $5x-2\gt3x+8$
4 $2(3-x)\le5x-8$
5 $-(x+4)\gt2(x-5)$
6 $\dfrac x3+1\ge\dfrac x2-1$
7 $\dfrac{x-1}{4}-\dfrac{x+3}{6}\lt1$
8 $\dfrac{2x+3}{5}-1\ge\dfrac{x-2}{2}$
9 $3(2x-1)-2(3x+1)\lt4$
10 $5(x-2)\ge5x+1$
11 $0.3x-1.2\le0.5x+0.4$
12 Solve $12x-5\lt31$ when (a) $x\in\mathbb{N}$ (b) $x\in\mathbb{Z}$.

💡 Show solutions▶

1 $4x\lt16$, $x\lt4$: $(-\infty,4)$.

2 $-3x\ge-15$; divide by $-3$ and flip: $x\le5$: $(-\infty,5]$.

3 $2x\gt10$, $x\gt5$: $(5,\infty)$.

4 $6-2x\le5x-8$, $14\le7x$, $x\ge2$: $[2,\infty)$.

5 $-x-4\gt2x-10$, $6\gt3x$, $x\lt2$: $(-\infty,2)$.

6 Multiply by $6$: $2x+6\ge3x-6$, so $12\ge x$: $(-\infty,12]$.

7 Multiply by $12$: $3(x-1)-2(x+3)\lt12$, $x-9\lt12$, $x\lt21$: $(-\infty,21)$.

8 Multiply by $10$: $2(2x+3)-10\ge5(x-2)$, $4x-4\ge5x-10$, $6\ge x$: $(-\infty,6]$.

9 $6x-3-6x-2\lt4$ gives $-5\lt4$, true for every $x$: $\mathbb{R}=(-\infty,\infty)$.

10 $5x-10\ge5x+1$ gives $-10\ge1$, false for every $x$: no solution, $\phi$.

11 $-1.2-0.4\le0.5x-0.3x$, i.e. $-1.6\le0.2x$, so $x\ge-8$: $[-8,\infty)$. (Or multiply by $10$ first: $3x-12\le5x+4$.)

12 $12x\lt36$, $x\lt3$. (a) $\{1,2\}$ (b) $\{\ldots,-2,-1,0,1,2\}$.

Systems and double inequalities. Solve and give the solution set as an interval (or $\phi$).

1 $x+4\gt1$ and $2x-1\le7$
2 $3x-2\lt2x+1$ and $4-x\le2x+10$
3 $5x+1\gt3(x+3)$ and $1-2x\gt-11$
4 $2x-3\ge7$ and $x+1\lt4$
5 $-4\lt2x+2\le10$
6 $1\le7-3x\lt13$
7 $-2\lt\dfrac{3x+4}{5}\le5$
8 How many integers $x$ satisfy $-7\le3-2x\lt11$?

Check: solve ax + b ? cx + d and see the ray

$x\,+$$x\,+$

collected—

solution—

interval—

💡 Show solutions▶

1 $x\gt-3$ and $x\le4$: $(-3,4]$.

2 $x\lt3$, and $-6\le3x$ gives $x\ge-2$: $[-2,3)$.

3 $5x+1\gt3x+9$ gives $x\gt4$; $-2x\gt-12$ gives $x\lt6$ (flip): $(4,6)$.

4 $x\ge5$ and $x\lt3$: no common value, $\phi$.

5 Subtract $2$: $-6\lt2x\le8$; divide by $2$: $-3\lt x\le4$: $(-3,4]$.

6 Subtract $7$: $-6\le-3x\lt6$; divide by $-3$, flipping both signs: $2\ge x\gt-2$, i.e. $-2\lt x\le2$: $(-2,2]$.

7 Multiply by $5$: $-10\lt3x+4\le25$; subtract $4$: $-14\lt3x\le21$; divide by $3$: $-\tfrac{14}{3}\lt x\le7$: $\left(-\tfrac{14}3,7\right]$.

