Linear Inequalities

Intuition

Two lessons ago you pinned down what makes a step on an equation trustworthy. A reversible step on an equation is guaranteed to produce an equivalent equation, one with the identical solution set, since it can neither add a solution nor drop one. Every legal move on an inequality faces the same demand to preserve the solution set, but here reversibility is no longer the test. An inequality also carries an order, a claim about which side is larger, and a step is safe when it keeps faith with that order. That is the real subject of this lesson, so meet the hazard before any rules. Take $x < 3$ , whose solutions are every number below $3$ , and multiply both sides by $x$ . Out comes $x^2 < 3x$ . That new inequality is true exactly on $0 < x < 3$ : the value $x = -1$ fails it, since $1 < -3$ is false, and $x = 4$ fails it too, since $16 < 12$ is false, while $x = 1$ passes. So every solution at or below $0$ has silently disappeared. One clean-looking step destroyed part of the solution set. This lesson is about which steps respect that order and are therefore honest, and why the famous negative flip is forced rather than decreed. The lesson is also about why multiplying by something whose sign you cannot name is not a move at all.

Learning goals

Keep the direction under addition or positive scaling, reverse it under negative scaling
Derive the reversal from $a < b$ meaning $b - a > 0$
Refuse to multiply by an expression of unknown sign in one step
Split $ax > b$ four ways on the parameters
Read and as intersection and or as union

Which moves keep the solution set

Solving an inequality means the same disciplined thing that solving an equation did. Replace the inequality, step by step, with a simpler one that has the identical solution set, until the set can be read off directly. Two inequalities are equivalent when they have exactly the same solution set, and the whole task is to travel through a chain of equivalent inequalities. So the governing question is the one from lesson 1, asked again. Which operations are guaranteed to turn an inequality into an equivalent one? Sorting the operations answers it.

Adding the same expression to both sides, as long as that expression is defined for every $x$ , gives an equivalent inequality. Adding slides both sides the same distance in the same direction, so it never changes which side is larger, and the order (and with it the symbol) is preserved. Adding a constant, or a whole term like $3x$ , is fine, because each is defined for every real number. Adding something like $\frac{1}{x}$ , undefined at $x = 0$ , would not qualify.

Multiplying or dividing both sides by a positive constant keeps both the solution set and the direction of the symbol. It is reversible, undone by dividing or multiplying by that same positive constant. And scaling by a positive factor stretches or shrinks every distance from zero without ever carrying a number across zero. Because nothing crosses zero, the left-to-right order of the two sides is preserved, so the direction survives.

Multiplying or dividing both sides by a negative constant preserves the solution set only if you also reverse the direction of the symbol. This is the move usually handed to students as a rule to memorize, and here is a one-line demonstration that the reversal is not optional. Take $x < 3$ and multiply both sides by $-1$ without touching the symbol, giving $-x < -3$ . That step is defined for every $x$ , and multiplying by $-1$ again returns you to $x < 3$ . So the step is perfectly reversible, yet its solution set is every number above $3$ , while the original was every number below $3$ . Reversible, and wrong. So reversibility cannot be the standard: the multiplier was negative, and holding the direction fixed silently changed the answer. Reverse the symbol instead, to $-x > -3$ , and the two sides agree with the original again. The reversal is a forced consequence, not a convention, and the derivation shows exactly why it has to happen.

The same rule solves an ordinary inequality, not just a labeled step. Solve $-3x > 12$ for $x$ : divide both sides by $-3$ and reverse the direction, giving $x < -4$ . Check both sides of that boundary: at $x = -5$ the original reads $15 > 12$ , true, while at $x = -3$ it reads $9 > 12$ , false, so the answer really does change sides exactly at $-4$ . And $-4$ itself is not a solution, since it gives $12 > 12$ , which is false, confirming the flip and why the boundary is not included in the solution set.

