Chapter 5 · Linear Inequalities

5.1 Inequalities and the Rules for Handling Them

An inequality compares two quantities with $\lt$, $\gt$, $\le$ or $\ge$. You may add the same number to both sides, or multiply both sides by the same positive number — but multiply by a negative number and the sign must turn round.

An equation such as $24x=100$ has one answer. Change it to "$24$ rupees per notebook, and I have at most $100$ rupees" and the question becomes $24x\le100$: now $x=1,2,3$ and $4$ all work, and the answer is a set of numbers. Comparisons like this are everywhere — a speed limit of $60$ km/h says $v\le60$, a minimum age of $18$ for a driving licence says $a\ge18$, a pass mark of $33$ says $m\ge33$, and a fever is a body temperature $T\gt37.5\,^\circ\text{C}$.

To solve such statements we need to know which operations keep a comparison true. Most of the rules are exactly the rules you use for equations: you may add or subtract the same number on both sides, and multiply or divide both sides by the same positive number. The one difference — and the source of most mistakes in this chapter — is multiplying or dividing by a negative number, which reverses the direction of the inequality. In this section we set up the language of inequalities and their solutions, prove the rules, and see why the answer to an inequality depends on what kind of number $x$ is allowed to be. Section 5.2 then uses the rules to solve linear inequalities systematically.

Inequalities and Their Solutions

📘 Definition 5.1 — Inequality

Two real numbers or two algebraic expressions related by one of the symbols $\lt$ (less than), $\gt$ (greater than), $\le$ (less than or equal to) or $\ge$ (greater than or equal to) form an inequality.

Inequalities with $\lt$ or $\gt$ are strict inequalities; those with $\le$ or $\ge$ are slack (non-strict) inequalities. The statement $a\le b$ means "$a\lt b$ or $a=b$".
An inequality containing only numbers, such as $3\lt5$ or $-7\ge-9$, is a numerical inequality; it is simply true or false. One containing a variable, such as $2x+3\le11$, is a literal (algebraic) inequality; it is true for some values of the variable and false for others.
A pair of inequalities written together, such as $2\lt x\le5$ (meaning $2\lt x$ and $x\le5$), is a double inequality.

Remember that the order of real numbers is the order of points on the number line: $a\lt b$ means that $a$ lies to the left of $b$. So $-7\lt-2$ (because $-7$ is further left), even though $7\gt2$. On the number line, the inequality sign always points towards the smaller number.

📘 Definition 5.2 — Linear inequality in one variable

A linear inequality in one variable $x$ is an inequality that can be written in one of the forms $$ax+b\lt0,\quad ax+b\gt0,\quad ax+b\le0,\quad ax+b\ge0,$$ where $a$ and $b$ are real numbers and $a\ne0$. Examples: $3x-4\le11$, $\ 5-x\gt0$, $\ \tfrac{x}{2}+1\ge x$.

By contrast, $x^2-9\ge0$ is a quadratic inequality (it contains $x^2$), and $3x+2y\le6$ is a linear inequality in two variables; inequalities in two variables return in Class 12 (Linear Programming) and are not studied in this chapter.

✏️ Example 1 — Recognising inequalities

Classify each inequality as numerical or literal and as strict or slack, and say whether it is a linear inequality in one variable.
(a) $-7\gt-9$ (b) $3x-4\le11$ (c) $x^2\ge9$ (d) $5-x\gt0$ (e) $(x+1)^2\lt x^2+7$

💡 Show solution▶

(a) Numerical and strict (and true: $-7$ lies to the right of $-9$). It has no variable, so it is not a linear inequality in one variable.

(b), (c) (b) Literal, slack, linear: it is $3x-15\le0$, with $a=3$. (c) Literal and slack, but not linear — it contains $x^2$.

(d) Literal, strict, linear: $-x+5\gt0$ has $a=-1\ne0$.

(e) It looks quadratic, but expanding gives $x^2+2x+1\lt x^2+7$. Subtracting $x^2$ from both sides (allowed — Theorem 5.1 below) leaves $2x+1\lt7$, that is $2x-6\lt0$. So it is a strict linear inequality after all. Always simplify before you classify.

