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| Primary source | OpenStax College Algebra 2e, Section 2.7: "Linear Inequalities and Absolute Value Inequalities" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/2-7-linear-inequalities-and-absolute-value-inequalities |
College Algebra 2e Section 2.7 begins on page 215 of the PDF.
An equation asks: "For which single value of $x$ does this statement become true?"
An inequality asks: "For which values of $x$ does this comparison hold?" The answer is a set -- usually an interval or a union of intervals -- not a single number.
Almost every algebraic rule from equations carries over to inequalities unchanged. There is exactly one exception:
When you multiply or divide both sides of an inequality by a negative number, reverse the direction of the inequality sign.
This rule is not arbitrary. When you multiply every number on the number line by $-1$, the line flips: what was on the right moves to the left, and order reverses. For example, $2 < 5$, but $-2 > -5$. The inequality sign must flip to describe the same truth after the flip.
If you understand why the rule exists, you will never forget it under pressure.
Inequality symbols.
| Symbol | Meaning | Interval | Number line |
|---|---|---|---|
| $x < a$ | $x$ is strictly less than $a$ | $(-\infty, a)$ | Open circle at $a$, arrow left |
| $x \leq a$ | $x$ is less than or equal to $a$ | $(-\infty, a]$ | Closed circle at $a$, arrow left |
| $x > a$ | $x$ is strictly greater than $a$ | $(a, \infty)$ | Open circle at $a$, arrow right |
| $x \geq a$ | $x$ is greater than or equal to $a$ | $[a, \infty)$ | Closed circle at $a$, arrow right |
Interval notation rules.
( or ) means the endpoint is excluded (strict inequality or $\pm\infty$).[ or ] means the endpoint is included ($\leq$ or $\geq$).The one extra rule: Multiplying or dividing by a negative number reverses the inequality sign.
Intervals are the natural way to write the solution to an inequality. Before solving, make sure you can read and write interval notation fluently.
| Set description | Interval | Example values |
|---|---|---|
| All $x$ greater than 2 | $(2, \infty)$ | $3, 5, 2.001, 1000$ |
| All $x$ at most $-1$ | $(-\infty, -1]$ | $-1, -2, -100, -1.0001$ |
| All $x$ between $-3$ and $5$ (exclusive) | $(-3, 5)$ | $0, 4.9, -2.99$ |
| All $x$ between $-3$ and $5$ (inclusive) | $[-3, 5]$ | $-3, 0, 5$ |
| All real numbers | $(-\infty, \infty)$ or $\mathbb{R}$ | any real number |
| No solution | $\emptyset$ | none |
Compound intervals (union): If the solution consists of two separate intervals, write them with the union symbol $\cup$.
Example. The solution to $x < -2$ or $x > 3$ is $(-\infty, -2) \cup (3, \infty)$.
Linear inequalities are solved using the same steps as linear equations, with one addition: reverse the sign when you multiply or divide by a negative.
Example 1. Solve $3x - 7 < 11$.
Step 1. Add 7 to both sides: \[ 3x < 18 \] Step 2. Divide both sides by 3 (positive, so no flip): \[ x < 6 \]
Solution: $(-\infty, 6)$.
Example 2. Solve $-5x + 3 \geq 18$.
Step 1. Subtract 3 from both sides: \[ -5x \geq 15 \] Step 2. Divide both sides by $-5$ (negative, so flip the sign): \[ x \leq -3 \]
Solution: $(-\infty, -3]$.
Why the sign flips: Before dividing, the left side $-5x$ is at least $15$. When $x = -4$: $-5(-4) = 20 \geq 15$. True. When $x = -2$: $-5(-2) = 10 \geq 15$. False. The solution contains $-4$ but not $-2$, which is $x \leq -3$. Dividing by $-5$ and flipping gives exactly this.
Example 3. Solve $4(2x - 1) \leq 3(x + 5)$.