Why multiplying by a negative constant reverses the direction#

Suppose $a < b$ , so on the number line $a$ sits to the left of $b$ . Multiplying every number by $-1$ reflects the whole line through $0$ , sending each point to its mirror image on the far side. A reflection turns left into right, so whichever of $a$ and $b$ started on the left comes out on the right: their order reverses, giving $-b < -a$ . That is exactly the picture below.

The same fact confirms itself algebraically. It is enough to understand multiplication by $-1$ , because any negative constant is $-1$ times a positive one, and multiplying by that positive factor is harmless, since a positive scaling keeps the order, as shown above. Saying $a < b$ is the same as saying the gap $b - a$ is a positive number, since $b$ lies above $a$ . Multiplying $a$ and $b$ by $-1$ produces $-a$ and $-b$ , and comparing them through their difference,

(-a) - (-b) = b - a > 0,

shows $-a$ is now the larger of the two, that is $-b < -a$ again. Nothing was chosen; the flip is forced by the single fact that $b - a$ was positive to begin with.

Dividing by a negative constant is no different, since dividing by $-k$ with $k > 0$ is multiplying by the positive number $\frac{1}{k}$ and then by $-1$ . Of those two factors only the final $-1$ does any flipping, so the direction reverses exactly once.

Multiplying every number by negative one reflects the line through 0. The points a and b, with a to the left of b, map to their mirror images minus a and minus b, and the reflection lands them in swapped order, so minus b sits to the left of minus a and the direction reverses.

Multiplying both sides by $0$ is a catastrophe of a different kind. It turns any inequality into a statement about $0$ and $0$ alone: $0 < 0$ , $0 > 0$ , or $0 \ne 0$ come out false no matter what $x$ was, while $0 \le 0$ or $0 \ge 0$ come out true no matter what $x$ was. Either way, the result no longer remembers anything about the numbers you started with. As with equations, where multiplying by $0$ collapsed everything to $0 = 0$ , the step keeps no record of the original and cannot be undone.

Now the hazard the opening promised. Multiplying or dividing by a constant is safe once you know its sign: positive keeps the direction, negative reverses it, and only $0$ is forbidden. But suppose the thing you multiply by is not a constant at all, but an expression like $x$ or $x - 2$ whose sign changes with $x$ . Then you genuinely cannot say what the step does, because the answer depends on a value you have not fixed. Where the expression is positive the direction should stay, where it is negative the direction should flip, and where it is $0$ the inequality is destroyed. A single stroke of the pen cannot be all three at once. So multiplying both sides of an inequality by an expression of unknown sign is not a legal single step, not until that sign has been pinned down. It is not one operation but a fork in the road, and writing a single line as though the fork were not there is what loses solutions.

This is the inequality version of the warning from the equation lessons, sharpened. For an equation the fatal question about a multiplier was “can this expression be zero,” because zero was the one value where multiplying could not be undone. For an inequality that question is still fatal, and a second one joins it: “can this expression be negative.” That second question matters because a negative multiplier silently reverses a direction you have left pointing the wrong way. An expression of unknown sign can be negative and can be zero, so it fails both tests at once. That is why an unknown-sign multiplier is doubly barred. The same double failure is why the honest way through, as the rest of the lesson shows, is to split into cases where the sign is known.

Worked example 1 What the opening step destroyed

Return to $x < 3$ multiplied by $x$ . The multiplier $x$ has no fixed sign, so the step should have forked. Where $x > 0$ , multiplying keeps the direction and gives $x^2 < 3x$ ; where $x < 0$ , multiplying reverses it and gives $x^2 > 3x$ ; and at $x = 0$ the inequality $0 < 0$ is simply false. Writing $x^2 < 3x$ for every $x$ quietly used the first branch everywhere, so it can be trusted only where $x > 0$ .

Solving $x^2 < 3x$ properly confirms exactly the boundary the opening found by testing: the inequality holds precisely on $0 < x < 3$ , the same interval where $x = -1$ and $x = 4$ failed and $x = 1$ passed.