8 $-10\le-2x\lt8$, so $5\ge x\gt-4$, i.e. $-4\lt x\le5$. The integers are $-3,-2,\ldots,5$: $9$ integers.

III

Word problems. Set up an inequality (or a system), solve it, and answer in words.

1 Lift. A goods lift can carry at most $680$ kg. Two operators with a total mass of $140$ kg ride in it with crates of $45$ kg each. How many crates can go in one trip?
2 Phone plans. Plan A costs $\text{₹}\,199$ per month plus $\text{₹}\,1.50$ per minute; Plan B costs a flat $\text{₹}\,399$. For which whole numbers of minutes is Plan A cheaper?
3 Scholarship. A student scored $58$ and $71$ (out of $100$) in two terms. A scholarship needs an average of at least $65$ over three terms. What marks in the third term qualify?
4 Consecutive even numbers. Find all pairs of consecutive even natural numbers, both greater than $5$, whose sum is less than $30$.
5 Triangle. In a triangle the longest side is twice the shortest side, and the third side is $3$ cm longer than the shortest. The perimeter must not exceed $39$ cm. What is the greatest possible length of the shortest side? (Make sure the "longest" side really is the longest.)
6 Water. At normal pressure water is liquid when $0\lt C\lt100$ (degrees Celsius). Using $F=\dfrac95C+32$, find the range in degrees Fahrenheit.
7 Saline. $1000$ mL of a $24\%$ salt solution is to be diluted with water so that the salt concentration is between $8\%$ and $12\%$ (ends excluded). How much water must be added?
8 Savings. Asha has $\text{₹}\,1200$ and saves $\text{₹}\,350$ a week. She needs at least $\text{₹}\,5000$ for a bicycle. What is the least number of weeks she must save?

💡 Show solutions▶

1 $140+45n\le680\iff45n\le540\iff n\le12$. At most $12$ crates.

2 $199+1.5m\lt399\iff1.5m\lt200\iff m\lt133.\overline3$. Plan A is cheaper for $0$ to $133$ minutes.

3 $\dfrac{58+71+x}{3}\ge65\iff129+x\ge195\iff x\ge66$; also $x\le100$. Any mark from $66$ to $100$: $66\le x\le100$.

4 Smaller number $x$ (even), $x\gt5$ and $x+(x+2)\lt30$, i.e. $x\lt14$. So $x\in\{6,8,10,12\}$: pairs $(6,8)$, $(8,10)$, $(10,12)$, $(12,14)$.

5 Sides $s$, $2s$, $s+3$: $4s+3\le39\iff s\le9$. For $2s$ to be the longest, $2s\ge s+3$, i.e. $s\ge3$ (and $s+(s+3)\gt2s$ always holds, so a triangle exists). So $3\le s\le9$: the greatest shortest side is $9$ cm (sides $9$, $18$, $12$ cm, perimeter $39$ cm).

6 Multiply by $\tfrac95$: $0\lt\tfrac95C\lt180$; add $32$: $32\lt F\lt212$ (degrees Fahrenheit).

7 Salt $=240$ mL; with $w$ mL of water: $0.08\lt\dfrac{240}{1000+w}\lt0.12$. Multiply by the positive $1000+w$: $80+0.08w\lt240$ gives $w\lt2000$, and $240\lt120+0.12w$ gives $w\gt1000$. So $1000\lt w\lt2000$ mL.

8 $1200+350w\ge5000\iff350w\ge3800\iff w\ge10.86$ (to 2 d.p.). She must save for at least $11$ weeks.

📘 School-assessment practice (graphical solution in two variables — see the 📘 box; not in the year-end board exam). Describe each solution region: the boundary (solid or dashed), the test point used and the side shaded.

1 $2x+y\ge6$
2 $3x-4y\lt12$
3 $y\gt x$
4 $x\lt2$
5 The system $x+y\le6$, $x\ge1$, $y\ge2$; find the corners of the region.
6 Show that the system $x+y\le2$, $x+y\ge5$ has no solution.

