Linear Inequalities

• Definitions and vocabulary — what an inequality is, the difference between strict ($<,>$) and slack ($\le,\ge$) inequalities, and between numerical, literal and double inequalities.
• The two solving rules — why addition/subtraction never disturbs an inequality, and why multiplication/division by a negative number must reverse it.
• Algebraic solutions in one variable — clearing fractions, collecting terms, and reading off the solution set depending on whether $x$ ranges over natural numbers, integers or real numbers.
• Interval notation and the number line — translating $x<a$, $x\ge a$, and compound inequalities like $a\le x<b$ into brackets and into open/closed-circle diagrams.
• Compound inequalities and systems in one variable — solving two inequalities simultaneously and taking the intersection of their solution sets.
• Graphical solution in two variables — the half-plane picture, the test-point method, and dashed-versus-solid boundary lines.
• Systems of inequalities in two variables — overlaying half-planes to find the common (feasible) region.
• Word problems — translating "at least," "at most," "more than," and ranges (marks, mixtures, temperature conversions) into inequalities and solving them.

Inequalities: definitions and types

Definition. Two real numbers or two algebraic expressions related by one of the symbols $<$ (less than), $>$ (greater than), $\le$ (less than or equal to) or $\ge$ (greater than or equal to) form an inequality. Statements like $3<5$ or $7>5$, built from pure numbers, are numerical inequalities; statements like $x<5$ or $y\ge3$, involving a variable, are literal inequalities. A statement such as $3<x<5$ — read "$x$ is greater than $3$ and less than $5$" — chains two inequalities about the same expression and is called a double or compound inequality. Inequalities that use the strict symbols $<$ or $>$ are called strict inequalities; those using $\le$ or $\ge$, which allow equality as a possibility, are called slack inequalities. An expression of the form

is a linear inequality in one variable, and

is a linear inequality in two variables. where: $a,b,c$ are fixed real constants (with $a\neq0$ in (1), and $a,b$ not both zero in (2)), and $x,y$ are the variables. An expression like $ax^2+bx+c\le0$ is a quadratic inequality — genuinely different, because the sign of a quadratic expression can change twice as $x$ varies, not once — and this chapter deliberately stays inside the linear case.

The two golden rules for solving inequalities

Solving an equation relies on two familiar moves: add or subtract the same quantity from both sides, and multiply or divide both sides by the same nonzero quantity. Inequalities obey a close cousin of these moves, with one crucial twist.

Rule 1. Equal numbers may be added to or subtracted from both sides of an inequality without affecting the direction of the inequality sign.

Rule 2. Both sides of an inequality may be multiplied or divided by the same positive number without changing the direction of the inequality. But if both sides are multiplied or divided by a negative number, the direction of the inequality is reversed: $<$ becomes $>$, $\le$ becomes $\ge$, and so on. The asymmetry is easy to see with pure numbers: $3>2$ is true, but multiplying both sides by $-1$ turns it into $-3$ and $-2$, and since $-3<-2$, the sign had to flip to stay true. The full logical justification — not just an example, but a proof that covers every case — is worked out in the first derivation below, because this is the single fact the entire chapter is built on.

Solving in one variable, and where the solution set lives

Given an inequality such as $30x<200$, "solving" it means finding every value of $x$ that makes the statement true — its solution set. Crucially, that set depends on which numbers $x$ is allowed to be. If $x$ must be a natural number, the solution set of $30x<200$ (i.e. $x<20/3$) is the finite set $\{1,2,3,4,5,6\}$; if $x$ must be an integer, it is the infinite set $\{\ldots,-3,-2,-1,0,1,2,3,4,5,6\}$; and if $x$ may be any real number, the solution is the entire ray of real numbers below $20/3$, written as the interval $\left(-\infty,\ \tfrac{20}{3}\right)$. Unless a problem restricts $x$ to naturals or integers, this chapter — like the rest of algebra — solves over the real numbers by default.

Interval notation is the standard shorthand for such rays and segments: a round bracket means the endpoint is excluded (strict inequality, open circle on a number line), and a square bracket means the endpoint is included (slack inequality, filled circle). So $x<a$ is $(-\infty,a)$, $x\ge a$ is $[a,\infty)$, and a compound inequality like $2\le x<6$ is the half-open interval $[2,6)$.

