Comprehensive Notes on Rational Expressions and Variations

Direct Variation
- A direct variation is an equation written in the form $y = kx$ , where $k \neq 0$ .
- It is expressed as: " $y$ varies directly as $x$ ."
- The graph of a direct variation is characteristic of a straight line with a slope equivalent to the constant $k$ . This line must pass through the origin $(0, 0)$ .
Inverse Variation
- An inverse variation is an equation represented as $xy = k$ or $y = \frac{k}{x}$ , where $x \neq 0$ .
- This relationship is expressed as: " $y$ varies inversely as $x$ ."
Joint Variation
- A joint variation is defined by the form $z = kxy$ , where $k \neq 0$ .
- It is expressed as: " $z$ varies jointly as $x$ and $y$ ."
Process for solving Joint Variation problems
- Start with the joint variation formula: $z = kxy$ .
- Replace known variables: For instance, if given that $z = 18$ , $x = 2$ , and $y = 3$ , substitute these into the formula: $18 = k(2)(3)$ .
- Solve for the constant $k$ : $18 = 6k \rightarrow k = 3$ .
- Establish the formula with the found constant: $z = 3xy$ .
- Use the new formula to find unknown values: For example, if given new values for $x$ and $y$ , substitute them to solve for $z$ . If solving for a different variable like $w$ , substitute and solve accordingly.

Automobile Stopping Distance
- The distance it takes for an automobile to stop varies directly as the square of its speed ( $d = ks^2$ ).
- Given: A car traveling at $40\,mph$ has a stopping distance of $320\,feet$ .
- Calculation of $k$ : $320 = k(40)^2$ $320 = 1600k$ $k = \frac{320}{1600} = 0.2$
- Solving for new speed: Find the stopping distance for a car traveling at $50\,mph$ . $d = 0.2 \times (50)^2$ $d = 0.2 \times 2500$ $d = 500\,ft$
Inverse Variation with Roots
- Scenario: $y$ varies inversely as $\sqrt{x}$ . Given $y = 12$ when $x = 16$ .
- Formula: $y = \frac{k}{\sqrt{x}}$ .
- Solving for $k$ : $12 = \frac{k}{\sqrt{16}}$ $12 = \frac{k}{4}$ $k = 48$
- Finding $y$ when $x = 100$ : $y = \frac{48}{\sqrt{100}}$ $y = \frac{48}{10} = 4.8$
Light Brightness and Distance (Inverse Square Law)
- The brightness of light ( $L$ ) varies inversely as the square of the distance ( $d$ ) from the light source ( $L = \frac{k}{d^2}$ ).
- $L$ is measured in lumens; $d$ is measured in meters.
- Example problem: At a distance of $2\,meters$ , the brightness is measured at $9\,lumens$ .
- Solving for $k$ : $9 = \frac{k}{2^2}$ $9 = \frac{k}{4}$ $k = 36$
- Ratio comparison: Brightness at distance $d$ compared to distance $1.5d$ .
- $L_1 = \frac{k}{d^2}$
- $L_2 = \frac{k}{(1.5d)^2} = \frac{k}{2.25d^2}$
- Ratio $L_1 : L_2 = \frac{k}{d^2} / \frac{k}{2.25d^2} = 2.25 : 1$

General Procedure
- Rational equations contain rational expressions.
- Multiply every term on both sides of the equation by the Least Common Denominator (LCD) of all the fractions to clear the denominators.
- Solve the resulting polynomial equation.
- Cross-multiplication may be used if both sides of the equation are single fractions.
Extraneous Solutions and Undefined Expressions
- A solution is invalid if it makes the denominator of the original equation equal to zero.
- Example: Solve $4x - 8 = 6x - 12$ . Solving leads to $2 = x$ . However, if substituting $2$ back into the original equation results in an undefined expression (e.g., division by zero), then the equation has no solution.
Undefined Functions
- A function is undefined at any value of $x$ that results in a denominator of zero.
- Case 1: $f(x) = \frac{1}{(x-a)^2 - 4(x-a) + 4}$ . If defined except where $x = 6$ , then the denominator must be zero at $x=6$ . $(6-a)^2 - 4(6-a) + 4 = 0$ which is $( (6-a) - 2 )^2 = 0 \rightarrow 4-a = 0 \rightarrow a = 4$ .
- Case 2: $g(x) = \frac{1}{(x+3)^2 - 24(x+3) + 144}$ . To find where $g(x)$ is undefined, set the denominator to zero: $( (x+3) - 12 )^2 = 0$ $x + 3 - 12 = 0$ $x - 9 = 0 \rightarrow x = 9$ .

Work Rate Formula
- Standard formula: $\text{Work rate} \times \text{Time} = \text{Work done}$ .
- Definition of work rate: The specific amount of a job completed over a set duration of time.

Factoring and Division
- When dividing rational expressions, rewrite as a multiplication sentence by using the reciprocal of the divisor.
- Example: $\frac{3x}{15} / \frac{x^2 - 9}{5x}$
- Reciprocal: $\frac{3x}{15} \times \frac{5x}{x^2 - 9}$
- Factoring: $\frac{3x}{15} \times \frac{5x}{(x+3)(x-3)}$
- Simplify by cancelling terms and factors.
Equivalent Expressions
- Complex Fraction: $1 + \frac{1}{y-1}$
- Common denominator: $\frac{y-1}{y-1} + \frac{1}{y-1} = \frac{y-1+1}{y-1} = \frac{y}{y-1}$ .
- Fraction addition with variable signs: $\frac{n^2}{n-4} + \frac{4n}{4-n}$
- Notice $4-n = -(n-4)$ .
- $\frac{n^2}{n-4} - \frac{4n}{n-4} = \frac{n^2 - 4n}{n-4} = \frac{n(n-4)}{n-4} = n$ (where $n \neq 4$ ).