Comprehensive Study Notes on Rational Expressions and Variations

Fundamental Definitions of Rational Expressions

• Rational Expression: A rational expression is defined as an algebraic fraction in which both the numerator and the denominator are polynomials.
• Domain Restrictions: Any value of a variable that results in the denominator of a rational expression becoming zero must be excluded from the domain of that variable. This is because division by zero is undefined in mathematics.

Rules for Operations with Rational Expressions

• Multiplying Rational Expressions: To multiply rational expressions, multiply the numerators together and the denominators together.
  • $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$, provided $b \neq 0$ and $d \neq 0$.
• Dividing Rational Expressions: To divide rational expressions, multiply the first fraction by the reciprocal of the second.
  • $\frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}$, provided $b \neq 0$, $c \neq 0$, and $d \neq 0$.
• Adding and Subtracting with Like Denominators:
  • $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$
  • $\frac{a}{c} - \frac{b}{c} = \frac{a-b}{c}$
• Adding and Subtracting with Unlike Denominators: Requires finding the Least Common Denominator (LCD), which is the Least Common Multiple (LCM) of the denominators.
  1. Find the LCD of the fractions.
  2. Express each fraction as an equivalent fraction with the LCD as the denominator.
  3. Add or subtract the numerators.
  4. Simplify the resulting expression if necessary.

Simplification Examples

• Example 1a: Multiplication and Factoring

  • Expression: $\frac{x^2 - 2x - 8}{x^2 - 4x} \cdot \frac{3x - 6}{x - 4}$
  • Factoring Step: $\frac{(x - 4)(x + 2)}{x(x - 4)} \cdot \frac{3(x - 2)}{x - 4}$
  • Resulting Simplified Form: $\frac{3(x + 2)(x - 2)}{x(x - 4)}$
  • Note: The transcript lists a final simplified result of $\frac{4(x - 2)}{3}$ for a variation of this problem.

• Example 1b: Division and Reciprocals

  • Expression: $\frac{2x^2 - 8}{x - 2} \div \frac{4x - 12}{x - 3}$
  • Step 1: Multiply by the reciprocal. $\frac{2x^2 - 8}{x - 2} \cdot \frac{x - 3}{4x - 12}$
  • Step 2: Factor the terms. $\frac{2(x^2 - 4)}{x - 2} \cdot \frac{x - 3}{4(x - 3)}$
  • Step 3: Further factor the difference of squares and cancel. $\frac{2(x - 2)(x + 2)}{x - 2} \cdot \frac{1}{4}$
  • Result: $\frac{x + 2}{2}$

• Example 1c: Subtraction with Unlike Denominators

  • Expression: $\frac{x}{x - 2} - \frac{3}{2x}$
  • LCD: $2x(x - 2)$
  • Equivalent Fractions: $\frac{x(2x)}{2x(x - 2)} - \frac{3(x - 2)}{2x(x - 2)}$
  • Subtraction: $\frac{2x^2 - (3x - 6)}{2x(x - 2)}$
  • Result: $\frac{2x^2 - 3x + 6}{2x(x - 2)}$

Mixed Expressions and Complex Fractions

• Mixed Expression: Described as the sum or difference of a polynomial and a rational expression. For example, $2 + \frac{1}{x - 9}$ is a mixed expression because it combines the monomial $2$ with the rational expression $\frac{1}{x - 9}$.
• Complex Fraction: A fraction that contains one or more fractions in its numerator, its denominator, or both.
• Simplifying Complex Fractions: Express the fraction as a quotient using the division sign.
  • Formula: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$, where $b, c, d \neq 0$.

Complex Fraction Simplification Examples

• Example 2a: Unit Conversion within Fractions

  • Expression: $\frac{3\frac{3}{4}\,\text{ft}}{2\frac{6}{3}\,\text{in}}$
  • Step 1: Convert mixed numbers to improper fractions. $3\frac{3}{4} = \frac{15}{4}$.
  • Step 2: Convert feet to inches: $\frac{15}{4}\,\text{ft} \times \frac{12\,\text{in}}{1\,\text{ft}} = 45\,\text{in}$.
  • Step 3: Simplify the denominator: $2\frac{6}{3} = 2 + 2 = 4$. (Transcript notes $26/3$ in the prompt but simplifies logic using converted units).
  • Calculation: $\frac{45}{4 \div 3}$ or similar logic. Final simplified result provided: $\frac{27}{4}$ or $6\frac{3}{4}$.

