3.1 Linear Equations & Linear Inequalities

3.1 Linear Equations & Linear Inequalities

Equations and inequalities form the fundamental bedrock of college algebra. Mastering single-variable linear equations, rational equations with domain restrictions, and linear inequalities prepares candidates for analyzing complex mathematical relationships throughout the CLEP College Algebra examination.

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1. Single-Variable & Multi-Step Linear Equations

A linear equation in one variable is a first-degree polynomial equation that can be written in the standard form:

$ax + b = 0$

where $a$ and $b$ are real numbers and $a \neq 0$. The exponent on the variable $x$ is implicitly $1$.

Fundamental Properties of Equality

To solve linear equations, we apply inverse operations governed by the axiomatic properties of equality:

• Addition/Subtraction Property of Equality: If $a = b$, then $a + c = b + c$ and $a - c = b - c$. Adding or subtracting the same quantity from both sides preserves equality.
• Multiplication/Division Property of Equality: If $a = b$ and $c \neq 0$, then $a \cdot c = b \cdot c$ and $\frac{a}{c} = \frac{b}{c}$. Multiplying or dividing both sides by a non-zero quantity preserves equality.

Classification of Equations

Linear equations are categorized into three distinct classes based on their solution sets:

1. Conditional Equation: An equation that is true for only a specific, finite set of values. Example: $2x + 3 = 7 \implies x = 2$.
2. Identity: An equation that is true for all real numbers in its domain. Simplifying an identity yields a statement that is universally true, such as $0 = 0$ or $5 = 5$. Solution set: $(-\infty, \infty)$.
3. Inconsistent Equation: An equation that has no solution because it contains a logical contradiction. Simplifying an inconsistent equation leads to a false statement, such as $-23 = -13$ or $0 = 8$. Solution set: empty set ($\emptyset$).

Systematic Multi-Step Solving Algorithm

1. Clear Fractions or Decimals: Multiply every term on both sides of the equation by the Least Common Denominator (LCD) or an appropriate power of 10.
2. Distribute & Remove Grouping Symbols: Apply the distributive property $a(b + c) = ab + ac$ to eliminate parentheses, brackets, or braces.
3. Combine Like Terms: Collect variable terms and constant terms independently on each side of the equation.
4. Isolate Variable Terms: Use addition or subtraction to transfer all terms containing the target variable to one side and all constant terms to the opposite side.
5. Isolate the Variable: Divide both sides by the variable's numerical coefficient to obtain $x = c$.
6. Verify the Solution: Substitute the resulting value back into the original equation to verify equality.

Worked Example 1: Multi-Step Linear Equation with Distributive Expansion

Problem: Solve for $x$: $3(2x - 5) - 4(x + 2) = 2(x - 7) + 1$

Step-by-Step Solution:

1. Distribute constants across parentheses: $6x - 15 - 4x - 8 = 2x - 14 + 1$
2. Combine like terms on each side: $(6x - 4x) + (-15 - 8) = 2x + (-14 + 1)$ $2x - 23 = 2x - 13$
3. Isolate variable terms by subtracting $2x$ from both sides: $2x - 2x - 23 = 2x - 2x - 13$ $-23 = -13$
4. Interpret Result: The variable terms cancel completely, leaving the false numerical statement $-23 = -13$. Therefore, this equation is an inconsistent equation, and its solution set is no solution ($\emptyset$).

Worked Example 2: Multi-Step Equation with Fractions

Problem: Solve for $x$: $\frac{2x - 1}{3} - \frac{x + 4}{5} = \frac{x - 2}{15} + 2$

Step-by-Step Solution:

1. Identify the LCD: The denominators are 3, 5, and 15. The LCD is $15$.
2. Clear fractions by multiplying every term by 15: $15 \cdot \left(\frac{2x - 1}{3}\right) - 15 \cdot \left(\frac{x + 4}{5}\right) = 15 \cdot \left(\frac{x - 2}{15}\right) + 15 \cdot (2)$ $5(2x - 1) - 3(x + 4) = 1(x - 2) + 30$
3. Distribute and expand terms: $10x - 5 - 3x - 12 = x - 2 + 30$
4. Combine like terms: $7x - 17 = x + 28$
5. Isolate $x$ on the left side: $7x - x = 28 + 17$ $6x = 45 \implies x = \frac{45}{6} = \frac{15}{2} = 7.5$

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2. Rational Linear Equations & Domain Restrictions

A rational equation contains one or more algebraic fractions with variables in the denominator.