📘 Definition 5.3 — Solution, solution set, equivalent inequalities

A solution of an inequality in $x$ is a value of $x$ that turns it into a true numerical statement. The set of all solutions is the solution set; to solve the inequality means to find it. The solution set depends on the set of values $x$ is allowed to take (for example $\mathbb{N}$, $\mathbb{Z}$ or $\mathbb{R}$); when nothing is said, $x$ is a real number.

Two inequalities with the same solution set are equivalent inequalities; we write $\iff$ between them. Solving an inequality means replacing it by simpler equivalent ones until the solution set can be read off, such as $x\lt4$.

✏️ Example 2 — Is it a solution?

Test each of $x=2$, $x=3$, $x=-5$ and $x=2.9$ in (a) $4x-3\lt9$ (b) $4x-3\le9$.

💡 Show solution▶

Substitute $4(2)-3=5$, $\ 4(3)-3=9$, $\ 4(-5)-3=-23$, $\ 4(2.9)-3=8.6$.

(a) $5\lt9$ ✓, $\ 9\lt9$ ✗, $\ -23\lt9$ ✓, $\ 8.6\lt9$ ✓. So $2$, $-5$ and $2.9$ are solutions; $3$ is not.

(b) The same, except that $9\le9$ is true: in the slack inequality $x=3$ is also a solution.

Lesson Strict and slack inequalities differ only at the boundary value, where the two sides are equal — here $x=3$. On the number line this is the difference between a hollow and a filled end point.

✏️ Example 3 — The solution set depends on the number system

Find the solution set of $24x\lt100$ when (a) $x\in\mathbb{N}$ (b) $x\in\mathbb{Z}$ (c) $x\in\mathbb{R}$.

💡 Show solution▶

Simplify Dividing both sides by the positive number $24$ (Theorem 5.1 below) gives the equivalent inequality $x\lt\dfrac{100}{24}=\dfrac{25}{6}=4.1\overline{6}$.

(a) The natural numbers less than $4.1\overline6$: $\{1,2,3,4\}$. Check: $24\times4=96\lt100$ ✓ but $24\times5=120$ ✗.

(b) Every integer up to $4$, including $0$ and all negative integers: $\{\ldots,-3,-2,-1,0,1,2,3,4\}$.

(c) Every real number less than $\tfrac{25}{6}$: the interval $\left(-\infty,\tfrac{25}{6}\right)$. The end point is excluded because the inequality is strict.

Pick an inequality and the set $x$ must come from. The green marks are the solution set: separate dots for $\mathbb{N}$ and $\mathbb{Z}$, a shaded ray for $\mathbb{R}$. The test point steps along the line in halves; for each value the readouts say whether it makes the inequality true and whether it is allowed at all.

Inequality:x ∈

simplified—

solution set—

test point—

solution?—

The Rules for Inequalities

Everything rests on one basic fact about real numbers: $a\lt b$ exactly when the difference $b-a$ is positive. Together with two facts you have known for years — the sum of two positive numbers is positive, and the product of two positive numbers is positive while the product of a positive and a negative number is negative — this is enough to prove every rule.

📐 Theorem 5.1 — The rules for inequalities

Let $a$, $b$, $c$ be real numbers with $a\lt b$. Then:

Rule 1 (adding or subtracting). For every real $c$: $\ a+c\lt b+c\ $ and $\ a-c\lt b-c$.
Rule 2 (multiplying or dividing by a positive number). If $c\gt0$: $\ ac\lt bc\ $ and $\ \dfrac ac\lt\dfrac bc$.
Rule 3 (multiplying or dividing by a negative number). If $c\lt0$: $\ ac\gt bc\ $ and $\ \dfrac ac\gt\dfrac bc$ — the sign is reversed.

The same rules hold with $\le$ in place of $\lt$ (Rule 3 then turns $\le$ into $\ge$), and with $\gt$ or $\ge$ (read from right to left). Each rule can be undone — subtract $c$ again, or multiply by $\tfrac1c$ — so each one turns an inequality into an equivalent inequality.