Compare the two sets. The original $x < 3$ has solution set $(-\infty, 3)$ , while the multiplied inequality has solution set $(0, 3)$ . Every solution with $x \le 0$ was thrown away. Those lost solutions are exactly the values where the multiplier $x$ was negative or zero, the very places the single step had no right to speak for. The destroyed solutions were lost at the fork that was never taken.

Check your understanding

You want to multiply both sides of $x - 1 < 5$ by the expression $x + 2$ . Why is that not a legal step on its own?

A It is legal, since

x + 2

is usually positive B Because only constants may ever be added to both sides C Because the sign of

x + 2

is not fixed, so you cannot tell whether to keep the direction, reverse it, or forbid the step D Because multiplying both sides always requires a positive number

The parameter case, solving ax > b

Lesson 2 took the equation $ax = b$ , let $a$ be a parameter you could not inspect, and found that solving it forced a split into cases at $a = 0$ . The strict inequality $ax > b$ is the direct sequel. Everything from lesson 2 returns, and one thing is added: the sign of $a$ now decides not only whether you may divide but which way the symbol points afterward. So the split is no longer just “can I divide” but “divide and keep, divide and flip, or cannot divide at all.”

The four cases of $ax > b$ #

Solve $ax > b$ for $x$ , with $a$ and $b$ fixed but unknown. Everything turns on the sign of $a$ .

If $a > 0$ , divide both sides by the positive number $a$ . Dividing by a positive keeps the direction, so the solution is $x > \frac{b}{a}$ , the interval $\left(\frac{b}{a}, \infty\right)$ .

If $a < 0$ , divide both sides by the negative number $a$ . Dividing by a negative reverses the direction, so the solution is $x < \frac{b}{a}$ , the interval $\left(-\infty, \frac{b}{a}\right)$ . The flip here is not a rule applied from memory; it is the forced consequence proved above, now triggered because this case placed a negative number under the division.

If $a = 0$ , there is nothing to divide by, and the left side $ax$ is $0$ for every $x$ . The inequality collapses to $0 > b$ , a statement with no $x$ in it, whose truth depends only on $b$ . When $b < 0$ , the claim $0 > b$ is true, so every real number is a solution and the solution set is all of $\mathbb{R}$ . When $b \ge 0$ , the claim $0 > b$ is false, so no value of $x$ works and the solution set is $\varnothing$ .

These cases are exhaustive, because $a$ is positive, negative, or zero, and in the zero case $b$ is negative or it is not. Notice where the dividing line falls. The boundary sits at $b = 0$ , and $b = 0$ lands on the empty-set side, because $0 > 0$ is false. This is a genuine difference from the equation $ax = b$ of lesson 2, where the $a = 0$ fork was $b = 0$ against $b \neq 0$ . In lesson 2’s $a = 0$ case the value $b = 0$ gave every real number, since $0 = 0$ is true. A strict inequality reads its boundary the other way: $b = 0$ passes an equals test but fails a strictly-greater test.

The two cases where the boundary itself is a genuine ray are worth seeing side by side, because what separates them is smaller than the algebra suggests. Take $\frac{b}{a} = -3$ and shade the figure below to the right: that is $x > -3$ , the answer when $a$ is positive. Now change nothing except the side. The boundary has not moved, the circle is still hollow, and the picture is now $x < -3$ , the answer when $a$ is negative. Dividing by a negative did not compute a different number. It kept the same number and turned the solution set around it inside out.

The interactive figure below only draws boundary-and-direction answers, so it cannot show the other two cases directly, but every one of the four outcomes still has its own number-line picture. The diagram after the table draws all four together.