The Critical Rule: Screening for Domain Restrictions

Before performing algebraic manipulations on a rational equation, you MUST identify all values of the variable that cause any denominator to evaluate to zero. Division by zero is undefined in real arithmetic. These restricted values must be excluded from the domain.

$\text{If } \frac{P(x)}{Q(x)} \text{ appears in an equation, we require } Q(x) \neq 0$

If an algebraic step yields a candidate solution that matches a restricted domain value, that candidate is an extraneous solution and must be discarded.

Worked Example 3: Rational Equation with Valid Solution

Problem: Solve for $x$: $\frac{2}{x - 3} + \frac{1}{x} = \frac{5}{x(x - 3)}$

Step-by-Step Solution:

1. Identify Domain Restrictions: Set each denominator equal to zero:
   • $x - 3 = 0 \implies x = 3$
   • $x = 0$
   • Restrictions: $x \neq 0$ and $x \neq 3$.
2. Determine the LCD: The LCD of the denominators is $x(x - 3)$.
3. Multiply every term by the LCD $x(x - 3)$: $x(x - 3) \cdot \frac{2}{x - 3} + x(x - 3) \cdot \frac{1}{x} = x(x - 3) \cdot \frac{5}{x(x - 3)}$ $2x + 1(x - 3) = 5$
4. Simplify and solve for $x$: $2x + x - 3 = 5$ $3x - 3 = 5 \implies 3x = 8 \implies x = \frac{8}{3}$
5. Screen against Domain Restrictions: Compare $x = \frac{8}{3}$ against restrictions ($x \neq 0, 3$). Since $\frac{8}{3}$ does not violate any restriction, $x = \frac{8}{3}$ is a valid solution.

Worked Example 4: Rational Equation That Produces an Extraneous Root

Clearing denominators is not a reversible step: multiplying both sides by an expression that can equal zero may manufacture a root the original equation never had. Any candidate that lands on a restricted value is extraneous.

Problem: Solve for $x$: $\frac{x}{x - 4} + \frac{4}{x + 1} = \frac{20}{(x - 4)(x + 1)}$

Step-by-Step Solution:

1. Identify Domain Restrictions before touching the algebra:
   • $x - 4 = 0 \implies x \neq 4$
   • $x + 1 = 0 \implies x \neq -1$
2. Multiply every term by the LCD $(x - 4)(x + 1)$: $x(x + 1) + 4(x - 4) = 20$
3. Expand and collect into standard quadratic form: $x^2 + x + 4x - 16 = 20 \implies x^2 + 5x - 36 = 0$
4. Factor and read off both candidates: $(x + 9)(x - 4) = 0 \implies x = -9 \quad \text{or} \quad x = 4$
5. Screen each candidate against the restrictions:
   • $x = 4$ is exactly an excluded value. It makes the denominator $x - 4$ zero in the original equation, so it is extraneous and must be discarded.
   • $x = -9$ violates nothing. Verify it: the left side is $\frac{-9}{-13} + \frac{4}{-8} = \frac{9}{13} - \frac{1}{2} = \frac{18 - 13}{26} = \frac{5}{26}$, and the right side is $\frac{20}{(-13)(-8)} = \frac{20}{104} = \frac{5}{26}$. They match.
6. Conclusion: the solution set is $\{-9\}$, not $\{-9, 4\}$.

Exam Trap: CLEP distractors on rational equations almost always include the full candidate list (here, $-9$ and $4$) and often the extraneous value alone. Solving correctly but skipping the domain screen lands you on a wrong choice that the test writers deliberately supplied.