Proof. We know $b-a\gt0$. Rule 1: $(b+c)-(a+c)=b-a\gt0$, so $a+c\lt b+c$; subtracting $c$ is adding $-c$. Rule 2: $bc-ac=(b-a)c$ is a product of two positive numbers, so it is positive and $ac\lt bc$. Rule 3: now $(b-a)c$ is positive times negative, so $bc-ac\lt0$, that is $ac-bc\gt0$, which says $ac\gt bc$. Finally, dividing by $c$ is multiplying by $\tfrac1c$, which has the same sign as $c$. For $\le$, the extra case $a=b$ gives $ac=bc$, which is allowed by both $\le$ and $\ge$. $\blacksquare$

Why the flip makes sense: multiplying by $-1$ reflects the number line about $0$. The point that was further left ends up further right. Since $2\lt5$, the reflections satisfy $-2\gt-5$. Multiplying by $0$ is never allowed when solving — $0\cdot a=0\cdot b$ throws away all the information.

✏️ Example 4 — Using the rules

(a) Starting from $-3\lt5$, write the true inequality obtained by: adding $4$; multiplying by $2$; multiplying by $-2$; dividing by $-1$.
(b) If $x\lt7$, what can you say about $1-2x$?
(c) Solve $-\dfrac{x}{3}\ge2$ using the rules.

💡 Show solution▶

(a) Add $4$: $1\lt9$. $\ $ Multiply by $2$: $-6\lt10$. $\ $ Multiply by $-2$ (Rule 3, flip): $6\gt-10$. $\ $ Divide by $-1$ (flip): $3\gt-5$. All four are true.

(b) Multiply Multiply $x\lt7$ by $-2$; the sign flips: $-2x\gt-14$.

(b) Add Add $1$ to both sides (no flip): $1-2x\gt-13$. Check with $x=0$: $1\gt-13$ ✓; with $x=7$ (just outside): $1-14=-13$, the boundary.

(c) Multiply both sides by $-3$ and reverse the sign: $x\le-6$. Check: $x=-9$ gives $-\tfrac{-9}{3}=3\ge2$ ✓; $x=0$ gives $0\ge2$ ✗, as expected since $0\gt-6$.

Two numbers with $a\lt b$ sit on the number line (faint). Choose an operation and drag $k$. The bold dots show the results. Adding $k$ slides both points the same way, so the order never changes. Multiplying by $k$ stretches the line — and the moment $k$ becomes negative, the line is reflected through $0$: the red point overtakes the blue one and the inequality sign turns round.

Operation:Numbers:$k$:2

result for $a$—

result for $b$—

the new inequality—

📐 Theorem 5.2 — Combining inequalities

For real numbers $a$, $b$, $c$, $d$:

(i) Transitivity. If $a\lt b$ and $b\lt c$, then $a\lt c$.
(ii) Adding. If $a\lt b$ and $c\lt d$, then $a+c\lt b+d$. (Inequalities pointing the same way may be added — but not subtracted.)
(iii) Reciprocals of positive numbers. If $0\lt a\lt b$, then $\dfrac1a\gt\dfrac1b$.
(iv) Squares of non-negative numbers. If $0\le a\lt b$, then $a^2\lt b^2$.

Proof. (i) $c-a=(c-b)+(b-a)$ is a sum of two positive numbers. (ii) $(b+d)-(a+c)=(b-a)+(d-c)$, again a sum of two positives. (iii) $\dfrac1a-\dfrac1b=\dfrac{b-a}{ab}$; the numerator is positive, and so is $ab$ because $a,b\gt0$. (iv) $b^2-a^2=(b-a)(b+a)$; here $b-a\gt0$, and $b+a\gt0$ because $b\gt a\ge0$. $\blacksquare$

✏️ Example 5 — From a range of $x$ to a range of an expression

Given $2\lt x\lt5$, find the range of values of (a) $3x-1$ (b) $4-2x$ (c) $\dfrac1x$ (d) $x^2$.

💡 Show solution▶

Method A double inequality is two inequalities at once, so apply each rule to all three parts.

(a) Multiply by $3$: $6\lt3x\lt15$. Subtract $1$: $5\lt3x-1\lt14$.

(b) Multiply by $-2$ and reverse both signs: $-4\gt-2x\gt-10$, i.e. $-10\lt-2x\lt-4$. Add $4$: $-6\lt4-2x\lt0$.

