01.4.1 Understanding Algebraic Inequalities

An inequality is a mathematical statement comparing two quantities that are not necessarily equal. Consider the following examples: \[ 5>3 \] This means that 5 is greater than 3. \[ 2<7 \] This means that 2 is less than 7. Inequalities can also contain variables. For example: \[ x+3>8 \] This inequality means that the value of \(x+3\) must be greater than 8. Subtracting 3 gives: \[ x>5 \] Therefore, every real number greater than 5 satisfies the inequality. For example, \(x=6\), \(x=7\), and \(x=10\) all satisfy the inequality. However, \(x=5\) does not satisfy it because: \[ 5+3=8 \] And 8 is not greater than 8.

The Four Main Inequality Symbols

Symbol Meaning Example
> Greater than x > 4
< Less than x < 4
≥ Greater than or equal to x ≥ 4
≤ Less than or equal to x ≤ 4

The distinction between strict inequalities (\(>,<\)) and non-strict inequalities (\(\geq,\leq\)) is important.

For example:

  • \(x>4\) does not include 4.

  • \(x\geq4\) includes 4.Equations vs. inequalitiesConsider:\[ x+2=7 \]This equation has exactly one solution:\[ x=5 \]Now consider:\[ x+2<7 \]The solution is:\[ x<5 \]This inequality has infinitely many real-number solutions, including 4, 0, −3, and 4.9.Key concept: Solving an inequality means finding every value of the variable that makes the inequality true.01.4.2 Fundamental Properties of InequalitiesThe properties of inequalities tell us which mathematical operations we can perform while maintaining a logically equivalent inequality.These rules form the foundation of inequality solving.Property 1: Addition PropertyAdding the same number to both sides of an inequality does not change its direction.If:\[ a<b \]Then:\[ a+c<b+c \]This rule also applies to \(>\), \(\leq\), and \(\geq\).Example 1Given:\[ 3<8 \]Add 5 to both sides:\[ 3+5<8+5 \]Therefore:\[ 8<13 \]The inequality remains true.Example 2: Solve an inequality\[ x-6<10 \]Add 6 to both sides:\[ x-6+6<10+6 \]Simplify:
  •  
  • x < 16
  •  
  • Solution in interval notation:\[ (-\infty,16) \]Property 2: Subtraction PropertySubtracting the same number from both sides of an inequality does not change its direction.If:\[ a>b \]Then:\[ a-c>b-c \]Example 3Given:\[ 12>7 \]Subtract 4 from both sides:\[ 12-4>7-4 \]Therefore:\[ 8>3 \]Example 4: Solve an inequality\[ x+9\geq15 \]Subtract 9 from both sides:\[ x+9-9\geq15-9 \]Simplify:
  • x ≥ 6
  •  
  • Solution in interval notation:\[ [6,\infty) \]Property 3: Multiplication PropertyMultiplying both sides of an inequality by the same positive number preserves the inequality’s direction.If:\[ a<b,\qquad c>0 \]Then:\[ ac<bc \]Example 5Given:\[ 2<5 \]Multiply both sides by 3:\[ 2(3)<5(3) \]Therefore:\[ 6<15 \]The inequality sign remains unchanged because 3 is positive.Example 6: Multiplying by a negative numberConsider:\[ 2<5 \]Multiply both sides by −3:\[ 2(-3)>5(-3) \]Therefore:\[ -6>-15 \]Notice that the inequality sign has reversed from \(<\) to \(>\).This happens because multiplication by a negative number reverses the order of real numbers.The general rule is:If:\[ a<b,\qquad c<0 \]Then:

    \[ ac>bc \]

     

Important rule: Multiplication by a negative number

Multiply by a positive number Multiply by a negative number

2 < 5
2 < 5

6 < 15
−6 > −15
✓
Sign stays the same		 ⇆
Sign reverses

Property 4: Division Property

Dividing both sides of an inequality by the same positive number preserves the inequality sign.

If:

\[ a>b,\qquad c>0 \]

Then:

\[ \frac{a}{c}>\frac{b}{c} \]

However, dividing both sides by a negative number reverses the inequality sign.

If:

\[ a>b,\qquad c<0 \]

Then:

\[ \frac{a}{c}<\frac{b}{c} \]

Example 7: Division by a positive number

Solve:

\[ 4x>20 \]

Divide both sides by 4:

\[ \frac{4x}{4}>\frac{20}{4} \]

Therefore:

x > 5

Example 8: Division by a negative number

Solve:

\[ -3x\leq12 \]

Divide both sides by −3 and reverse the sign:

\[ \frac{-3x}{-3}\geq\frac{12}{-3} \]

Therefore:

x ≥ −4

Why did the sign change?

