6.1 Solving One-Step & Two-Step Linear Equations

6.1 Solving One-Step & Two-Step Linear Equations

Linear equations form the cornerstone of algebraic reasoning across TABE 13&14 Mathematics Levels M, D, and A. Whether calculating hourly labor rates on a job site, budgeting material costs in manufacturing, or converting units of measurement, linear equations allow you to express unknown quantities symbolically and determine their exact values with mathematical precision.

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1. The Anatomy of an Algebraic Equation & The Balance Principle

An algebraic expression is a mathematical phrase combining numbers, variables, and operational symbols without an equality sign (e.g., $3x + 7$). In contrast, an algebraic equation is a complete mathematical sentence asserting that two expressions represent the exact same numerical quantity, connected by an equal sign ($=$):

$$\underbrace{3x + 7}_{\text{Left Expression}} = \underbrace{22}_{\text{Right Expression}}$$

Key Algebraic Vocabulary

• Variable: A letter or symbol (such as $x, y, m, t$) representing an unknown numerical quantity.
• Term: A single number, a single variable, or the product of numbers and variables separated by addition or subtraction signs ($3x$ and $7$ are terms).
• Constant: A term with a fixed numerical value that contains no variable ($7$ and $22$).
• Coefficient: The numerical factor multiplying a variable ($3$ in $3x$; if a variable appears alone as $x$, its implicit coefficient is $1$; if $-x$, its coefficient is $-1$).
• Solution: Any value assigned to the variable that makes the equation a true statement ($x = 5$ because $3(5) + 7 = 22$).

The Two-Pan Balance Scale Model

Think of an equation as a perfectly balanced two-pan balance scale. The equal sign represents the fulcrum. To maintain balance, whatever operation is applied to one side of the scale must be applied identically to the other side.

         [ Left Expression ]        [ Right Expression ]
         \=================/        \==================/
                 \                          /
                  \                        /
                   \                      /
                    +--------------------+
                              /\
                             /  \
                            /____\
                            Fulcrum (=)

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2. Properties of Equality & Inverse Operations

The fundamental strategy for solving any equation is isolation of the variable—manipulating the equation so the variable stands alone on one side with a coefficient of positive $1$ ($1x = k$ or $x = k$). This is accomplished by applying inverse operations governed by the four fundamental Properties of Equality:

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3. Solving One-Step Linear Equations

A one-step linear equation requires exactly one inverse operation to isolate the variable.

Equations Involving Signed Integers and Decimals

Example 1 (Addition/Subtraction with Signed Decimals): Solve $y - 14.6 = -6.2$.

$$\begin{aligned} y - 14.6 + 14.6 &= -6.2 + 14.6 \quad \text{(Apply Addition Property of Equality)} \\ y &= 8.4 \end{aligned}$$

Example 2 (Division with Signed Integers): Solve $-8w = 72$.

$$\begin{aligned} \frac{-8w}{-8} &= \frac{72}{-8} \quad \text{(Apply Division Property of Equality)} \\ w &= -9 \end{aligned}$$

The Reciprocal Multiplication Technique for Fractional Coefficients

When a variable is multiplied by a proper or improper fraction $\frac{a}{b}$, you can isolate the variable in one step by multiplying both sides by its multiplicative inverse (reciprocal) $\frac{b}{a}$, because $\frac{b}{a} \times \frac{a}{b} = 1$.

Example 3: Solve $-\frac{4}{7}k = 28$.

$$\begin{aligned} \left(-\frac{7}{4}\right) \cdot \left(-\frac{4}{7}k\right) &= \left(-\frac{7}{4}\right) \cdot 28 \quad \text{(Multiply both sides by reciprocal } -\frac{7}{4}\text{)} \\ 1k &= -\frac{7 \times 28}{4} = -\frac{196}{4} = -49 \\ k &= -49 \end{aligned}$$

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4. Solving Two-Step Linear Equations

A two-step linear equation contains two operations applied to the variable, typically in the form $ax + b = c$ or $\frac{x}{a} + b = c$.