(c), (d) All parts are positive, so Theorem 5.2(iii) reverses the order: $\dfrac15\lt\dfrac1x\lt\dfrac12$. Theorem 5.2(iv) keeps it: $4\lt x^2\lt25$.

Check Take $x=3$: $3x-1=8\in(5,14)$, $\ 4-2x=-2\in(-6,0)$, $\ \tfrac13\in\left(\tfrac15,\tfrac12\right)$, $\ 9\in(4,25)$ ✓.

✏️ Example 6 — A proof with the rules

Prove that if $a\lt b$, then $a\lt\dfrac{a+b}{2}\lt b$ — the average of two different numbers lies strictly between them.

💡 Show solution▶

Left half Add $a$ to both sides of $a\lt b$ (Rule 1): $2a\lt a+b$. Divide by the positive number $2$ (Rule 2): $a\lt\dfrac{a+b}2$.

Right half Add $b$ to both sides of $a\lt b$: $a+b\lt2b$, so $\dfrac{a+b}2\lt b$.

Combine The two halves together form the double inequality $a\lt\dfrac{a+b}2\lt b$. $\blacksquare$ Check: $a=-3$, $b=8$: the average $2.5$ lies between them ✓. A consequence: between any two different real numbers there is always another one, so a strict inequality such as $x\lt4$ has no "largest" solution.

⚠️ Common mistakes with the rules

1. Forgetting the flip. From $-3x\lt12$, dividing by $-3$ gives $x\gt-4$, not $x\lt-4$. Test a value: $x=0$ satisfies $-3x\lt12$, and $0\gt-4$ ✓.

2. Flipping when you should not. Only multiplying or dividing by a negative number flips the sign. From $x-5\gt-2$, adding $5$ gives $x\gt3$ (no flip). From $3x\lt-12$, dividing by the positive $3$ gives $x\lt-4$ — a negative answer is not a reason to flip.

3. Multiplying by something whose sign you do not know. From $\dfrac1x\lt2$ you may not multiply by $x$ to get $1\lt2x$: if $x$ is negative the sign should flip. ($x=-1$ satisfies $\tfrac1x\lt2$ but not $1\lt2x$.)

4. Subtracting two inequalities. $5\lt7$ and $1\lt6$ are true, but $5-1\lt7-6$ says $4\lt1$ — false. Only adding is safe (Theorem 5.2(ii)).

5. Squaring or taking reciprocals of numbers with mixed signs. $-3\lt2$, yet $9\gt4$; and $-2\lt3$ gives $-\tfrac12\lt\tfrac13$ (no flip). Theorem 5.2(iii) and (iv) need the numbers to be non-negative.

Exercises

The language of inequalities. In parts 1–4 write the statement as an inequality.

1 The age $a$ (in years) of a voter is at least $18$.
2 The cost $c$ of a recharge, in rupees, is not more than $500$.
3 The time $t$ (in seconds) is positive but less than $10$.
4 A number $n$ exceeds twice the number $m$ by more than $3$.
5 Which of these are linear inequalities in one variable? (a) $3-4x\ge0$ (b) $x^2\lt4$ (c) $2x+3y\gt1$ (d) $5\le9$ (e) $(x-1)^2\le x^2+7$
6 Decide which of $x=-2$, $0$, $1.5$, $3$ are solutions of $5-2x\gt2$.
7 Repeat part 6 for $5-2x\ge2$. Which value changes its answer, and why?
8 Find the solution set of $7x\lt30$ when (a) $x\in\mathbb{N}$ (b) $x\in\mathbb{Z}$ (c) $x\in\mathbb{R}$.

💡 Show solutions▶

1 $a\ge18$.

2 $c\le500$ ("not more than" includes $500$ itself).

3 $0\lt t\lt10$.

4 $n-2m\gt3$.

5 (a) Yes ($a=-4$). (b) No — quadratic. (c) No — two variables. (d) No — numerical. (e) Yes: expanding, $x^2-2x+1\le x^2+7$ simplifies to $-2x-6\le0$.

6 $5-2x$ equals $9$, $5$, $2$, $-1$. Only $9\gt2$ and $5\gt2$ hold: the solutions are $-2$ and $0$.