Case	What you may do	Solution	What the picture looks like
$a > 0$	divide by positive $a$ , keep the direction	$x > \dfrac{b}{a}$	ray shaded right of $\dfrac{b}{a}$ , hollow circle
$a < 0$	divide by negative $a$ , reverse the direction	$x < \dfrac{b}{a}$	ray shaded left of $\dfrac{b}{a}$ , hollow circle
$a = 0,\ b < 0$	no division is possible; $0 > b$ is true	all of $\mathbb{R}$	the whole line shaded
$a = 0,\ b \ge 0$	no division is possible; $0 > b$ is false	$\varnothing$	nothing shaded

All four outcomes of ax greater than b, drawn as number lines. A positive a shades a ray to the right of the boundary and a negative a shades a ray to the left, both with a hollow circle at the boundary. When a is 0 the boundary disappears: b less than 0 shades the entire line, and b at least 0 leaves the line completely unshaded.

Dividing by a negative turns the set around, it does not move the boundary

Endpoint Side Endpoint is

x > -3. Every number to the right of -3 is a solution. -3 itself is not, so its circle is hollow: the sign is strict.

A number line carrying the solution set of an inequality, starting strict with a hollow endpoint. Try only the Side control: the shading flips to the other side of the boundary while the boundary itself does not move, the whole visible effect of dividing by a negative number. The Endpoint and Endpoint-is controls let you explore other inequalities, but are not part of this demonstration.

Worked example 2 Solve $ax > 2$ as a case analysis on $a$

Here $b = 2$ , a positive number. Run the three sign cases for $a$ .

If $a > 0$ , divide by the positive $a$ and keep the direction:

x > \frac{2}{a}.

If $a < 0$ , divide by the negative $a$ and reverse the direction:

x < \frac{2}{a}.

If $a = 0$ , the inequality becomes $0 > 2$ , which is false, so there is no solution.

Because $b = 2$ is positive, the $a = 0$ branch can only produce the false statement $0 > 2$ , so this family never has every real number as its answer. Its three possible outcomes are a ray to the right, a ray to the left, or nothing at all.

Worked example 3 Solve $ax > -3$ as a case analysis on $a$

The only change from the previous example is the sign of $b$ : now $b = -3$ , a negative number. The two nonzero cases look the same, but the $a = 0$ case turns over.

If $a > 0$ , dividing by the positive $a$ keeps the direction:

x > -\frac{3}{a}.

If $a < 0$ , dividing by the negative $a$ reverses it:

x < -\frac{3}{a}.

If $a = 0$ , the inequality becomes $0 > -3$ , which is true, so every real number is a solution and the solution set is all of $\mathbb{R}$ .

Set this beside the previous example. The coefficient $a$ steers the two nonzero cases identically in both, so it is not what separates them. At the boundary $a = 0$ it is the sign of $b$ that decides everything: $b = 2 > 0$ gave no solution, while $b = -3 < 0$ gives every real number. That is the very fork lesson 2 pointed out, where once the coefficient vanished it was the constant that split no solution from all of $\mathbb{R}$ .

Check your understanding

For which values of the parameter $a$ does the inequality $ax > 7$ have no solution at all?

A every

a > 0

B every

a < 0

C no value of

a

D only

a = 0

The illegal move, up close

The unknown-sign warning from the first section has a classic trap all its own. Consider $\frac{1}{x} < 1$ . One caution before we touch it: this is not a linear inequality, and this section is not here to teach a general method for solving inequalities like it, only to show the illegal move in full daylight and the one habit that fixes it. The tempting step is to clear the fraction by multiplying both sides by $x$ , turning $\frac{1}{x} < 1$ into $1 < x$ and concluding $x > 1$ . Test that conclusion against the original at $x = -1$ : the original reads $\frac{1}{-1} = -1 < 1$ , which is true, so $x = -1$ is a genuine solution, yet $x = -1 > 1$ is false. The step multiplied by $x$ as though $x$ were positive, and every negative solution fell through the gap. Multiplying by $x$ is the unknown-sign move, and the cure is to split by the sign of $x$ , exactly as the refusal rule from earlier in this lesson demands.