Because −3 is negative, the direction of the inequality must be reversed to preserve its truth.

Summary of arithmetic properties

Operation on both sides Inequality direction
Add any real number Unchanged
Subtract any real number Unchanged
Multiply by a positive number Unchanged
Multiply by a negative number Reversed
Divide by a positive number Unchanged
Divide by a negative number Reversed

Special caution: Never divide an inequality by zero. Multiplying by zero also destroys the original comparison, so it is not a valid equivalent step in solving an inequality.

If you want to multiply or divide by an expression containing a variable, its sign must be known first. For example, you cannot simply divide by \(x\) without determining whether \(x\) is positive, negative, or zero.

01.4.3 Additional Properties of Inequalities

Property 5: Transitive Property

The transitive property allows us to compare two quantities through a third quantity.

If:

\[ a<b \]

and:

\[ b<c \]

Then:

\[ a<c \]

Example 9

Given:

\[ 3<7 \]

and:

\[ 7<12 \]

We can conclude:

\[ 3<12 \]

The property also works with compatible strict and non-strict inequalities.

For instance, if:

\[ a\leq b,\qquad b<c \]

Then:

\[ a<c \]

Property 6: Reversing the Order of Comparison

If two sides of an inequality exchange positions, the comparison symbol must also be reversed.

For example:

\[ 4<9 \]

is equivalent to:

\[ 9>4 \]

Likewise:

\[ x\geq6 \]

is equivalent to:

\[ 6\leq x \]

These statements describe exactly the same relationship.

Property 7: Negation Property

Multiplying both sides of an inequality by −1 reverses the inequality sign.

If:

\[ a<b \]

Then:

\[ -a>-b \]

Example 10

Given:

\[ -7<-2 \]

Multiply both sides by −1:

\[ 7>2 \]

The reversed inequality remains true.

This property is particularly important when the variable has a negative coefficient.

01.4.4 Solving Basic Linear Inequalities

A linear inequality in one variable is an inequality that can be written in one of the following forms:

\[ ax+b<c \]

\[ ax+b\leq c \]

\[ ax+b>c \]

\[ ax+b\geq c \]

where \(a\), \(b\), and \(c\) are real numbers and \(a\neq0\).

The main objective is to isolate the variable on one side of the inequality.

General solving procedure

    • Simplify both sides by removing brackets and combining like terms.

    • Move variable terms to one side.

    • Move constant terms to the other side.

    • Divide by the coefficient of the variable.

    • Reverse the inequality sign if dividing by a negative number.

    • Express the solution using inequality or interval notation.

    • Check the solution, especially the boundary value.

      Example 11: Two-step linear inequality

      Solve:

      \[ 5x-7\leq18 \]

      Step 1. Add 7 to both sides.

      \[ 5x-7+7\leq18+7 \]
       
      \[ 5x\leq25 \]

      Step 2. Divide both sides by 5.

      \[ \frac{5x}{5}\leq\frac{25}{5} \]
       

      Final answer:

      x ≤ 5

      Interval notation:

      \[ (-\infty,5] \]

      The solution includes all real numbers less than or equal to 5.

      Example 12: Solving with a negative coefficient

      Solve:

      \[ -4x+9>21 \]

      Step 1. Subtract 9 from both sides.

      \[ -4x>12 \]

      Step 2. Divide both sides by −4.

      Since we are dividing by a negative number, reverse \(>\) to \(<\).

      \[ x<\frac{12}{-4} \]
       

      Final answer:

      x < −3

      Interval notation:

      \[ (-\infty,-3) \]

      Verification

      Choose a value inside the solution set, such as \(x=-4\).

      \[ -4(-4)+9>21 \]
      \[ 25>21 \]

      This is true.

      Now choose \(x=-2\), which is outside the solution set.

      \[ -4(-2)+9>21 \]
      \[ 17>21 \]

      This is false.

      Therefore, our solution is consistent.

      Example 13: Variables on both sides

      Solve:

      \[ 3x+5>2x+11 \]

      Step 1. Subtract \(2x\) from both sides.

      \[ 3x-2x+5>11 \]
       
      \[ x+5>11 \]

      Step 2. Subtract 5 from both sides.

      \[ x>6 \]
       

      Final answer:

      x > 6

      Interval notation:

      \[ (6,\infty) \]

      An important point is that moving terms between sides is simply a shortcut for performing the same addition or subtraction on both sides.

       

      01.4.5 Solving Inequalities with Parentheses

      Some inequalities contain expressions inside brackets or parentheses.

      To solve these inequalities, first apply the distributive property.

