Solving Literal Equations and Formula Rearrangement

Definition of Literal Equations:
- A literal equation is an equation that contains two or more variables.
- Rearranging formulas involves using inverse mathematical operations to isolate a target variable on one side of the equal sign.
Core Principles of Variable Isolation:
- Identify the target variable that needs to be isolated.
- Apply inverse operations (addition/subtraction, multiplication/division) symmetrically to both sides of the equation to maintain balance.
- Treat non-target variables as known constants or standard numbers throughout the algebraic manipulation steps.

Example 1: Isolating $L$ in the Area Formula ( $A = L \times W$ )
- Given formula: $A = L \times W$
- Target variable: $L$
- Step 1: Identify that $W$ is multiplied by $L$ .
- Step 2: Divide both sides of the equation by $W$ : $\frac{A}{W} = \frac{L \times W}{W}$
- Step 3: Simplify to obtain the final isolated formula: $L = \frac{A}{W}$
Example 2: Isolating $W$ in the Area Formula ( $A = L \times W$ )
- Given formula: $A = L \times W$
- Target variable: $W$
- Step 1: Identify that $L$ is multiplied by $W$ .
- Step 2: Divide both sides of the equation by $L$ : $\frac{A}{L} = \frac{L \times W}{L}$
- Step 3: Simplify to obtain the final isolated formula: $W = \frac{A}{L}$
Example 3: Isolating $W$ in the Perimeter Formula ( $P = 2(L + W)$ )
- Given formula: $P = 2(L + W)$
- Target variable: $W$
- Step 1: Divide both sides of the equation by $2$ to eliminate the coefficient outside the parentheses: $\frac{P}{2} = L + W$
- Step 2: Subtract $L$ from both sides to isolate $W$ : $\frac{P}{2} - L = W$
- Final Solution: $W = \frac{P}{2} - L$
Example 4: Manipulation of Surface Area / Height Equations
- Context terms and expressions: $S$ , $4 w$ , $4 W$ , $S - 2 w h$ , $40$
- Target variable: $H$ or $h$
- Procedure: Subtract terms not containing $h$ from the total $S$ , then divide by the remaining coefficients to isolate $h$ .

Example 5: Solving $a x + b y = c$ for $y$
- Given equation: $a x + b y = c$
- Target variable: $y$
- Step 1: Subtract the term $a x$ from both sides of the equation: $b y = c - a x$
- Step 2: Divide every term on both sides by the coefficient $b$ : $\frac{b y}{b} = \frac{c - a x}{b}$
- Step 3: State the isolated result: $1 y = \frac{c - a x}{b}$ (or $y = \frac{c - a x}{b}$ )
Example 6: Solving $S = R w^3 + 4 h$ for $h$
- Given equation: $S = R w^3 + 4 h$
- Target variable: $h$
- Step 1: Subtract $R w^3$ from both sides: $S - R w^3 = 4 h$
- Step 2: Divide both sides by $4$ : $\frac{S - R w^3}{4} = h$
- Final Solution: $h = \frac{S - R w^3}{4}$
Example 7: Solving $(5 F) - 9 c = 160$ for $c$
- Given equation: $(5 F) - 9 c = 160$
- Target variable: $c$
- Step 1: Subtract $5 F$ from both sides of the equation: $-9 c = 160 - 5 F$
- Step 2: Divide both sides by $-9$ : $c = \frac{160 - 5 F}{-9}$
- Alternative notation / equivalent expressions: $C = \frac{160 + 5 F}{-9}$ or $C = 160 + 5 F$

Arithmetic and Algebraic Warm-Up Scratchpad Work:
- Expression 1: $6 - 2 \times 2 - 4$
- Intermediate difference: $1 - 6$
- Subtraction calculation: $-24 - 10$
- Recorded values/operations: $X45$ , $23456$
- Extended algebraic combination: $2 - 6 x - 5 - -78 + 4$
- Variable consolidation step: $+7 x$ , yielding $13 x$
- Multiplication step: $\times 4$
- Result: $15$
Student Do Problem 1: Solving for $m$ in a Multi-Variable Product Equation
- Target variable: $m$
- Given equation/terms: $K = 2 m v$
- Step 1: Divide both sides by $2 v$ : $m = \frac{K}{2 v}$
Student Do Problem 2: Solving $P = a + b + m$ for $m$
- Given equation: $P = a + b + m$
- Target variable: $m$
- Step 1: Subtract $a$ from both sides: $P - a = b + m$
- Step 2: Subtract $b$ from both sides: $P - a - b = m$
- Final Isolated Expression: $m = P - a - b$