Worked example 4 Solve $\dfrac{1}{x} < 1$ by cases on the sign of $x$

The denominator forbids $x = 0$ outright, since $\frac{1}{x}$ is undefined there, so $x = 0$ cannot be a solution for any reason. That leaves two honest cases, and in each the sign of $x$ is known, so multiplying by $x$ is a legal, single-direction step.

Case $x > 0$ . Multiplying both sides by the positive number $x$ keeps the direction:

\frac{1}{x} < 1 \;\Rightarrow\; 1 < x.

So $x > 1$ . Every such value is already positive, so the whole ray $x > 1$ survives this case.

Case $x < 0$ . Multiplying both sides by the negative number $x$ reverses the direction:

\frac{1}{x} < 1 \;\Rightarrow\; 1 > x.

This says $x < 1$ , which every negative number already satisfies, so the entire case $x < 0$ works.

Collect the two cases. The solutions are the negative numbers together with the numbers above $1$ :

x < 0 \quad\text{or}\quad x > 1,

that is, the union $(-\infty, 0) \cup (1, \infty)$ . The naive step kept only the $x > 1$ half and lost the whole $(-\infty, 0)$ ray, precisely the case where multiplying by $x$ should have flipped the symbol.

Check your understanding

Solve $\dfrac{5}{x} < 1$ by splitting on the sign of $x$ . What is the full solution set?

x > 5

x < 0

x > 5

0 < x < 5

x < 5

Compound inequalities as set operations

You already solve compound inequalities mechanically. Here is the single idea that ties them to this lesson. A compound joins two conditions, and each condition has its own solution set; the compound’s solution set is built from those two by a set operation. An and takes the intersection: a value solves ” $P$ and $Q$ ” exactly when it lies in both solution sets, so you keep only what the two share. An or takes the union: a value solves ” $P$ or $Q$ ” when it lies in at least one, so you sweep together everything either piece allows. The chained form $a < x < b$ is simply the “and” of $x > a$ and $x < b$ , its solution the overlap of two rays.

Seeing a compound as a set operation makes its degenerate answers obvious, and they echo lesson 1’s identity and contradiction exactly. If two “and” conditions share nothing, their intersection is empty and the compound has no solution. That is why a chained statement like $3 < x < 1$ names no numbers at all: it asks for $x > 3$ and $x < 1$ at once. Those two rays $(3, \infty)$ and $(-\infty, 1)$ do not overlap, so the intersection is $\varnothing$ . A union runs the other way, since two “or” pieces can together cover the whole line, and then every real number is a solution. And if solving either piece cancels the variable and leaves a bare statement, that piece is all of $\mathbb{R}$ when the statement is true and $\varnothing$ when it is false. That is the same identity-or-contradiction verdict from the equation lesson, now feeding into the intersection or union.

Worked example 5 An 'and' compound whose pieces do not overlap

Solve $x - 2 > 3$ and $x + 4 < 5$ together. Solve each piece on its own first. The left piece gives

x - 2 > 3 \;\Rightarrow\; x > 5,

with solution set $(5, \infty)$ . The right piece gives

x + 4 < 5 \;\Rightarrow\; x < 1,

with solution set $(-\infty, 1)$ .

Now intersect, because the two are joined by “and.” A solution must be greater than $5$ and less than $1$ at the same time, and no number is both, so the intersection is empty. The compound has no solution, $\varnothing$ . Written as a chain it would read $5 < x < 1$ , ends pointing in conflict, which is the visible signature of an empty “and.”

For contrast, the same two pieces joined by “or” would ask for $x > 5$ or $x < 1$ . And their union $(-\infty, 1) \cup (5, \infty)$ is a large set, everything except the band from $1$ to $5$ . The word between the pieces, “and” or “or,” is the entire difference between an empty answer and a nearly full line.