7 Now $2\ge2$ is true, so $-2$, $0$ and $1.5$ are solutions. Only $x=1.5$ changes — it is the boundary value where both sides are equal.

8 Dividing by $7$: $x\lt\tfrac{30}{7}\approx4.29$. (a) $\{1,2,3,4\}$ (b) $\{\ldots,-2,-1,0,1,2,3,4\}$ (c) $\left(-\infty,\tfrac{30}{7}\right)$.

Using the rules. In parts 1–4 start from $-4\lt9$ and write the true inequality obtained. In parts 5–6 use the given fact to make a statement about the expression. In parts 7–10 solve, using a single rule.

1 Add $6$ to both sides.
2 Multiply both sides by $3$.
3 Multiply both sides by $-3$.
4 Divide both sides by $-1$.
5 $x\ge-2$; the expression $-5x$.
6 $x\lt3$; the expression $7-2x$.
7 $-3x\ge12$
8 $\dfrac{x}{-2}\lt5$
9 $x-7\gt-3$
10 $-x\le-6$

Check: apply one rule to a true inequality p < q

$p$:$q$:operation:$k$:

start—

result—

sign—

💡 Show solutions▶

1 $2\lt15$.

2 $-12\lt27$.

3 $12\gt-27$ (Rule 3: flip).

4 $4\gt-9$ (flip).

5 Multiply by $-5$ and flip: $-5x\le10$.

6 Multiply by $-2$: $-2x\gt-6$; add $7$: $7-2x\gt1$.

7 Divide by $-3$ and flip: $x\le-4$. Check: $x=-5$ gives $15\ge12$ ✓.

8 Multiply by $-2$ and flip: $x\gt-10$. Check: $x=0$ gives $0\lt5$ ✓.

9 Add $7$ (no flip): $x\gt4$.

10 Multiply by $-1$ and flip: $x\ge6$.

III

Ranges, counter-examples and proofs.

1 If $-1\le x\le4$, find the range of $2x+3$.
2 If $-1\le x\le4$, find the range of $5-3x$.
3 If $3\lt x\lt6$, find the range of $\dfrac1x$.
4 If $1\lt x\lt3$ and $2\lt y\lt5$, find the range of $x+y$ and of $x-y$.
5 Give a counter-example to the claim "if $a\lt b$, then $a^2\lt b^2$".
6 Give a counter-example to the claim "if $a\lt b$ and $c\lt d$, then $a-c\lt b-d$".
7 Prove that if $0\lt a\lt1$, then $a^2\lt a$.
8 Prove that if $a\lt b$, then $a\lt\dfrac{2a+b}{3}\lt b$.

💡 Show solutions▶

1 $-2\le2x\le8$, so $1\le2x+3\le11$.

2 Multiply by $-3$ and flip: $3\ge-3x\ge-12$, i.e. $-12\le-3x\le3$. Add $5$: $-7\le5-3x\le8$.

3 All parts are positive, so reciprocals reverse the order: $\dfrac16\lt\dfrac1x\lt\dfrac13$.

4 Adding: $3\lt x+y\lt8$. For $x-y$, first $-5\lt-y\lt-2$ (multiply by $-1$, flip), then add to $1\lt x\lt3$: $-4\lt x-y\lt1$.

5 $a=-3$, $b=1$: $-3\lt1$ but $a^2=9\gt1=b^2$.

6 $a=1$, $b=2$, $c=0$, $d=10$: $1\lt2$ and $0\lt10$, but $a-c=1$ and $b-d=-8$, and $1\lt-8$ is false.

7 Multiply $a\lt1$ by the positive number $a$ (Rule 2): $a\cdot a\lt1\cdot a$, i.e. $a^2\lt a$. $\blacksquare$

8 $\dfrac{2a+b}3-a=\dfrac{b-a}3\gt0$ and $b-\dfrac{2a+b}3=\dfrac{2(b-a)}3\gt0$, because $b-a\gt0$. $\blacksquare$ (The number $\tfrac{2a+b}3$ is one third of the way from $a$ to $b$.)

Applications.