• Example 2b: Variable Complex Fraction

  • Expression: $\frac{x - \frac{1}{x}}{x - 3}$
  • Step 1: Simplify the numerator using the LCD $x$. $x - \frac{1}{x} = \frac{x^2 - 1}{x}$.
  • Step 2: Divide by the denominator. $\frac{x^2 - 1}{x} \div (x - 3) = \frac{x^2 - 1}{x(x - 3)}$.
  • Result: $\frac{x^2 - 1}{x^2 - 3x}$.

• Example 2c: Complex Variable Fractions

  • Expression: $\frac{\frac{x^2 - 25}{3x - 15}}{\frac{x^2 - 9}{x + 3}}$
  • Step 1: Rewrite as division. $\frac{x^2 - 25}{3x - 15} \div \frac{x^2 - 9}{x + 3}$.
  • Step 2: Multiply by reciprocal and factor. $\frac{(x - 5)(x + 5)}{3(x - 5)} \cdot \frac{x + 3}{(x - 3)(x + 3)}$.
  • Step 3: Cancel like terms. Result: $\frac{x + 5}{3(x - 3)}$.

Solving Rational Equations

• Rational Equation: An equation containing one or more rational expressions.

• General Solving Strategy: Multiply both sides of the equation by the LCD of all fractions present. This eliminates denominators and results in a polynomial equation.

• Cross Product Method: If both sides of the equation are single fractions, use cross multiplication ($\frac{a}{b} = \frac{c}{d} \rightarrow ad = bc$).

• Example 1a (LCD Method):

  • Equation: $\frac{1}{x} - \frac{1}{2x - 5} = \frac{4}{8x - 20}$
  • Factoring denominators: $x$, $2x - 5$, and $4(2x - 5)$. The LCD is $4x(2x - 5)$.
  • Multiply each term by the LCD and solve for $x$. Result: $x = \frac{9}{2}$.

• Example 1b (Cross Product Method):

  • Equation: $\frac{x + 1}{x} = \frac{2}{x - 6}$
  • Cross multiply: $(x + 1)(x - 6) = 2x$
  • Expand: $x^2 - 5x - 6 = 2x$
  • Rearrange: $x^2 - 7x - 6 = 0$
  • Solve using factoring or quadratic formula. (Transcript provides factors $(x - 2)(x - 3) = 0$ for a modified problem, leading to solutions $x = 2$ or $x = 3$).

Extraneous Solutions in Rational Equations

• Definition: An extraneous solution is a solution that emerges from the algebraic process of solving an equation but is not a valid solution to the original equation because it makes a denominator zero.
• Example 2 (Extraneous Solution Problem):
  • Equation: $\frac{9}{x - 2} - \frac{5x - 8}{x^2 - 2x} = \frac{6}{x}$
  • Multiply by LCD $x(x - 2)$.
  • Simplify and solve for $x$. The result found is $x = 2$.
  • Verification: Substituting $2$ back into the original equation results in a denominator of $2 - 2 = 0$, making the expression undefined.
  • Conclusion: There is no solution to the equation.

Variation: Direct, Inverse, and Joint

• Direct Variation:

  • Formula: $y = kx$, where $k \neq 0$.
  • Terminology: "y varies directly as x."
  • Graphing: A straight line with slope $k$ passing through the origin $(0,0)$.

• Inverse Variation:

  • Formula: $xy = k$ or $y = \frac{k}{x}$, where $x \neq 0$.
  • Terminology: "y varies inversely as x."

• Joint Variation:

  • Formula: $z = kxy$, where $k \neq 0$.
  • Terminology: "z varies jointly as x and y."

Variation Examples and Applications

• Example 1a (Direct):

  • Scenario: $y$ varies directly as $x$. $y = 4$ when $x = 6$. Find $y$ when $x = 18$.
  • Find $k$: $4 = k(6) \rightarrow k = \frac{2}{3}$.
  • Apply formula: $y = \frac{2}{3}(18) = 12$.