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Worked Example 5: Every Candidate Extraneous

Problem: Solve $\dfrac{x}{x - 3} = \dfrac{3}{x - 3} + 4$.

Step-by-Step Solution:

1. Restriction: $x \neq 3$.
2. Multiply by $(x - 3)$: $x = 3 + 4(x - 3)$.
3. Solve: $x = 3 + 4x - 12 \implies -3x = -9 \implies x = 3$.
4. Screen: the sole candidate is the excluded value, so it is extraneous.
5. Conclusion: the equation has no solution ($\emptyset$). Note that "no solution" here comes from a domain conflict, not from a contradiction such as $-23 = -13$; both outcomes report an empty solution set but arise for different reasons.

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3. Linear Inequalities & Interval Notation

A linear inequality in one variable is structured similarly to a linear equation, but uses an inequality symbol ($<, \le, >, \ge$).

The Negative Multiplication/Division Reversal Rule

CRITICAL RULE: When multiplying or dividing both sides of an inequality by a negative real number, you MUST flip (reverse) the direction of the inequality sign ($<$ becomes $>$, $\le$ becomes $\ge$, and vice versa).

Intuitive Proof: We know that $2 < 5$ is a true statement. Multiplying both sides by $-1$ yields $-2$ and $-5$. On the real number line, $-2$ lies to the right of $-5$, so $-2 > -5$. Thus, multiplying by a negative number reverses order relationships.

Worked Example 6: Linear Inequality with Sign Reversal

Problem: Solve the inequality and write the solution set in interval notation: $-4(2x - 3) \ge 3x + 23$

Step-by-Step Solution:

1. Distribute on the left side: $-8x + 12 \ge 3x + 23$
2. Subtract $3x$ from both sides: $-11x + 12 \ge 23$
3. Subtract $12$ from both sides: $-11x \ge 11$
4. Divide both sides by $-11$ and REVERSE the inequality symbol: $x \le \frac{11}{-11} \implies x \le -1$
5. Write in Interval Notation: The solution set contains all real numbers less than or equal to $-1$, written as $(-\infty, -1]$.

Summary Table: Inequality Notation & Interval Representations

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4. Compound Inequalities (AND vs. OR)

A compound inequality combines two inequalities using the logical operators AND (conjunction) or OR (disjunction).

Conjunctions (AND / Intersection $\cap$)

An AND compound inequality requires both component statements to be satisfied simultaneously. The resulting solution set is the intersection ($\cap$) of the individual solution sets.

Worked Example 7: Bounded Compound Inequality (AND)

Problem: Solve for $x$ and express in interval notation: $-5 \le 3x + 1 < 13$

Step-by-Step Solution: Perform operations simultaneously across all three parts of the compound inequality:

1. Subtract $1$ from all three sections: $-5 - 1 \le 3x + 1 - 1 < 13 - 1$ $-6 \le 3x < 12$
2. Divide all three sections by $3$: $\frac{-6}{3} \le \frac{3x}{3} < \frac{12}{3}$ $-2 \le x < 4$
3. Interval Notation: $[-2, 4)$.

Disjunctions (OR / Union $\cup$)

An OR compound inequality requires at least one of the component statements to be true. The solution set is the union ($\cup$) of the individual solution sets.

Worked Example 8: Disjunction Compound Inequality (OR)

Problem: Solve for $x$: $4x - 1 < -9 \quad \text{or} \quad 2 - 3x \le -10$

Step-by-Step Solution:

1. Solve the left inequality: $4x - 1 < -9 \implies 4x < -8 \implies x < -2$ Interval notation: $(-\infty, -2)$.
2. Solve the right inequality: $2 - 3x \le -10 \implies -3x \le -12 \implies x \ge 4 \quad \text{(flipping inequality)}$ Interval notation: $[4, \infty)$.
3. Combine solution sets using Union ($\cup$): $(-\infty, -2) \cup [4, \infty)$