Distribute: \[ 8x - 4 \leq 3x + 15 \] Subtract $3x$: \[ 5x - 4 \leq 15 \] Add 4: \[ 5x \leq 19 \] Divide by 5 (positive): \[ x \leq \frac{19}{5} \]
Solution: $\left(-\infty, \dfrac{19}{5}\right]$.
A compound inequality imposes two conditions on $x$ simultaneously. There are two types.
Type A: "and" (intersection). Both conditions must hold. The solution is the overlap (intersection) of the two solution sets.
Standard form: $a < x < b$ (shorthand for $a < x$ and $x < b$).
Example 4. Solve $-2 \leq 3x + 1 < 10$.
Apply the same operation to all three parts at once.
Subtract 1 from all three parts: \[ -3 \leq 3x < 9 \] Divide all three parts by 3: \[ -1 \leq x < 3 \]
Solution: $[-1, 3)$.
Common Error -- Operating on Only Two Parts. When you have a three-part inequality $a < f(x) < b$, every operation must be applied to all three expressions simultaneously. Students who subtract from only the left and middle parts break the chain.
Type B: "or" (union). At least one condition must hold. The solution is the union of the two solution sets.
Example 5. Solve $2x + 1 > 9$ or $3x - 2 < -8$.
Solve each separately.
Left: $2x + 1 > 9 \implies 2x > 8 \implies x > 4$.
Right: $3x - 2 < -8 \implies 3x < -6 \implies x < -2$.
Solution: $(-\infty, -2) \cup (4, \infty)$.
Check: Is $x = 5$ in the solution? $2(5)+1 = 11 > 9$. Yes. Is $x = 0$ in the solution? $0 > 4$? No. $0 < -2$? No. Correctly excluded.
A quadratic inequality like $x^2 - x - 6 < 0$ asks: "For which $x$-values is the parabola below the $x$-axis?"
Method (sign analysis).
Example 6. Solve $x^2 - x - 6 < 0$.
Step 1. Already in the form $f(x) < 0$ with $f(x) = x^2 - x - 6$.
Step 2. Factor: $x^2 - x - 6 = (x - 3)(x + 2)$. Zeros: $x = 3$ and $x = -2$.
Step 3. The zeros split the number line into three intervals: \[ (-\infty, -2), \quad (-2, 3), \quad (3, \infty) \]
Step 4. Test each interval:
| Interval | Test point | $f(\text{test}) = (x-3)(x+2)$ | Sign |
|---|---|---|---|
| $(-\infty, -2)$ | $x = -4$ | $(-7)(-2) = 14$ | $+$ |
| $(-2, 3)$ | $x = 0$ | $(-3)(2) = -6$ | $-$ |
| $(3, \infty)$ | $x = 5$ | $(2)(7) = 14$ | $+$ |
Step 5. We need $f(x) < 0$, which occurs in $(-2, 3)$. The endpoints are excluded (strict inequality).
Solution: $(-2, 3)$.
Example 7. Solve $x^2 - 4x + 4 \geq 0$.
Factor: $(x-2)^2 \geq 0$.
A square is always $\geq 0$. So this is true for all real $x$, with equality at $x = 2$.
Solution: $(-\infty, \infty)$ (all real numbers).
This result makes geometric sense: the parabola $(x-2)^2$ is tangent to the $x$-axis at $x = 2$ and lies above it everywhere else. It never goes below zero.
Example 8. Solve $x^2 + 2x + 5 \leq 0$.
Check the discriminant: $\Delta = 4 - 20 = -16 < 0$. No real zeros.
Since $a = 1 > 0$, the parabola opens upward and has no $x$-intercepts. It is strictly above the $x$-axis for all real $x$. There is no $x$ where it is $\leq 0$.
Solution: $\emptyset$ (no solution).
Discriminant shortcut for quadratic inequalities: If $\Delta < 0$ and $a > 0$, then $ax^2 + bx + c > 0$ for all real $x$ (parabola entirely above axis). If $\Delta < 0$ and $a < 0$, then $ax^2 + bx + c < 0$ for all real $x$. No sign analysis needed.