      Recall:

      \[ a(b+c)=ab+ac \]

      Example 14: Inequality with parentheses

      Solve:

      \[ 2(3x-1)\leq4x+6 \]

      Step 1. Expand the brackets.

      \[ 6x-2\leq4x+6 \]

      Step 2. Subtract \(4x\) from both sides.

      \[ 2x-2\leq6 \]

      Step 3. Add 2 to both sides.

      \[ 2x\leq8 \]

      Step 4. Divide both sides by 2.

      x ≤ 4

       

    • Interval notation:
      \[ (-\infty,4] \]

      Example 15: Parentheses with a negative coefficient

      Solve:

      \[ -2(x+3)\geq8 \]

      Step 1. Expand the brackets.

      \[ -2x-6\geq8 \]

      Step 2. Add 6 to both sides.

      \[ -2x\geq14 \]

      Step 3. Divide both sides by −2 and reverse the sign.

x ≤ −7

Interval notation:

\[ (-\infty,-7] \]

Common mistake: Students sometimes forget to distribute the negative coefficient to every term inside the parentheses.

For example:

\[ -2(x+3)=-2x-6 \]

not:

\[ -2x+6 \]
 

01.4.6 Solving Inequalities with Fractions

Some linear inequalities contain fractions.

A useful strategy is to eliminate the denominators by multiplying both sides by their least common multiple (LCM).

For inequalities with positive constant denominators, the LCM is positive, so the inequality sign does not change.

Example 16: Inequality containing fractions

Solve:

\[ \frac{x-1}{4}\leq\frac{x+2}{3} \]

Step 1. Find the LCM of the denominators.

The denominators are 4 and 3.

LCM(4, 3) = 12
 

Step 2. Multiply both sides by 12.

\[ 12\left(\frac{x-1}{4}\right) \leq 12\left(\frac{x+2}{3}\right) \]

Simplify:

\[ 3(x-1)\leq4(x+2) \]

Step 3. Expand the brackets.

\[ 3x-3\leq4x+8 \]

Step 4. Subtract \(3x\) from both sides.

\[ -3\leq x+8 \]

Step 5. Subtract 8 from both sides.

\[ -11\leq x \]

Rewrite with the variable on the left.

Final answer:

x ≥ −11

Interval notation:

\[ [-11,\infty) \]

Important: Multiplying by a positive LCM is safe. If a denominator contains a variable, its sign and whether it can equal zero must be considered separately. Those more advanced cases require additional techniques.

 

01.4.7 Inequalities with All Real Numbers or No Solution

Not every inequality produces a solution such as \(x>3\) or \(x\leq7\).

Sometimes the variable disappears after simplification.

Two special situations can occur.

Case 1: All real numbers are solutions

Consider:

\[ 3(x+2)\geq3x+6 \]

Expand:

\[ 3x+6\geq3x+6 \]

Subtract \(3x\) from both sides:

\[ 6\geq6 \]

This statement is always true.

Therefore, every real number satisfies the original inequality.

Final answer:

x ∈ ℝ

Interval notation:

\[ (-\infty,\infty) \]

Case 2: No solution

Consider:

\[ 4(x-1)<4x-7 \]

Expand:

\[ 4x-4<4x-7 \]

Subtract \(4x\) from both sides:

\[ -4<-7 \]

This statement is false.

Therefore, there is no real number that satisfies the inequality.

Final answer:

∅

Here, ∅
   represents the empty set.

Result after simplification Meaning Solution set
Always true, such as 5 < 8 All real numbers satisfy it ℝ
Always false, such as 9 < 4 No real number satisfies it ∅

01.4.8 Basic Compound Inequalities

A compound inequality combines two inequalities.

There are two common types:

  • Inequalities connected by AND.

  • Inequalities connected by OR.

Type 1: Compound inequalities using AND

Consider:

\[ 2\leq3x-1<11 \]

This means:

\[ 2\leq3x-1 \]

and simultaneously:

\[ 3x-1<11 \]

Both conditions must be satisfied.

Step 1. Add 1 to all three parts.

\[ 3\leq3x<12 \]

Step 2. Divide all three parts by 3.

\[ 1\leq x<4 \]

Final answer:

1 ≤ x < 4

Interval notation:

\[ [1,4) \]

This includes 1 but excludes 4.

Type 2: Compound inequalities using OR

Consider:

\[ x+1<-2 \]

OR:

\[ 2x\geq8 \]

Solve each inequality separately.

First:

\[ x<-3 \]

Second:

\[ x\geq4 \]

Because the conditions are connected by OR, a value needs to satisfy at least one of them.

Final answer:

x < −3      or        x ≥ 4

 

Interval notation:

\[ (-\infty,-3)\cup[4,\infty) \]

Here, \(\cup\) represents the union of sets.

