2.3 Solving Algebraic Inequalities

Key Takeaways

  • Algebraic inequalities ($<, \le, >, \ge$) define a continuous range of solutions on a number line rather than a single discrete point.
  • Crucial Sign-Flip Rule: Whenever you multiply or divide an inequality by a negative number, you MUST reverse the direction of the inequality sign.
  • Compound inequalities combine two conditions using AND (conjunction, bounded intervals) or OR (disjunction, separate rays).
  • National Electrical Code (NEC) continuous load rules state that continuous loads (running 3+ hours) must not exceed 80% of circuit breaker capacity ($I_{\text{continuous}} \le 0.80 \times I_{\text{breaker}}$).
  • Inequality modeling ensures electrical circuits, overcurrent protective devices, and conductor thermal limits are properly sized to prevent hazards.
Last updated: August 2026

2.3 Solving Algebraic Inequalities

Quick Summary: Algebraic inequalities express mathematical relationships where one expression is greater than, less than, or bounded relative to another. While solving inequalities follows steps similar to solving linear equations, multiplying or dividing an inequality by a negative number requires reversing the direction of the inequality sign. In electrical trades, inequalities are critical for sizing circuit breakers, determining maximum continuous load currents under the National Electrical Code (NEC), and enforcing safety thresholds.


Principles & Symbols of Algebraic Inequalities

Unlike equations that yield discrete, exact solutions (e.g., $x = 5$), an algebraic inequality defines a continuous range of valid solutions (a solution set).

Inequality Symbols & Number Line Representation

SymbolMeaningBoundary Circle on Number LineInterval Notation Bracket
$<$Strictly Less ThanOpen Circle ($\circ$)Parentheses $( \dots )$
$>$Strictly Greater ThanOpen Circle ($\circ$)Parentheses $( \dots )$
$\le$Less Than or Equal ToClosed Circle ($\bullet$)Square Bracket $[ \dots ]$
$\ge$Greater Than or Equal ToClosed Circle ($\bullet$)Square Bracket $[ \dots ]$
                     Number Line Solutions
                     
  x < 5  :  <======o------------>  (Open circle at 5, shaded left)
                   5
  x >= -3:  <------•=======>  (Closed circle at -3, shaded right)
                  -3

Core Operational Rules & The Critical Negative Sign-Flip Rule

Solving a linear inequality involves isolating the variable using properties similar to linear equations:

  • Addition & Subtraction Property: Adding or subtracting the same quantity from both sides preserves the inequality direction. If x4>10    x>14\text{If } x - 4 > 10 \implies x > 14
  • Positive Multiplication & Division Property: Multiplying or dividing both sides by a positive number preserves the inequality direction. If 3x18    x6\text{If } 3x \le 18 \implies x \le 6

The Critical Negative Multiplier/Divisor Sign-Flip Rule

The most frequent source of error on the EIAT occurs when multiplying or dividing an inequality by a negative number.

CRITICAL RULE: Whenever you multiply or divide both sides of an inequality by a negative quantity, you MUST reverse (flip) the inequality direction ($>$ becomes $<$, $\le$ becomes $\ge$, etc.).

Geometric & Numerical Proof of the Sign-Flip Rule

Consider the true numerical statement: 2<52 < 5

If we multiply both sides by positive 2: 2(2)<5(2)    4<10(Still True)2(2) < 5(2) \implies 4 < 10 \quad \text{(Still True)}

However, if we multiply both sides of $2 < 5$ by negative 1 ($-1$): 2(1)vs.5(1)    2vs.52(-1) \quad \text{vs.} \quad 5(-1) \implies -2 \quad \text{vs.} \quad -5

On a standard number line, $-2$ lies to the right of $-5$, meaning $-2$ is greater than $-5$. Thus, to maintain truth, the inequality sign must flip: 2>5-2 > -5

When solving $-4x > -20$, dividing both sides by $-4$ forces the sign to reverse: x<204    x<5x < \frac{-20}{-4} \implies x < 5


Solving Compound Inequalities

A compound inequality combines two inequalities into a single statement using the words AND (conjunction) or OR (disjunction).

Bounded Conjunction Inequalities (AND)

Bounded inequalities express a variable restricted between lower and upper boundaries, written as: 113x2<13-11 \le 3x - 2 < 13

To solve a three-part compound inequality, perform operations simultaneously across all three sections:

  1. Add 2 to All Three Parts: 11+23x2+2<13+2-11 + 2 \le 3x - 2 + 2 < 13 + 2 93x<15-9 \le 3x < 15
  2. Divide All Three Parts by Positive 3: 933x3<153\frac{-9}{3} \le \frac{3x}{3} < \frac{15}{3} 3x<5-3 \le x < 5

The solution set includes all real numbers $x$ that are simultaneously greater than or equal to $-3$ AND strictly less than $5$, expressed in interval notation as $[-3, 5)$.