Sign analysis extends to any polynomial inequality.
Example 9. Solve $(x + 1)(x - 2)(x - 4) > 0$.
Zeros: $x = -1$, $x = 2$, $x = 4$. These create four intervals.
| Interval | Test point | $(x+1)$ | $(x-2)$ | $(x-4)$ | Product | Sign |
|---|---|---|---|---|---|---|
| $(-\infty, -1)$ | $x = -2$ | $-$ | $-$ | $-$ | $(-)(-)(-) = -$ | $-$ |
| $(-1, 2)$ | $x = 0$ | $+$ | $-$ | $-$ | $(+)(-)(-) = +$ | $+$ |
| $(2, 4)$ | $x = 3$ | $+$ | $+$ | $-$ | $(+)(+)(-) = -$ | $-$ |
| $(4, \infty)$ | $x = 5$ | $+$ | $+$ | $+$ | $(+)(+)(+) = +$ | $+$ |
Need product $> 0$: intervals $(-1, 2)$ and $(4, \infty)$.
Solution: $(-1, 2) \cup (4, \infty)$.
Many functions require constraints on the input. Inequalities are how those constraints are written.
Example 10. For what values of $x$ is $f(x) = \sqrt{4 - x^2}$ defined?
The square root requires $4 - x^2 \geq 0$.
Rearrange: $x^2 \leq 4$.
Solve: $x^2 - 4 \leq 0 \implies (x-2)(x+2) \leq 0$.
Sign analysis: the quadratic $(x-2)(x+2)$ is $\leq 0$ on $[-2, 2]$ (endpoints included because the inequality is non-strict).
Domain of $f$: $[-2, 2]$.
| Error | Example | Correction |
|---|---|---|
| Forgetting to flip sign | $-3x < 9 \to x < -3$ | Divide by $-3$ and flip: $x > -3$ |
| Flipping sign when dividing by a positive | $3x > 9 \to x < 3$ | Dividing by $3$ (positive) does not flip: $x > 3$ |
| Using $[$ for a strict endpoint | Solution $x > 2$ written as $[2, \infty)$ | Strict inequality: $(2, \infty)$ |
| Applying operation to only two parts of a compound inequality | $-2 \leq 3x+1 < 10$, subtracting 1 from only left and middle | Subtract from all three: $-3 \leq 3x < 9$ |
| Stopping at zeros without testing signs | Concluding $(x-3)(x+2) < 0$ has solution $\{-2, 3\}$ | Zeros are boundaries, not solutions; test each interval |
Problem 1. Solve: $-4x + 3 > 15$. Write the solution in interval notation.
Subtract 3: $-4x > 12$.
Divide by $-4$ (flip sign): $x < -3$.
Solution: $(-\infty, -3)$.
Problem 2. Solve the compound inequality: $-1 \leq 2x + 5 \leq 11$.
Subtract 5 from all three parts: $-6 \leq 2x \leq 6$.
Divide all three by 2: $-3 \leq x \leq 3$.
Solution: $[-3, 3]$.
Problem 3. Solve: $5x + 1 < 16$ and $3x - 2 > 4$.
Left: $5x < 15 \implies x < 3$.
Right: $3x > 6 \implies x > 2$.
Both must hold: $2 < x < 3$.
Solution: $(2, 3)$.
Problem 4. Solve: $x^2 - 5x + 6 \leq 0$.
Factor: $(x-2)(x-3) \leq 0$.
Zeros: $x = 2$, $x = 3$.
Test intervals:
Need $\leq 0$: the interval $[2, 3]$ (endpoints included; the expression equals zero there).
Solution: $[2, 3]$.
Problem 5. Solve: $2x^2 + x - 6 > 0$.
Factor: $2x^2 + x - 6$. AC method: $ac = -12$, $m + n = 1$: try $4$ and $-3$. Rewrite: $2x^2 + 4x - 3x - 6 = 2x(x+2) - 3(x+2) = (x+2)(2x-3)$.
Zeros: $x = -2$, $x = \frac{3}{2}$.