The "Reverse PEMDAS" Order

When evaluating arithmetic expressions, we follow the standard order of operations (PEMDAS: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). However, when unraveling (solving) an equation, we perform the inverse operations in reverse order (SADMEP):

1. Step 1: Undo addition or subtraction first to isolate the variable term.
2. Step 2: Undo multiplication or division second to isolate the variable itself.

Solving Sequence (SADMEP):
[Two-Step Equation: ax + b = c]
       │
       ├── 1. Subtraction/Addition: ax = c - b  (Eliminate Constant)
       │
       └── 2. Division/Multiplication: x = (c - b) / a  (Eliminate Coefficient)

Step-by-Step Worked Examples

Worked Example 1 (Standard Form $ax + b = c$): Solve $6x - 19 = 47$.

• Step 1 (Undo Subtraction): Add $19$ to both sides: $$6x - 19 + 19 = 47 + 19 \implies 6x = 66$$
• Step 2 (Undo Multiplication): Divide both sides by $6$: $$\frac{6x}{6} = \frac{66}{6} \implies x = 11$$

Worked Example 2 (Division Form $\frac{x}{a} + b = c$): Solve $\frac{m}{4} + 13 = 5$.

• Step 1 (Undo Addition): Subtract $13$ from both sides: $$\frac{m}{4} + 13 - 13 = 5 - 13 \implies \frac{m}{4} = -8$$
• Step 2 (Undo Division): Multiply both sides by $4$: $$\frac{m}{4} \cdot 4 = -8 \cdot 4 \implies m = -32$$

Worked Example 3 (Grouped Numerators $\frac{x + b}{a} = c$): Solve $\frac{2x + 10}{5} = 8$.

• Step 1 (Clear Denominator): Multiply both sides by $5$: $$\left(\frac{2x + 10}{5}\right) \cdot 5 = 8 \cdot 5 \implies 2x + 10 = 40$$
• Step 2 (Undo Addition): Subtract $10$ from both sides: $$2x + 10 - 10 = 40 - 10 \implies 2x = 30$$
• Step 3 (Undo Multiplication): Divide both sides by $2$: $$\frac{2x}{2} = \frac{30}{2} \implies x = 15$$

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5. Translating Word Problems into Linear Equations

On the TABE assessment, problems are frequently presented in narrative workplace and consumer contexts. Mastering keyword translations is essential for correct equation setup:

Applied Workplace Word Problem

Problem: An HVAC service company charges a flat emergency diagnostic call-out fee of $\text{\textdollar}85$ plus $\text{\textdollar}65$ per hour of labor. If a customer's total repair bill comes to $\text{\textdollar}345$, how many hours of labor $h$ were performed?

1. Define Variable: Let $h = \text{number of labor hours}$.
2. Set Up Equation: $\text{Fixed Call-Out Fee} + (\text{Hourly Rate} \times \text{Hours}) = \text{Total Bill}$ $$85 + 65h = 345$$
3. Solve Equation: $$\begin{aligned} 65h + 85 - 85 &= 345 - 85 \\ 65h &= 260 \\ \frac{65h}{65} &= \frac{260}{65} \\ h &= 4 \text{ hours} \end{aligned}$$

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6. Rigorous Solution Verification by Substitution

Always verify your calculated solution by substituting the numerical result back into the original equation in place of the variable:

$$\begin{aligned} \text{Original Equation:} &\quad 85 + 65h = 345 \\ \text{Substitute } h = 4: &\quad 85 + 65(4) \stackrel{?}{=} 345 \\ \text{Simplify Left Side:} &\quad 85 + 260 = 345 \\ &\quad 345 = 345 \quad \checkmark \text{ (True Statement)} \end{aligned}$$

Because both sides evaluate to identical values, $h = 4$ is confirmed to be the exact and correct solution.