Trade Application: Continuous Load Sizing & Circuit Protection

In electrical contracting, circuit protection components (such as branch circuit conductors and overcurrent protective devices / circuit breakers) are governed by strict safety limits set by the National Electrical Code (NEC).

The NEC 125% / 80% Continuous Load Rule

The NEC defines a continuous load as any electrical load where the maximum current is expected to continue for 3 hours or more (e.g., commercial office lighting, space heaters, storefront signage).

  • 80% Safety Rating: Standard circuit breakers must not be loaded continuously beyond 80% of their nominal ampere rating. Imax continuous0.80×IbreakerI_{\text{max continuous}} \le 0.80 \times I_{\text{breaker}}
  • 125% Breaker Sizing Rule: Alternatively, the breaker ampere rating must be at least 125% of the continuous load current: Ibreaker1.25×IcontinuousI_{\text{breaker}} \ge 1.25 \times I_{\text{continuous}}

Example: Sizing a 20-Amp Circuit

For a standard 20-Amp branch circuit breaker: Imax continuous0.80×20 A=16 AmpsI_{\text{max continuous}} \le 0.80 \times 20\text{ A} = 16\text{ Amps}

If an electrician connects multiple electrical space heaters drawing $3.2\text{ Amps}$ each, the continuous load inequality is: 3.2×h163.2 \times h \le 16

Where $h$ represents the number of space heaters. Solving for $h$: h163.2    h5 heatersh \le \frac{16}{3.2} \implies h \le 5\text{ heaters}

Connecting 6 heaters would draw $6 \times 3.2 = 19.2\text{ Amps}$ continuously, exceeding the $16\text{ A}$ safety limit and tripping the breaker due to cumulative thermal buildup.


Step-by-Step Worked Examples

Worked Example 1: Linear Inequality with Negative Division

Problem: Solve the linear inequality for $x$: 4x+7>13-4x + 7 > -13

Solution:

  1. Subtract 7 from Both Sides: 4x>137    4x>20-4x > -13 - 7 \implies -4x > -20
  2. Divide Both Sides by Negative 4 and Flip the Inequality Sign ($>$ becomes $<$): x<204x < \frac{-20}{-4}
  3. Simplify Fraction: x<5x < 5

Final Answer: $x < 5$


Worked Example 2: Bounded Compound Inequality

Problem: Solve the three-part compound inequality: 113x2<13-11 \le 3x - 2 < 13

Solution:

  1. Add 2 across All Three Parts: 11+23x<13+2-11 + 2 \le 3x < 13 + 2 93x<15-9 \le 3x < 15
  2. Divide All Three Parts by Positive 3: 933x3<153\frac{-9}{3} \le \frac{3x}{3} < \frac{15}{3} 3x<5-3 \le x < 5

Final Answer: $-3 \le x < 5$


Worked Example 3: Industrial Lighting Circuit Load Inequality

Problem: An electrician is installing commercial LED high-bay fixtures on a 15-Amp circuit breaker. The maximum continuous current rating for the breaker is $12\text{ Amps}$ ($80%$ of $15\text{ A}$). Each LED fixture draws $0.75\text{ Amps}$. Write and solve an inequality to determine the maximum number of fixtures $f$ that can be safely installed.

Solution:

  1. Set Up the Load Inequality: Total Continuous CurrentMax Rating\text{Total Continuous Current} \le \text{Max Rating} 0.75f120.75 \cdot f \le 12
  2. Divide Both Sides by Positive $0.75$ (or $\frac{3}{4}$): f120.75=12×43=483=16f \le \frac{12}{0.75} = 12 \times \frac{4}{3} = \frac{48}{3} = 16
  3. Interpret Result: The electrician can install at most 16 LED fixtures ($f \le 16$).

Final Answer: $f \le 16\text{ fixtures}$

Test Your Knowledge

Solve the linear inequality for $x$: $-4x + 7 > -13$

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Test Your Knowledge

Solve the compound inequality: $-11 \le 3x - 2 < 13$

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Test Your Knowledge

An electrician is sizing a continuous load circuit on a 20-Amp breaker (max continuous rating 16 Amps). Each electric space heater draws 3.2 Amps. Write and solve an inequality for the maximum number of heaters $h$.

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D