Test intervals:
Need $> 0$: $(-\infty, -2) \cup \left(\frac{3}{2}, \infty\right)$.
Problem 6. Solve: $x^2 + 6x + 9 < 0$.
Factor: $(x+3)^2 < 0$.
A square is never negative. No real $x$ satisfies this.
Solution: $\emptyset$.
Problem 7. Find all $x$ for which $f(x) = \sqrt{3x - x^2}$ is defined.
Require $3x - x^2 \geq 0$.
Factor: $x(3 - x) \geq 0$.
Zeros: $x = 0$ and $x = 3$.
Test intervals:
Need $\geq 0$: $[0, 3]$ (endpoints included; expression equals zero there, and $\sqrt{0} = 0$ is defined).
Domain: $[0, 3]$.
Problem 8. Solve: $(x^2 - 1)(x + 3) \leq 0$.
Factor completely: $(x-1)(x+1)(x+3) \leq 0$.
Zeros: $x = -3$, $x = -1$, $x = 1$.
Four intervals:
Need $\leq 0$: $(-\infty, -3]$ and $[-1, 1]$. Include endpoints (non-strict).
Solution: $(-\infty, -3] \cup [-1, 1]$.
Problem 9. For what values of $k$ does $kx^2 - 2x + k > 0$ hold for all real $x$?
A quadratic $f(x) = kx^2 - 2x + k$ is positive for all real $x$ when:
Compute $\Delta = (-2)^2 - 4(k)(k) = 4 - 4k^2$.
Require $\Delta < 0$: $4 - 4k^2 < 0 \implies k^2 > 1 \implies |k| > 1$.
Combining with $k > 0$: $k > 1$.
Check $k = 2$: $2x^2 - 2x + 2$. Discriminant $= 4 - 16 = -12 < 0$ and $a = 2 > 0$: positive for all $x$. $\checkmark$
Solution: $k > 1$, or $(1, \infty)$.
Think of a number line as a street with addresses. Solving an inequality is like finding which addresses satisfy a rule.
A single zero divides the street into two neighborhoods. A sign analysis test is like visiting one house in each neighborhood and checking whether the rule holds there. If it holds for the test house, it holds for the whole neighborhood (because the sign of the expression does not change between zeros).
This is the key insight behind sign analysis:
A polynomial's sign can only change at its zeros. Between two consecutive zeros, the polynomial is either entirely positive or entirely negative.
This fact means you only need one test per interval -- one representative address -- to know the entire interval. You do not need to check infinitely many values; you just need one.
The rule about flipping the inequality sign when multiplying by a negative fits the same picture: if you multiply the address of every house by $-1$, the house at $2$ moves to $-2$, and the house at $5$ moves to $-5$. The neighborhood that was "to the right of zero" is now "to the left of zero." All comparisons reverse.
For students who find math intimidating: The sign-flip rule is the only new rule in this entire lesson. Everything else is the same algebra you already know from solving equations. Write the rule on a card: "Dividing by a negative flips the sign." Learn one example by heart (like Example 2 above). That is all you need to get started.
For students interested in proof: The fact that a polynomial's sign cannot change between its zeros follows from the Intermediate Value Theorem (a consequence of continuity). A polynomial is continuous everywhere; if it were positive at some point and negative at another, it would have to equal zero somewhere in between. So the only sign changes occur at zeros.
For students interested in careers: Inequalities appear in constraint optimization: "maximize profit subject to $x \geq 0$ and cost $\leq$ budget." Linear programming (operations research, supply chains, scheduling) is an entire field built on systems of linear inequalities. The feasible region is the intersection of finitely many half-planes.
For gifted and curious students: Sign analysis generalizes to rational functions. $\dfrac{(x-1)(x+3)}{x-2}$ changes sign at $x = 1$, $x = -3$ (zeros of the numerator) and at $x = 2$ (zero of the denominator, which is excluded from the domain). The same interval-testing method applies, with the excluded point creating an interval boundary.
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