1 Weather. On a summer day in Chennai the temperature ranged over $59\le F\le95$ degrees Fahrenheit. Using $C=\dfrac59(F-32)$, find the range in degrees Celsius.
2 Travel time. A car keeps its speed $v$ between $40$ km/h and $60$ km/h (inclusive) on a $90$ km stretch of highway. Find the range of the time $t=\dfrac{90}{v}$ it takes.
3 Medicine. A syrup is prescribed at $15$ mg per kg of body mass. Find the range of doses $D$ for children whose mass $m$ satisfies $12\le m\le20$ kg.
4 Geometry. A rectangle has length $8$ cm and width $w$ cm with $2\lt w\lt5$. Find the range of its perimeter $P$ and of its area $A$.
5 Taxi fare. A taxi charges a fixed $\text{₹}\,50$ plus $\text{₹}\,18$ per km. A trip cost from $\text{₹}\,230$ to $\text{₹}\,410$ (inclusive). Use the rules to find the range of the distance $d$ travelled.

💡 Show solutions▶

1 Subtract $32$: $27\le F-32\le63$. Multiply by the positive number $\tfrac59$: $15\le C\le35$ (degrees Celsius).

2 $40\le v\le60$ with all parts positive, so $\dfrac1{60}\le\dfrac1v\le\dfrac1{40}$; multiply by $90$: $1.5\le t\le2.25$ hours (from $1$ h $30$ min to $2$ h $15$ min).

3 $D=15m$: multiply by $15$: $180\le D\le300$ mg.

4 $P=16+2w$: $4\lt2w\lt10$ gives $20\lt P\lt26$ cm. $A=8w$: $16\lt A\lt40$ cm².

5 $230\le50+18d\le410$. Subtract $50$: $180\le18d\le360$. Divide by $18$: $10\le d\le20$ km.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marks

How this is examined: Board papers test the rules directly — a 1-mark MCQ or Assertion–Reason on the sign flip, short questions on the range or the greatest value of an expression, and statements that must hold "for all real $a$, $b$", where one counter-example is enough to reject an option. Linear inequalities are not a listed topic in JEE Main or JEE Advanced 2026, although the same rules are used inside many JEE problems.

E1. Board · 1 mark If $a\lt b$ and $c\lt0$, which of the following is true? (A) $ac\lt bc$ (B) $\dfrac ac\lt\dfrac bc$ (C) $a-c\gt b-c$ (D) $ac\gt bc$

💡 Show solution▶

Idea Multiplying or dividing by the negative number $c$ reverses the sign (Rule 3); subtracting $c$ does not change it (Rule 1).

Trap (A) and (B) forget the flip. (C) wrongly flips on subtraction: in fact $a-c\lt b-c$. Test with $a=1$, $b=2$, $c=-1$: $ac=-1\gt-2=bc$.

Answer (D).

E2. Board · 1 mark Assertion (A): If $-2x\gt6$, then $x\lt-3$. Reason (R): When both sides of an inequality are divided by the same negative number, the direction of the inequality is reversed. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check A Divide $-2x\gt6$ by $-2$ and flip: $x\lt-3$. True. (Test: $x=-4$ gives $8\gt6$ ✓.)

Check R R is Rule 3 of Theorem 5.1 — true, and it is exactly the step used to get A.

Answer (A).

E3. Board · 2 marks If $-2\le x\le3$, find the least and the greatest values of $4-3x$, and the values of $x$ at which they occur.

💡 Show solution▶

Multiply Multiply by $-3$ and reverse both signs: $6\ge-3x\ge-9$, i.e. $-9\le-3x\le6$. (1 mark)

Add Add $4$: $-5\le4-3x\le10$. (½ mark)

Answer Least value $-5$ (at $x=3$), greatest value $10$ (at $x=-2$) — the flip means the largest $x$ gives the smallest value. (½ mark)

E4. Board · 3 marks (i) Prove that if $0\lt a\lt b$, then $\dfrac1a\gt\dfrac1b$. (ii) A cyclist rides $24$ km at a steady speed $v$ km/h, where $12\le v\le16$. Find the range of the time taken.