The same two rays, x greater than 5 and x less than 1, joined two ways. Joined by and, they never overlap, so the intersection is empty. Joined by or, together they cover every number except the band between 1 and 5.

Check your understanding

Two solution sets are $(-\infty, 4)$ and $(1, \infty)$ . Since $1 < 4$ , the two rays overlap on $(1, 4)$ . If the two pieces are joined by "or", what is the solution set of the compound?

(1, 4)

, just the overlap B All of

\mathbb{R}

, the union of the two rays C

\varnothing

, since the rays disagree D

(-\infty, 1) \cup (4, \infty)

, everything outside the overlap

Common mistakes

Multiplying by the variable to clear a fraction. Multiplying both sides of $\frac{1}{x} < 1$ by $x$ treats $x$ as positive and loses every negative solution, because the multiplier $x$ has no fixed sign here. Not every expression holding a variable is like this; $x^2 + 1$ , for instance, is positive for every $x$ and may be multiplied freely. Check whether the sign of the multiplier is actually known before you multiply, and split into cases only when it is not.
Treating the flip as a rule to memorize. Reversing the symbol is not an arbitrary law; it is forced, because $a < b$ makes $b - a$ positive and taking opposites turns that into $-b < -a$ . If you ever doubt the direction, fall back on that fact rather than a slogan.
Flipping for a negative that is not the multiplier. Only the sign of the number you multiply or divide both sides by can flip the symbol. A negative constant sitting elsewhere in the problem, or a subtraction, never does.
Forgetting the $a = 0$ branches of $ax > b$ . Writing only $x > \frac{b}{a}$ or $x < \frac{b}{a}$ quietly assumes $a \neq 0$ . When $a = 0$ the inequality is $0 > b$ , which gives every real number if $b < 0$ and no solution if $b \ge 0$ .
Reading a strict boundary like an equals boundary. In the $a = 0$ branch, $ax = b$ with $b = 0$ gave every real number, but $ax > b$ with $a = 0$ and $b = 0$ gives no solution, because $0 = 0$ is true while $0 > 0$ is false. Switch to the non-strict $ax \ge b$ and $b = 0$ gives every real number again, since $0 \ge 0$ is true.
Confusing “and” with “or”. An “and” compound is the intersection of the two solution sets, and an “or” compound is their union. A chain like $3 < x < 1$ is an empty “and,” not a typo to be repaired by swapping the ends.

Key takeaways

Two inequalities are equivalent when they have the same solution set, but a safe step is not merely a reversible one: it must respect order. Keep the direction under a strictly increasing operation like adding a constant or scaling by a positive; reverse it under a strictly decreasing one like scaling by a negative.
Adding any expression defined for all $x$ , and multiplying or dividing by a positive constant, preserve the solution set and the direction. Multiplying or dividing by a negative constant preserves the set but reverses the direction, while multiplying by $0$ destroys the inequality.
The reversal is a consequence, not a rule: $a < b$ means $b - a > 0$ , and since $(-a) - (-b) = b - a > 0$ you get $-b < -a$ , which is just the reflection of the line through $0$ .
Multiplying by an expression of unknown sign is not a legal move, because it can be positive, negative, or zero. For equations the fatal question was “can it be zero,” and for inequalities “can it be negative” joins it.
The parameter inequality $ax > b$ splits four ways: $x > \frac{b}{a}$ if $a > 0$ , $x < \frac{b}{a}$ if $a < 0$ , all of $\mathbb{R}$ if $a = 0$ and $b < 0$ , and $\varnothing$ if $a = 0$ and $b \ge 0$ .
Solve an unknown-sign multiplication like $\frac{1}{x} < 1$ by cases on the sign, giving $x < 0$ or $x > 1$ . A compound inequality is a set operation, with “and” the intersection and “or” the union of the two solution sets.

Linear Inequalities

Learning goals

Which moves keep the solution set

The parameter case, solving ax > b

The illegal move, up close

Compound inequalities as set operations

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice