Equations and Inequalities (1.2): Linear and Rational Equations

Chapter 1: Equations and Inequalities - 1.2 Linear and Rational Equations

Objectives

• Solve linear equations in one variable.

• Solve linear equations containing fractions.

• Solve rational equations with variables in the denominators.

• Recognize identities, conditional equations, and inconsistent equations.

Definition of a Linear Equation

A linear equation in one variable $x$ is an equation that can be written in the form $ax + b = 0$, where $a$ and $b$ are real numbers, and $a \neq 0$.

Generating Equivalent Equations

An equation can be transformed into an equivalent equation by one or more of the following operations:

1. Simplify Expressions: Remove grouping symbols and combine like terms.

2. Addition/Subtraction Property of Equality: Add or subtract the same real number or variable expression on both sides of the equation.

3. Multiplication/Division Property of Equality: Multiply or divide by the same nonzero quantity on both sides of the equation.

4. Interchange Sides: Interchange the two sides of the equation without altering its equality.

Solving a Linear Equation: Steps

1. Simplify: Simplify the algebraic expression on each side of the equation by removing grouping symbols (e.g., parentheses) and combining like terms.

2. Collect Variable Terms: Collect all variable terms on one side of the equation and all constant terms (numbers) on the other side.

3. Isolate and Solve: Isolate the variable term and solve for the variable by dividing by its coefficient.

4. Check Solution: Check the proposed solution in the original equation to ensure its validity.

Examples: Solving Linear Equations

Example 1: Basic Linear Equation

Solve and check: $4x + 5 = 29$

• Subtract $5$ from both sides: $4x + 5 - 5 = 29 - 5 \Rightarrow 4x = 24$

• Divide both sides by $4$: $\frac{4x}{4} = \frac{24}{4} \Rightarrow x = 6$

• Check: $4(6) + 5 = 29 \Rightarrow 24 + 5 = 29 \Rightarrow 29 = 29$ (The solution is correct).

Example 2: Linear Equation with Grouping Symbols

Solve and check: $2(x - 3) - 17 = 13 - 3(x + 2)$

• Distribute: $2x - 6 - 17 = 13 - 3x - 6$

• Combine like terms: $2x - 23 = -3x + 7$

• Collect variable terms (add $3x$ to both sides): $2x + 3x - 23 = -3x + 3x + 7 \Rightarrow 5x - 23 = 7$

• Collect constant terms (add $23$ to both sides): $5x - 23 + 23 = 7 + 23 \Rightarrow 5x = 30$

• Isolate $x$ (divide by $5$): $\frac{5x}{5} = \frac{30}{5} \Rightarrow x = 6$

Example 3: Complex Linear Equation

Solve and check: $25 - [2 + 5y - 3(y + 2)] = -3(2y - 5) - [5(y - 1) - 3y + 3]$

• Simplify within innermost grouping symbols first:
  $25 - [2 + 5y - 3y - 6] = -6y + 15 - [5y - 5 - 3y + 3]$

• Combine like terms within brackets:
  $25 - [2y - 4] = -6y + 15 - [2y - 2]$

• Remove brackets by distributing negative signs:
  $25 - 2y + 4 = -6y + 15 - 2y + 2$

• Combine like terms on each side:
  $-2y + 29 = -8y + 17$

• Collect variable terms (add $8y$ to both sides): $-2y + 8y + 29 = -8y + 8y + 17 \Rightarrow 6y + 29 = 17$

• Collect constant terms (subtract $29$ from both sides): $6y + 29 - 29 = 17 - 29 \Rightarrow 6y = -12$

• Isolate $y$ (divide by $6$): $\frac{6y}{6} = \frac{-12}{6} \Rightarrow y = -2$

Solving Linear Equations Involving Fractions

Steps for Equations with Constant Denominators

1. Find the Least Common Denominator (LCD) of all fractions in the equation.

2. Multiply every term on both sides of the equation by the LCD. This clears the fractions.

3. Solve the resulting linear equation using the steps outlined above.

4. Check the solution.

Example: Linear Equation with Fractions (Equation 1)

Solve and check: $\frac{x + 2}{4} - \frac{x - 1}{3} = 2$

• The LCD of $4$ and $3$ is $12$. Multiply both sides by $12$:
  $12 \left( \frac{x + 2}{4} - \frac{x - 1}{3} \right) = 12(2)$

• Distribute $12$ to each term:
  12 \left( \frac{x + 2}{4}
  ight) - 12 \left( \frac{x - 1}{3}
  ight) = 24

• Simplify by dividing out common factors:
  $3(x + 2) - 4(x - 1) = 24$

• Distribute:
  $3x + 6 - 4x + 4 = 24$

• Combine like terms:
  $-x + 10 = 24$

• Subtract $10$ from both sides: $-x + 10 - 10 = 24 - 10 \Rightarrow -x = 14$

• Multiply by $-1$ (or divide by $-1$) to make $x$ positive: $x = -14$

Example: Linear Equation with Fractions (Equation 2)

Solve and check: $\frac{x - 3}{4} = \frac{5}{14} - \frac{x + 5}{7}$

• The LCD of $4, 14, 7$ is $28$. Multiply both sides by $28$:
  28 \left( \frac{x - 3}{4}
  ight) = 28 \left( \frac{5}{14} \right) - 28 \left( \frac{x + 5}{7}
  ight)

• Simplify:
  $7(x - 3) = 2(5) - 4(x + 5)$

• Distribute:
  $7x - 21 = 10 - 4x - 20$

• Combine like terms on the right:
  $7x - 21 = -4x - 10$

• Collect variable terms (add $4x$ to both sides): $7x + 4x - 21 = -4x + 4x - 10 \Rightarrow 11x - 21 = -10$

• Collect constant terms (add $21$ to both sides): $11x - 21 + 21 = -10 + 21 \Rightarrow 11x = 11$

• Isolate $x$: $\frac{11x}{11} = \frac{11}{11} \Rightarrow x = 1$

• Check: Substitute $x = 1$ into the original equation.
  $\frac{1 - 3}{4} = \frac{5}{14} - \frac{1 + 5}{7}$
  $\frac{-2}{4} = \frac{5}{14} - \frac{6}{7}$
  $-\frac{1}{2} = \frac{5}{14} - \frac{12}{14}$
  $-\frac{1}{2} = \frac{-7}{14}$
  $-\frac{1}{2} = -\frac{1}{2}$ (The solution is correct).

Example: Linear Equation with Fractions (Equation 3)

Solve: $\frac{3}{5}x - x = 10 - \frac{5}{2}x$

• The LCD of $5$ and $2$ is $10$. Multiply both sides by $10$:
  $10 \left( \frac{3}{5}x \right) - 10x = 10(10) - 10 \left( \frac{5}{2}x \right)$

• Simplify:
  $2(3x) - 10x = 100 - 5(5x)$
  $6x - 10x = 100 - 25x$

• Combine like terms on the left:
  $-4x = 100 - 25x$

• Collect variable terms (add $25x$ to both sides): $-4x + 25x = 100 - 25x + 25x \Rightarrow 21x = 100$

• Isolate $x$: $x = \frac{100}{21}$

Rational Equations

• A rational equation is an equation that includes at least one variable in the denominator of a fraction.

• Unlike linear equations with constant denominators, rational equations require identifying restrictions on the variable, as denominators cannot be zero.

Steps for Solving Rational Equations

1. Identify Restrictions: Determine all values of the variable that make any denominator zero. These are restricted values and cannot be solutions to the equation. Explicitly write down the restrictions.

2. Find LCD: Find the Least Common Denominator (LCD) of all terms in the equation.

3. Clear Fractions: Multiply every term on both sides of the equation by the LCD. This will eliminate all denominators.

4. Solve: Solve the resulting linear or polynomial equation.

5. Check for Restrictions: Compare the proposed solution(s) against the restricted values. Any solution that matches a restricted value must be rejected. If all proposed solutions are restricted, then there is no solution.

Example 1: Basic Rational Equation

Solve: $\frac{5}{2x} = \frac{17}{18} - \frac{1}{3x}$

• Restrictions: Set denominators to zero: $2x = 0 \Rightarrow x = 0$ and $3x = 0 \Rightarrow x = 0$. So, $x \neq 0$.

• LCD: The LCD of $2x, 18, 3x$ is $18x$. Multiply both sides by $18x$:
  18x \left( \frac{5}{2x}
  ight) = 18x \left( \frac{17}{18}
  ight) - 18x \left( \frac{1}{3x}
  ight)

• Simplify:
  $9(5) = x(17) - 6(1)$
  $45 = 17x - 6$

• Collect constant terms (add $6$ to both sides): $45 + 6 = 17x - 6 + 6 \Rightarrow 51 = 17x$

• Isolate $x$: $\frac{51}{17} = \frac{17x}{17} \Rightarrow x = 3$

• Check Restrictions: The solution $x = 3$ is not equal to the restricted value $0$. Therefore, $x = 3$ is a valid solution.

Example 2: Rational Equation with No Solution

Solve: $\frac{5}{x + 2} + \frac{3}{x - 2} = \frac{12}{x^2 - 4}$

• Restrictions: Factor the denominator of the right side: $x^2 - 4 = (x + 2)(x - 2)$.
  Set denominators to zero:
  $x + 2 = 0 \Rightarrow x = -2$
  $x - 2 = 0 \Rightarrow x = 2$
  So, $x \neq -2$ and $x \neq 2$.

• LCD: The LCD is $(x + 2)(x - 2)$. Multiply both sides by the LCD:
  (x + 2)(x - 2) \left( \frac{5}{x + 2} \right) + (x + 2)(x - 2) \left( \frac{3}{x - 2}
  ight) = (x + 2)(x - 2) \left( \frac{12}{(x + 2)(x - 2)}
  ight)

• Simplify:
  $(x - 2)(5) + (x + 2)(3) = 12$

• Distribute:
  $5x - 10 + 3x + 6 = 12$

• Combine like terms:
  $8x - 4 = 12$

• Collect constant terms (add $4$ to both sides): $8x - 4 + 4 = 12 + 4 \Rightarrow 8x = 16$

• Isolate $x$: $\frac{8x}{8} = \frac{16}{8} \Rightarrow x = 2$

• Check Restrictions: The proposed solution $x = 2$ is a restricted value. Because $x$ cannot be $2$, this equation has no solution.

Example 3: Finding Values where Two Rational Expressions are Equal

Find all values of $x$ for which $y<em>1 = y</em>2$, where
$y<em>1 = \frac{1}{x + 4} + \frac{1}{x - 4}$ and $y</em>2 = \frac{22}{x^2 - 16}$

• Set $y<em>1 = y</em>2$:
  $\frac{1}{x + 4} + \frac{1}{x - 4} = \frac{22}{x^2 - 16}$

• Restrictions: Factor $x^2 - 16 = (x + 4)(x - 4)$.
  Set denominators to zero:
  $x + 4 = 0 \Rightarrow x = -4$
  $x - 4 = 0 \Rightarrow x = 4$
  So, $x \neq -4$ and $x \neq 4$.

• LCD: The LCD is $(x + 4)(x - 4)$. Multiply both sides by the LCD:
  (x + 4)(x - 4) \left( \frac{1}{x + 4} \right) + (x + 4)(x - 4) \left( \frac{1}{x - 4}
  ight) = (x + 4)(x - 4) \left( \frac{22}{(x + 4)(x - 4)}
  ight)

• Simplify:
  $(x - 4)(1) + (x + 4)(1) = 22$

• Distribute and combine:
  $x - 4 + x + 4 = 22$
  $2x = 22$

• Isolate $x$: $\frac{2x}{2} = \frac{22}{2} \Rightarrow x = 11$

• Check Restrictions: The solution $x = 11$ is not a restricted value. Therefore, $x = 11$ is the valid solution.

Types of Equations

Equations can be categorized based on their solution sets:

1. Conditional Equation: An equation that is true for at least one (but not all) real number(s). Most equations solved in algebra (e.g., $x = 6$ or $y = -2$) are conditional equations.

2. Identity: An equation that is true for all real numbers for which both sides of the equation are defined. When solved, an identity results in a true statement, such as $9 = 9$ or $0 = 0$. The solution set is the set of all real numbers.

3. Inconsistent Equation: An equation that is not true for any real number. When solved, an inconsistent equation results in a false statement, such as $-7 = -1$ or $0 = 5$. There is no solution to an inconsistent equation; its solution set is the empty set $\emptyset$

Examples: Categorizing Equations

Example 1: Inconsistent Equation

Solve and categorize: $4x - 7 = 4(x - 1) + 3$

• Distribute:
  $4x - 7 = 4x - 4 + 3$

• Combine like terms on the right:
  $4x - 7 = 4x - 1$

• Subtract $4x$ from both sides:
  $4x - 4x - 7 = 4x - 4x - 1$
  $-7 = -1$

• This is a false statement. Therefore, the equation is an inconsistent equation, and there is no solution.

Example 2: Identity

Solve and categorize: $7x + 9 = 9(x + 1) - 2x$

• Distribute:
  $7x + 9 = 9x + 9 - 2x$

• Combine like terms on the right:
  $7x + 9 = 7x + 9$

• Subtract $7x$ from both sides:
  $7x - 7x + 9 = 7x - 7x + 9$
  $9 = 9$

• This is a true statement. Therefore, the equation is an identity, and the solution set is the set of all real numbers.