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(i) $\dfrac1a-\dfrac1b=\dfrac{b-a}{ab}$. Here $b-a\gt0$ and $ab\gt0$ (both positive), so the difference is positive and $\dfrac1a\gt\dfrac1b$. (1½ marks)

(ii) Reciprocals Time $t=\dfrac{24}{v}$. By (i), $12\le v\le16$ gives $\dfrac1{16}\le\dfrac1v\le\dfrac1{12}$. (1 mark)

Answer Multiply by $24$: $1.5\le t\le2$ hours. (½ mark)

E5. Board · 2 marks Real numbers $x$ and $y$ satisfy $1\le x\le4$ and $-3\le y\le2$. Find the greatest value of $2x-3y$.

💡 Show solution▶

Separate $2\le2x\le8$, and multiplying $-3\le y\le2$ by $-3$ (flip): $-6\le-3y\le9$.

Add Adding the two double inequalities (Theorem 5.2(ii)): $-4\le2x-3y\le17$. The value $17$ is reached at $x=4$, $y=-3$: $8+9=17$.

Trap Taking the largest $y$ ($y=2$) because "largest values give the largest answer" gives only $8-6=2$: the minus sign in front of $3y$ wants the smallest $y$.

Answer $17$.

E6. Board · 1 mark Which statement is true for all real numbers $a$ and $b$ with $a\lt b$? (A) $a^2\lt b^2$ (B) $\dfrac1a\gt\dfrac1b$ (whenever $a,b\ne0$) (C) $a^3\lt b^3$ (D) $|a|\lt|b|$

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Reject (A) and (D): $a=-3$, $b=1$ gives $9\gt1$ and $3\gt1$. (B): $a=-1$, $b=1$ gives $-1\gt1$, false.

Prove (C) $b^3-a^3=(b-a)(a^2+ab+b^2)$ and $a^2+ab+b^2=\left(a+\tfrac b2\right)^2+\tfrac34b^2$, which is positive unless $a=b=0$ — impossible since $a\lt b$. Both factors are positive, so $a^3\lt b^3$.

Quick test One counter-example kills an option; the only survivor of quick tests with negative numbers is (C), since cubing keeps signs.

Answer (C).

🔬 Research–Teaching NexusMini project · about 40 minutes

Research question. Every measurement is really an inequality. If you measure a sheet of paper with a ruler, how sure can you be of its area?

Why it matters. Engineers, surveyors and lab scientists never know a length exactly — they know it lies in an interval. A length read as $29.7$ cm on a ruler marked in millimetres means $29.65\le l\le29.75$. Working out how such intervals spread through a calculation ("interval arithmetic") is how bridges get safety margins and how laboratories report results such as $623.7\pm2.6\ \text{cm}^2$.

What to do.

Measure the length $l$ and width $w$ of an A4 sheet (or your notebook) to the nearest millimetre. Write each as a double inequality, $l_0-0.05\le l\le l_0+0.05$ (in cm).
Use Theorem 5.1 and Theorem 5.2(ii) to find the range of the perimeter $P=2(l+w)$.
Show that if $0\lt a\le b$ and $0\lt c\le d$, then $ac\le bd$ (multiply by positives twice and use transitivity), and use it to find the range of the area $A=lw$.
Collect the measurements of five classmates. How many of their computed areas fall inside your interval? How wide would the interval be with a ruler marked only in centimetres?

Tools: A ruler, a calculator or spreadsheet

Hand in: Your measurements as inequalities, the ranges of $P$ and $A$ with every rule you used, and a sentence on what the class data showed.

Going further. The width of the area interval is roughly $0.05(l_0+w_0)\times2$. Explain where this comes from, and why a percentage error in each length roughly adds up to the percentage error in the area.

Instructor note: for A4 ($29.7$ cm by $21.0$ cm) the ranges are $101.2\le P\le101.6$ cm and $29.65\times20.95=621.1675\le A\le29.75\times21.05=626.2375$ cm², around the nominal $623.7$ cm². The width $\approx5.07$ cm² $\approx0.1\,(l_0+w_0)$.

🧪 Self-check quiz

0/5 correct

Q1. Which of these is a slack inequality?

Q2. Multiplying both sides of $-2\lt7$ by $-3$ gives

Q3. The solution set of $4x\lt13$ with $x\in\mathbb{N}$ is

Q4. If $x\gt5$, which statement must be true?

Q5. If $0\lt a\lt b$, then

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