03/14 3. Inequalities

Understanding Inequality Statements

An inequality compares two expressions using one of four basic symbols: $<$, $>$, $\le$, or $\ge$. Unlike an equation, which may have one or more specific solutions, an inequality often describes a range of values.

The complete collection of values that makes the statement true is the solution set. For example, $x<4$ describes every real number less than $4$, while $x\le4$ also includes $4$.

Reading comparison symbols

• $x<a$: $x$ is less than $a$.

• $x>a$: $x$ is greater than $a$.

• $x\le a$: $x$ is less than or equal to $a$.

• $x\ge a$: $x$ is greater than or equal to $a$.

The equality bar determines whether the boundary value belongs to the solution set.

Takeaway: An inequality describes acceptable values, and its symbol indicates the relationship and whether a boundary is included.

Solving Linear Inequalities

A linear inequality has a variable raised only to the first power. Examples include $3x-5>10$, $-2x+7\le15$, and $\frac{x}{4}+1\ge3$.

Solve a linear inequality using the same operations used for equations:

1. Distribute and combine like terms when necessary.

2. Move variable terms to one side and constants to the other.

3. Isolate the variable.

4. If you multiply or divide both sides by a negative number, reverse the inequality symbol.

For example, solve $-3x+5>14$:

The final step divides by $-3$, so $>$ becomes $<$. The solution is every real number less than $-3$. A check with $x=-4$ confirms the result, while a value such as $x=0$ does not satisfy the original inequality.

When the variable disappears

Sometimes simplification leaves only a statement about constants:

• A true statement, such as $2<5$, means all real numbers are solutions: $(-\infty,\infty)$.

• A false statement, such as $7<3$, means there are no solutions: $\varnothing$.

Takeaway: Solve linear inequalities like equations, but reverse the comparison when multiplying or dividing by a negative number.

Combining Inequalities with And and Or

A compound inequality combines two inequalities. The words and and or determine how the conditions are combined.

Conditions joined by “and”

Both conditions must be true, so the result is an intersection of solution sets. For a three-part inequality, perform the same operation on all three parts.

Solve:

Subtract $1$ throughout and then divide throughout by $3$:

The solution is $[-1,3)$. The left endpoint is included and the right endpoint is excluded.

Conditions joined by “or”

At least one condition must be true, so the result is a union of solution sets. Solve each inequality separately, then combine the results.

The two parts give $x<-2$ and $x\ge4$. Therefore, the combined solution is

Takeaway: “And” identifies the overlap between conditions; “or” combines the regions where at least one condition is satisfied.

Using Absolute Value as Distance

The expression $|u|$ represents the distance of $u$ from zero. Because distance is nonnegative, a positive distance usually corresponds to two positions.

Absolute-value equations

For $a\ge0$,

To solve $|2x-3|=7$, create two equations:

This gives $x=5$ or $x=-2$, so the solution set is $\{-2,5\}$.

Absolute-value inequalities

First isolate the absolute-value expression. For $a>0$, the comparison determines whether solutions lie inside or outside the boundary values:

• $|u|<a$ means $-a<u<a$.

• $|u|\le a$ means $-a\le u\le a$.

• $|u|>a$ means $u<-a$ or $u>a$.

• $|u|\ge a$ means $u\le-a$ or $u\ge a$.

For example, solve $|x-4|\le3$:

Adding $4$ throughout gives $1\le x\le7$, so the solution is $[1,7]$.

For the opposite pattern, solve $|2x+1|>5$:

Thus, $x<-3$ or $x>2$, which is $(-\infty,-3)\cup(2,\infty)$.

Edge cases

• $|u|<0$ has no solution because absolute value is never negative.

• $|u|\ge0$ is true for every real number.

• If the right side is negative, check the statement before applying the standard positive-bound rules.

Takeaway: “Less than” absolute-value inequalities describe values between two boundaries; “greater than” inequalities describe values outside them.

Writing Solutions in Interval Notation

Interval notation gives a compact description of a solution set. Its endpoint symbols have precise meanings:

• Parentheses exclude an endpoint, as in $(a,b)$.

• Brackets include an endpoint, as in $[a,b]$.

• A mixed interval includes one endpoint and excludes the other, as in $[a,b)$ or $(a,b]$.

• Infinity always uses parentheses because it is not a real number.

• The symbol $\cup$ joins separate intervals.

Common translations are:

• $x>a$ becomes $(a,\infty)$.

• $x\ge a$ becomes $[a,\infty)$.

• $x<a$ becomes $(-\infty,a)$.

• $x\le a$ becomes $(-\infty,a]$.

• $a<x<b$ becomes $(a,b)$.

• $a\le x\le b$ becomes $[a,b]$.

• $x<a$ or $x>b$ becomes $(-\infty,a)\cup(b,\infty)$.

• All real numbers become $(-\infty,\infty)$.

• No real numbers become $\varnothing$.

Always compare each endpoint with the original inequality. A strict symbol such as $<$ or $>$ requires a parenthesis, while a non-strict symbol such as $\le$ or $\ge$ requires a bracket.

Takeaway: Parentheses mean “not included,” brackets mean “included,” and $\cup$ combines separate solution regions.

Graphing Solution Sets

A number-line graph gives a visual representation of the solution set.

1. Locate every boundary value.

2. Draw an open circle when the boundary is excluded by $<$ or $>$.

3. Draw a closed circle when the boundary is included by $\le$ or $\ge$.

4. Shade the direction or interval containing the solutions.

Examples:

• $x>3$: open circle at $3$, shading to the right.

• $x\ge3$: closed circle at $3$, shading to the right.

• $x<3$: open circle at $3$, shading to the left.

• $x\le3$: closed circle at $3$, shading to the left.

• $-2\le x<4$: closed circle at $-2$, open circle at $4$, shading between them.

• $x<-2$ or $x\ge4$: open circle at $-2$, closed circle at $4$, shading outward.

A graph should agree with both the algebra and the interval notation. For example, $[-1,3)$ must show a closed endpoint at $-1$, an open endpoint at $3$, and shading between those values.

To check a graph, test one shaded value and one unshaded value in the original inequality. The shaded value should make the statement true, while the unshaded value should make it false.

Takeaway: Open and closed circles show endpoint inclusion, while shading shows which values satisfy the inequality.

Applying Inequalities to Real Situations

Inequalities model limits, ranges, minimums, maximums, and requirements. Translate the wording into a mathematical comparison before solving, then apply any restrictions on the variable.

Budget constraint

A student has $\$80$ available. After paying $\$20$ for admission, each notebook costs $\$6$. If $n$ is the number of notebooks, the total cannot exceed $\$80$:

Solving gives $n\le10$. Because a number of notebooks cannot be negative and must be a whole number, the realistic solutions are

Distance requirement

Suppose a delivery center serves locations within $12$ miles of mile marker $30$. If $x$ is the address mile marker, “within $12$ miles” means

Rewrite the absolute-value inequality:

Adding $30$ throughout gives

Thus, the delivery range is $[18,42]$.

A reliable checklist

1. Simplify both sides.

2. Move variable terms and constants to appropriate sides.

3. Isolate the variable, reversing the symbol when multiplying or dividing by a negative number.

4. For a compound inequality, identify whether the connector is and or or.

5. For absolute value, isolate the absolute-value expression first.

6. Check an endpoint and a test value from the proposed solution region.

7. State the answer in inequality notation and interval notation when appropriate.

8. Graph the result on a number line.

9. Apply context restrictions, such as whole-number counts or nonnegative quantities.

Takeaway: A correct applied solution must satisfy both the algebraic inequality and the restrictions imposed by the situation.