💡 Show solutions▶

1 Solid line through $(3,0)$ and $(0,6)$. Test $(0,0)$: $0\ge6$ is false, so shade the side away from the origin (line included).

2 Dashed line through $(4,0)$ and $(0,-3)$. Test $(0,0)$: $0\lt12$ is true, so shade the origin's side (line excluded).

3 Dashed line $y=x$ through the origin, so test another point, e.g. $(0,1)$: $1\gt0$ is true. Shade the side containing $(0,1)$ — above the line.

4 Dashed vertical line $x=2$. Test $(0,0)$: $0\lt2$ is true. Shade everything to the left of the line.

5 Solid lines $x+y=6$, $x=1$, $y=2$; $(0,0)$ satisfies the first only, so shade below $x+y=6$, right of $x=1$ and above $y=2$. The region is the triangle with corners $(1,2)$, $(4,2)$ and $(1,5)$ ($y=2$ meets $x+y=6$ at $x=4$; $x=1$ meets it at $y=5$).

6 The boundaries $x+y=2$ and $x+y=5$ are parallel. The first inequality keeps the side of $x+y=2$ containing the origin; the second keeps the side of $x+y=5$ away from it. No point can have $x+y$ both $\le2$ and $\ge5$, so the solution set is $\phi$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: Board papers ask you to solve an inequality or a system and represent the solution on the number line (keep the drawing — it carries marks), and set word problems as case studies; MCQs often ask you to count the integers in a solution set, where a wrong bracket at one end changes the answer. Linear inequalities are not a listed topic in JEE Main or JEE Advanced 2026.

E1. Board · 1 mark The solution set of $7-3x\ge1$ is (A) $(-\infty,2]$ (B) $[2,\infty)$ (C) $(-\infty,2)$ (D) $(-2,\infty)$

💡 Show solution▶

Solve $-3x\ge-6$; dividing by $-3$ reverses the sign: $x\le2$.

Trap (B) forgets the flip; (C) loses the end point $2$, which is included because of $\ge$ ($7-6=1\ge1$ ✓).

Answer (A).

E2. Board · 2 marks Solve $2(2x+3)-10\lt6(x-2)$ and represent the solution on the number line.

💡 Show solution▶

Expand $4x+6-10\lt6x-12$, i.e. $4x-4\lt6x-12$. (½ mark)

Collect $-4+12\lt6x-4x$, so $8\lt2x$ and $x\gt4$. (1 mark)

Answer $x\in(4,\infty)$: a hollow dot at $4$ with the ray shaded to the right. (½ mark) Check: $x=5$: $16\lt18$ ✓.

E3. Board · 3 marks Solve the system $2(x-1)\lt x+5$ and $3(x+2)\ge2-x$, and show the solution set on the number line.

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First $2x-2\lt x+5\iff x\lt7$. (1 mark)

Second $3x+6\ge2-x\iff4x\ge-4\iff x\ge-1$. (1 mark)

Answer Both: $-1\le x\lt7$, the interval $[-1,7)$ — filled dot at $-1$, hollow dot at $7$, shaded between. (1 mark)

E4. Board · Case study An event organiser books a hall for a school reunion. The hall charges a fixed $\text{₹}\,8000$ plus $\text{₹}\,350$ per guest, and the total budget is $\text{₹}\,43\,000$. (i) Write an inequality for the number of guests $g$ and find the greatest number of guests the budget allows. (2 marks) (ii) The hall insists on at least $60$ guests. Write the possible numbers of guests as a double inequality. (1 mark) (iii) If the charge rises to $\text{₹}\,400$ per guest, how many guests can the same budget pay for? (1 mark)

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(i) $8000+350g\le43\,000\iff350g\le35\,000\iff g\le100$. At most $100$ guests. (2 marks)

(ii) $60\le g\le100$, with $g$ a whole number. (1 mark)

Answer (iii) $8000+400g\le43\,000\iff g\le87.5$, so at most $87$ guests (rounding down — $88$ guests would cost $\text{₹}\,43\,200$). (1 mark)

E5. Board · 1 mark The number of integers $x$ satisfying $-5\le\dfrac{2-3x}{4}\lt\dfrac72$ is (A) $10$ (B) $11$ (C) $12$ (D) $13$

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Solve Multiply by $4$: $-20\le2-3x\lt14$. Subtract $2$: $-22\le-3x\lt12$. Divide by $-3$, reversing both signs: $\tfrac{22}{3}\ge x\gt-4$, i.e. $-4\lt x\le7\tfrac13$.

Count The integers are $-3,-2,\ldots,7$: that is $7-(-3)+1=11$.

Trap (C) $12$ wrongly includes $-4$: at $x=-4$, $\tfrac{2+12}{4}=\tfrac72$, and $\tfrac72\lt\tfrac72$ is false.

Answer (B).

E6. Board · 2 marks Find all the integers $x$ that satisfy both $5x-3\lt2x+18$ and $\dfrac{x+4}{3}\ge2$.

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First $5x-3\lt2x+18$ gives $3x\lt21$, so $x\lt7$.

Second Multiply by $3$ (positive, no flip): $x+4\ge6\iff x\ge2$.

Intersect $2\le x\lt7$, i.e. $x\in[2,7)$: the integers $2,3,4,5,6$. Check $x=7$: $35-3=32$ is not less than $14+18=32$, so $7$ is excluded ✓.

Answer $2,3,4,5,6$ (five integers).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. An electric scooter costs more to buy than a petrol one but much less to run. After how many kilometres does it become the cheaper choice — and is that distance reached within a few years of ordinary use?

Why it matters. Cities from Delhi to Amsterdam are switching to electric two-wheelers to cut air pollution, and governments offer subsidies to speed this up. A buyer's decision is a linear inequality: total cost = purchase price + running cost per km × distance, and the question is when one total is at most the other.

What to do.

Find current prices (online or at a showroom) for one petrol and one electric scooter of similar size. Find the petrol price per litre, the petrol scooter's mileage (km per litre), the electricity price per unit (kWh) and the e-scooter's range per full charge and battery size.
Work out the running cost per km of each scooter, $c_p$ and $c_e$ (in rupees per km).
With purchase prices $P_p$ and $P_e$, solve $P_e+c_ed\le P_p+c_pd$ for the distance $d$. Write the answer as an interval.
Survey five people: how many km do they ride per day? Convert your break-even distance into years of use for each of them. Repeat with a subsidy of $\text{₹}\,10\,000$ on the e-scooter.

Tools: A phone browser for prices, a calculator or spreadsheet

Hand in: Your data with sources, the inequality and its solution, a number-line sketch, and a recommendation for each surveyed rider.

Going further. The battery may need replacing after about $50\,000$ km. Add its cost to the electric total once $d\gt50\,000$, and find out whether the answer changes.

Instructor note: with sample figures $P_p=\text{₹}\,85\,000$, $c_p=\text{₹}\,2.40$/km, $P_e=\text{₹}\,1\,25\,000$, $c_e=\text{₹}\,0.25$/km: $40\,000\le2.15d$, so $d\ge18\,604.65$, i.e. from about $18\,605$ km on, $[18\,605,\infty)$ in whole km — roughly $620$ days at $30$ km per day. Students' figures will differ; the method is the point.

🧪 Self-check quiz

0/5 correct

Q1. The solution of $-2x\gt8$ is

Q2. In interval notation, $-1\lt x\le3$ is

Q3. The solution set of the system $x\ge2$ and $x\lt-1$ is

Q4. The solution set of $3(x-1)\lt3x+5$ is

Q5. A taxi charges $\text{₹}\,40$ plus $\text{₹}\,15$ per km. With at most $\text{₹}\,250$, the greatest whole number of kilometres you can travel is

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