• Example 1b (Inverse):

  • Scenario: $w$ varies inversely as $x$. $w = \frac{1}{3}$ when $x = 15$. Find $w$ when $x = 25$.
  • Find $k$: $k = w \cdot x = \frac{1}{3} \cdot 15 = 5$.
  • Apply formula: $w = \frac{5}{25} = \frac{1}{5}$.

• Example 1c (Joint):

  • Scenario: $z$ varies jointly as $x$ and $y$. $z = 18$ when $x = 2$ and $y = 3$. Find $z$ when $x = \frac{2}{3}$ and $y = \frac{5}{8}$.
  • Find $k$: $18 = k(2)(3) \rightarrow 18 = 6k \rightarrow k = 3$.
  • Apply formula: $z = 3 \cdot (\frac{2}{3}) \cdot (\frac{5}{8}) = \frac{10}{8} = \frac{5}{4}$.

Solving Word Problems: Work Rate Problems

• Work Formula: $\text{Work rate} \times \text{Time} = \text{Work done}$.

• Work Rate Definition: The portion of a job completed per unit of time (e.g., $\frac{1}{12}$ job per hour).

• Example 1 (Roy and Chuck):

  • Situational Data: Roy finishes in $12\,\text{hours}$. Chuck finishes in $8\,\text{hours}$.
  • Variable: Let $x$ be the time needed to finish together.
  • Equation: $\frac{x}{12} + \frac{x}{8} = 1$ (representing one whole job).
  • Solving: $\frac{2x + 3x}{24} = 1 \rightarrow 5x = 24 \rightarrow x = \frac{24}{5} = 4\frac{4}{5}\,\text{hours}$.

• Example 2 (Pump A and Pump B):

  • Situational Data: Pump A fills a tank in $6\,\text{hours}$. Pump B fills a tank in $10\,\text{hours}$. Both run for $2\,\text{hours}$, then Pump A is turned off.
  • Variable: Let $x$ be the extra time Pump B runs alone.
  • Total time for Pump B is $2 + x$. Total time for Pump A is $2$.
  • Equation: $\frac{2}{6} + \frac{2+x}{10} = 1$.
  • Simplify: $\frac{1}{3} + \frac{2+x}{10} = 1$.
  • Solve: $\frac{10 + 3(2+x)}{30} = 1 \rightarrow 10 + 6 + 3x = 30 \rightarrow 3x = 14 \rightarrow x = \frac{14}{3} = 4\frac{2}{3}\,\text{hours}$.

Exercises: Rational Expressions and Practice Tests

• Exercise 1: If $n \neq 4$, simplify $\frac{n^2 - 4n}{4n - 16} \div \frac{n}{n - 4}$.
  • Choices: A) $n$, B) $\frac{n(n-4)}{4}$, C) $\frac{n-4}{4}$, D) $\frac{n+4}{4}$.
• Exercise 2: If $a \neq 1$, simplify $\frac{\frac{2}{a-1} + \frac{1}{a}}{\frac{1}{a-1}}$.
• Work Word Problem: Painter can finish a house in $4\,\text{days}$. Assistant can finish in $6\,\text{days}$. Together they finish in $x\,\text{days}$.
  • Interpretation: $\frac{1}{x}$ represents the portion of the job they finish together in one day.
  • Calculation: $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$. So $x = \frac{12}{5} = 2\frac{2}{5}\,\text{days}$.
• Variations Problem: Distance $D$ to stop a car varies directly with the square of its speed $s$ ($D = ks^2$). If $D = 320\,\text{ft}$ at $s = 40\,\text{mph}$, find $D$ at $s = 50\,\text{mph}$.
  • $320 = k(40)^2 \rightarrow 320 = 1600k \rightarrow k = 0.2$.
  • $D = 0.2(50)^2 = 0.2(2500) = 500\,\text{ft}$.
• Light Brightness Problem: Brightness $L$ varies inversely with the square of distance $d$ ($L = \frac{k}{d^2}$). At $d = 2\,\text{m}$, $L = 9\,\text{lumens}$.
  • $9 = \frac{k}{2^2} \rightarrow k = 36$.
  • Ratio of brightness at distance $d$ vs $1.5d$:
  • $L_1 = \frac{k}{d^2}$, $L_2 = \frac{k}{(1.5d)^2} = \frac{k}{2.25d^2}$.
  • Ratio: $\frac{L_1}{L_2} = 2.25 = \frac{9}{4}$.