Linear inequalities in two variables: the half-plane picture

An inequality like $x+y<5$ has two independent variables, so a "solution" is now an ordered pair $(x,y)$ — a point in the plane — and there are infinitely many such points satisfying it. The boundary case $x+y=5$ is a straight line, and that line splits the entire Cartesian plane into exactly two half-planes. Every point on one side of the line makes $x+y-5$ positive, and every point on the other side makes it negative (the proof is the third derivation below); the line itself is where $x+y-5=0$. So the solution region of $x+y<5$ is precisely one of those two half-planes, and finding which one is a one-point check away. This is the test-point method. Draw the boundary line $ax+by=c$: dashed if the inequality is strict ($<,>$, boundary excluded), solid if it is slack ($\le,\ge$, boundary included). Pick any convenient point not on the line — the origin $(0,0)$ is the usual choice, unless the line itself passes through the origin, in which case pick another simple point such as $(1,0)$. Substitute that point into the inequality: if it is satisfied, shade the half-plane containing the test point; if not, shade the other half-plane.

Systems of linear inequalities in two variables

When several linear inequalities in $x$ and $y$ must all hold simultaneously — a system of inequalities — the solution is the set of points that lie in every one of the individual half-planes at once. Graphically, shade each half-plane in turn (or just its boundary and direction) and read off the region where all the shadings overlap; this common region is sometimes called the feasible region, a term that becomes central two chapters later in linear programming. If the system also carries $x\ge0$ and $y\ge0$ (as it does whenever $x,y$ represent physical quantities like litres or rupees), the feasible region is trapped inside the first quadrant.

Board level - a double inequality with number-line graph (NCERT Example 9 pattern)

Solve $-8\le5x-3<7$ and represent the solution on a number line.

Step 1. The statement is a compound inequality, shorthand for $-8\le5x-3$ and $5x-3<7$ together. Add $3$ to all three parts (Rule 1): $-8+3\le5x-3+3<7+3$, which gives $-5\le5x<10$.

Step 2. Divide all three parts by $5$ (Rule 2, positive divisor, no sign change): $\dfrac{-5}{5}\le x<\dfrac{10}{5}$, giving $-1\le x<2$.

Step 3. Write the answer in interval notation and describe the number-line picture: a filled (solid) circle at $-1$ because it is included, an open circle at $2$ because it is excluded, and a solid segment joining them. Answer: the solution set is $[-1,\,2)$, i.e. all real $x$ with $-1\le x<2$.

Board level - graphing the solution region of a linear inequality in two variables

Draw the graph of the solution set of $3x+4y\le12$.

Step 1. Find where the boundary line $3x+4y=12$ crosses the axes: setting $y=0$ gives $x=4$, and setting $x=0$ gives $y=3$. Draw the line through $(4,0)$ and $(0,3)$; because the inequality is slack ($\le$), the boundary itself is part of the solution, so draw it as a solid line.

Step 2. Pick a test point not on the line — the origin $(0,0)$ works here, since the line does not pass through it. Substitute into the inequality: $3(0)+4(0)=0$, and $0\le12$ is true.

Step 3. Because the test point satisfies the inequality, shade the half-plane that contains the origin — the side of the line closer to $(0,0)$, including the line itself. Answer: the solution region is the closed half-plane on the origin's side of the line $3x+4y=12$, i.e. every point $(x,y)$ with $3x+4y\le12$.

Competitive level (JEE / NEET) - an acid-mixture inequality with an integer-count twist (extends NCERT Example 13)

A manufacturer has $600$ litres of a $12\%$ acid solution. Let $x$ litres of a $30\%$ acid solution be added so that the acid content of the resulting mixture is more than $15\%$ but less than $18\%$. Find the range of $x$, and hence the number of integer litre-values of $x$ that satisfy the requirement, given the mixing must also keep the total volume $x+600$ under $1000$ litres for tank capacity.