Remember: AND means the common solutions (intersection), while OR means all values satisfying at least one condition (union).

 

01.4.9 Representing Solutions on a Number Line

A number line provides a visual representation of an inequality’s solution set.

There are two important conventions:

    • An open circle means the boundary value is excluded.

    • A filled circle means the boundary value is included.

      Example A: x > 2

       

       


      →

       
       
       
       
       

       


      0
      1
      2
      3
      4

      Open circle at 2, arrow to the right.
      Interval: (2, ∞).

      Example B: x ≤ −1

       

       


      ←

       
       
       
       
       

       


      −3
      −2
      −1
      0
      1

      Filled circle at −1, arrow to the left.
      Interval: (−∞, −1].

       

      Interval Notation Summary

      Inequality Interval notation
      x > a (a, ∞)
      x ≥ a [a, ∞)
      x < a (−∞, a)
      x ≤ a (−∞, a]
      a < x < b (a, b)
      a ≤ x ≤ b [a, b]

      Important:
      Infinity is never included as an endpoint, so it always uses a parenthesis.

    • 01.4.10 Applying Inequalities to Word Problems

      Inequalities are used whenever a problem involves a range, minimum requirement, maximum limit, or comparison.

      01.4.10 Applying Inequalities to Word Problems

      Inequalities are used whenever a problem involves a range,
      minimum requirement, maximum limit, or comparison.

      Common phrases include:

      Word or phrase Mathematical meaning
      More than >
      Less than <
      At least ≥
      At most ≤
      No more than ≤
      No less than ≥
      Exceeds >

    • Example 17: Budget problem

      A student has a budget of $38 for learning materials. A required textbook costs $10, and each additional practice workbook costs $4.

      How many practice workbooks can the student buy without exceeding the budget?

      Step 1. Define the variable.

      Let \(n\) represent the number of practice workbooks.

      Step 2. Construct an inequality.

      \[ 10+4n\leq38 \]

      We use \(\leq\) because the total amount cannot exceed $38.

      Step 3. Subtract 10.

      \[ 4n\leq28 \]

      Step 4. Divide by 4.

      \[ n\leq7 \]

      Since the number of workbooks must be a nonnegative integer:

      \[ n\in\{0,1,2,\ldots,7\} \]

      Final answer: The student can buy at most 7 practice workbooks.

       

      01.4.11 Common Mistakes to Avoid

      Understanding these mistakes is essential for solving inequalities accurately.

      Mistake 1: Forgetting to reverse the sign

      ⊗
      Incorrect      		 ✓
      Correct
      −2x > 8 −2x > 8
      x > −4 x < −4
    • The correct solution reverses the sign when dividing by −2.

      Mistake 2: Treating strict inequalities as inclusive

      \[ x<5 \]

      does not mean \(x\leq5\).

      The value 5 is excluded from the solution set.

      Mistake 3: Incorrect distribution

      \[ -3(x-2)=-3x+6 \]

      Both terms inside the parentheses must be multiplied by −3.

      Mistake 4: Multiplying by a variable without checking its sign

      Suppose:

      \[ \frac{1}{x}>2 \]

      It is not valid to multiply both sides by \(x\) while automatically keeping the inequality sign. The variable \(x\) may be positive or negative, and \(x=0\) is undefined.

      This type of inequality requires sign analysis.

      Mistake 5: Giving only one number as the solution

      For example:

      \[ x>3 \]

      has infinitely many real-number solutions, not just \(x=4\).

      When the problem asks for the solution set, the answer must describe all satisfying values.

      Mistake 6: Forgetting the context of a word problem

      An algebraic solution might allow any real number, but quantities such as the number of books, questions, or students must generally be whole numbers.

      Always apply the restrictions given by the problem.

       

      01.4.12 Lesson Summary

      This lesson introduced the essential rules for working with inequalities.

      Students should remember the following principles:

      1. Adding or subtracting the same quantity on both sides preserves the inequality direction.

      2. Multiplying or dividing by a positive number preserves the inequality direction.

      3. Multiplying or dividing by a negative number reverses the inequality direction.

      4. Solving a linear inequality means isolating the variable using equivalent operations.

      5. Compound inequalities require identifying the intersection (AND) or union (OR) of solution sets.

      6. A simplified inequality that is always true has all real numbers as solutions; one that is always false has no solution.

      7. Interval notation and number lines are useful ways to represent solution sets.

      Essential formula to memorize:


      For c ≠ 0,

       

    • a < b



      ⟹



      {

      ac < bc, c > 0
      ac > bc, c < 0

    • The same sign rule applies to division and to non-strict inequalities.

       
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