Step 1. The total mixture is $(x+600)$ litres, and its acid content, by volume, is $30\%$ of $x$ plus $12\%$ of $600$. Translate "more than $15\%$" and "less than $18\%$" of the new total into two simultaneous inequalities: $0.30x+0.12(600)>0.15(x+600)$ and $0.30x+0.12(600)<0.18(x+600)$.

Step 2. Clear the first inequality: $30x+7200>15x+9000$, so $15x>1800$, giving $x>120$. Clear the second inequality the same way: $30x+7200<18x+10800$, so $12x<3600$, giving $x<300$. Combining both, $120<x<300$.

Step 3. Apply the extra tank-capacity constraint $x+600<1000$, i.e. $x<400$, which is automatically satisfied by every $x$ already below $300$, so it adds nothing new here — a useful check rather than a further restriction; always verify whether an "extra" condition is genuinely binding before folding it in.

Step 4. Count the integers strictly between $120$ and $300$: these are $121,122,\ldots,299$, which is $299-121+1=179$ values. Answer: $120<x<300$, and there are $179$ integer litre-values of $x$ that satisfy the requirement. Timing note: about ninety seconds. Set up both percentage inequalities together rather than sequentially, clear decimals by multiplying through by $100$ immediately, and solve them as one linear pass rather than restarting the algebra twice. The integer count is the classic JEE add-on to an NCERT-style inequality — always subtract-and-add-one for a strict interval $(m,n)$ with integer endpoints; a slack interval $[m,n]$ instead would count $n-m+1$ directly with no adjustment, so check open versus closed before counting.

• Forgetting to flip the sign when multiplying or dividing both sides by a negative number — by far the single most common error in the whole chapter, and the reason Derivation 1 exists.
• Flipping only one part of a compound inequality. In $l\le E(x)<u$, if the coefficient of $x$ is negative, both inequality signs must flip, and the numbers must also swap order so the smaller sits on the left.
• Confusing open and closed on the number line or in interval notation — a strict inequality ($<,>$) gives an open circle and a round bracket; a slack inequality ($\le,\ge$) gives a filled circle and a square bracket. Mixing these changes whether the boundary point is actually a solution.
• Losing solutions by restricting the domain silently. The same inequality $30x<200$ has a different solution set depending on whether $x$ is a natural number, an integer, or a real number — always re-read the question for which set $x$ is drawn from before finalising the answer.
• Drawing the boundary line the wrong style. Strict inequalities ($<,>$) need a dashed/dotted boundary because the line itself is excluded; slack inequalities ($\le,\ge$) need a solid boundary because the line is included. Using the wrong style is a common careless mark-loss in graphical questions.
• Testing a point that lies on the boundary line, which gives no information about either side — always check first whether the line passes through the chosen test point (commonly the origin) before substituting.
• Shading the wrong side after a correct test. If the test point satisfies the inequality, shade its side; a surprising number of students shade the opposite side by habit.
• Treating the two-variable chapter like the one-variable chapter. A "solution" to $ax+by<c$ is a whole region of points, not a single ray of numbers — there is no single number line to fall back on.

• State the definition of an inequality and correctly classify a given statement as strict or slack, and as numerical, literal or double.
• Apply Rule 1 and Rule 2 to solve a linear inequality in one variable, remembering to reverse the sign when multiplying or dividing by a negative number.
• Solve a compound (double) inequality by operating on all three parts simultaneously, and handle the sign-flip correctly when the coefficient of $x$ is negative.
• Convert fluently between inequality notation, interval notation, and a number-line diagram with correctly open or closed endpoints.
• Translate a word problem (marks, mixtures, temperature ranges, lengths) into one or more linear inequalities and solve them.
• Graph the boundary line of a two-variable linear inequality with the correct dashed/solid style and use the test-point method to shade the correct half-plane.
• Find and describe the common (feasible) region of a system of two or more linear inequalities in two variables.
• Explain, using the order properties of real numbers, why multiplying by a negative number must reverse an inequality — not just state the rule from memory.

Ask the tutor: if flipping the inequality sign, choosing open versus closed circles, or shading the correct half-plane still feels shaky, ask the AI tutor for a fresh compound inequality and a two-variable system to solve step by step until the pattern clicks.