5.1 Solving Single-Variable Linear Equations

Fundamentals of Linear Equations & Properties of Equality

A linear equation in one variable is an algebraic statement asserting that two first-degree expressions are equal. On the ACCUPLACER Quantitative Reasoning, Algebra, and Statistics (QAS) test, single-variable linear equations form the foundational bedrock for algebra, coordinate geometry, and applied modeling.

The Standard Form of a Linear Equation

Any single-variable linear equation can be transformed into the standard form:

$ax + b = c$

where:

• $x$ is the unknown variable (with an implicit exponent of $1$).
• $a$, $b$, and $c$ are real constants, with the strict condition that $a \neq 0$.

Because the exponent on the variable is strictly $1$, linear equations produce straight-line graphs on a Cartesian coordinate plane and possess at most one unique numerical solution in standard conditions.

---

The Golden Rule of Algebra & Properties of Equality

An equation behaves like a balanced scale. To preserve the equality of both sides, any mathematical operation applied to one side must be applied identically to the other side. Four core axioms govern these transformations:

---

Systematic Algorithm for Solving Multi-Step Linear Equations

When confronting multi-step linear equations involving parentheses, fractions, decimals, and variables distributed across both sides, execute the following structured 6-step algorithm:

Step 1: Clear Fractions and Decimals
        Multiply EVERY term on both sides by the Least Common Denominator (LCD)
        or by a power of 10 (10, 100, 1000) to eliminate all fractional parts.

Step 2: Clear Parentheses (Distributive Property)
        Apply a(b + c) = ab + ac and a(b - c) = ab - ac, carefully distributing
        negative signs across all grouped terms.

Step 3: Combine Like Terms on Each Side
        Independently simplify the left-hand and right-hand expressions by summing
        variable terms with variable terms, and constant terms with constant terms.

Step 4: Collect Variable Terms on One Side
        Use Addition or Subtraction Properties to move all variable terms to one side
        of the equation (preferring the side that yields a positive coefficient).

Step 5: Collect Constant Terms on the Opposite Side & Isolate the Variable
        Use Addition or Subtraction Properties to isolate the variable term, then use
        the Division or Multiplication Property to reduce the coefficient to 1.

Step 6: Check / Verify the Solution
        Substitute the calculated value back into the ORIGINAL equation to ensure equality.

---

Clearing Fractions and Decimals

1. Clearing Fractions via the LCD

Instead of finding common denominators and manipulating rational fractions at each intermediate stage, eliminate all denominators in a single operation by multiplying every term on both sides by the Least Common Denominator (LCD):

$\frac{3x}{4} - \frac{5}{6} = \frac{x}{3} + \frac{1}{2}$

1. Identify denominators: $4, 6, 3, 2$. The $\text{LCD}(4, 6, 3, 2) = 12$.
2. Multiply every individual term by $12$: $12 \cdot \left(\frac{3x}{4}\right) - 12 \cdot \left(\frac{5}{6}\right) = 12 \cdot \left(\frac{x}{3}\right) + 12 \cdot \left(\frac{1}{2}\right)$ $3(3x) - 2(5) = 4(x) + 6(1)$ $9x - 10 = 4x + 6$
3. Isolate the variable: $9x - 4x = 6 + 10 \implies 5x = 16 \implies x = \frac{16}{5}$

2. Clearing Decimals

If an equation contains decimals, multiply every term by $10^k$, where $k$ is the maximum number of decimal places appearing in any term:

$0.25x + 1.4 = 0.05x - 0.2$

Multiply every term by $100$ (since $0.25$ and $0.05$ have two decimal places): $25x + 140 = 5x - 20$ $20x = -160 \implies x = -8$

---

Classification of Solutions

When simplifying a linear equation, all variable terms may cancel out entirely, leaving a statement containing only constants. Linear equations are classified into three distinct categories based on their solution sets:

Analytical Examples of Solution Types

• Conditional: $3(x - 2) = 2x + 1 \implies 3x - 6 = 2x + 1 \implies x = 7$. Exactly one solution.
• Inconsistent: $2(3x + 1) = 6x - 5 \implies 6x + 2 = 6x - 5 \implies 2 = -5$ (False). No real number $x$ can satisfy this equation; solution is $\emptyset$.
• Identity: $4(2x - 3) + 7 = 8x - 5 \implies 8x - 12 + 7 = 8x - 5 \implies 8x - 5 = 8x - 5 \implies 0 = 0$ (True). Every real number is a valid solution; solution is $\mathbb{R} = (-\infty, \infty)$.

---

Applied Linear Word Problems

On the ACCUPLACER QAS exam, linear equations frequently appear embedded in verbal scenarios. Success requires setting up a clear dictionary of variables and translating English phrases into algebraic equations.

Verbal-to-Algebraic Translation Key

---

Step-by-Step Worked Examples

Worked Example 1: Multi-Step Equation with Rational Fractions & Distribution

Solve the equation for $x$:

$\frac{2}{3}(2x - 1) - \frac{1}{4}(x + 3) = \frac{5}{12}x + 2$

Solution Plan & Execution:

1. Identify the LCD of all denominators ($3, 4, 12$): $\text{LCD} = 12$.

2. Multiply both sides of the entire equation by $12$: $12 \cdot \left[\frac{2}{3}(2x - 1)\right] - 12 \cdot \left[\frac{1}{4}(x + 3)\right] = 12 \cdot \left(\frac{5}{12}x\right) + 12 \cdot (2)$ $4 \cdot 2(2x - 1) - 3 \cdot 1(x + 3) = 5x + 24$ $8(2x - 1) - 3(x + 3) = 5x + 24$

3. Apply the Distributive Property: $16x - 8 - 3x - 9 = 5x + 24$

4. Combine like terms on the left side: $(16x - 3x) + (-8 - 9) = 5x + 24$ $13x - 17 = 5x + 24$

5. Isolate the variable terms and constant terms: $13x - 5x = 24 + 17$ $8x = 41$ $x = \frac{41}{8}$

---

Worked Example 2: Consecutive Integer Problem

The sum of three consecutive odd integers is $177$. What is the value of the largest integer?

Solution Plan & Execution:

1. Define the variables:

   • Let the first (smallest) odd integer be $n$.
   • Consecutive odd integers differ by $2$: second integer $= n + 2$, third (largest) integer $= n + 4$.

2. Set up the linear equation: $n + (n + 2) + (n + 4) = 177$

3. Combine like terms and solve: $3n + 6 = 177$ $3n = 177 - 6 = 171$ $n = \frac{171}{3} = 57$

4. Find the largest integer: $\text{Largest integer} = n + 4 = 57 + 4 = 61$ (Check: $57 + 59 + 61 = 177$. Correct.)

---

Worked Example 3: Geometric Perimeter Relation

The length of a rectangular sports field is $8\text{ meters}$ less than three times its width. If the total perimeter of the field is $112\text{ meters}$, what are the dimensions of the field and its total area?

Solution Plan & Execution:

1. Define variables:

   • Let width $= w$.
   • Length $= l = 3w - 8$.

2. Use the perimeter formula $P = 2l + 2w$: $2(3w - 8) + 2w = 112$

3. Distribute, combine like terms, and solve for $w$: $6w - 16 + 2w = 112$ $8w - 16 = 112$ $8w = 128 \implies w = 16\text{ meters}$

4. Compute the length and area: $l = 3(16) - 8 = 48 - 8 = 40\text{ meters}$ $\text{Area} = l \cdot w = 40 \times 16 = 640\text{ square meters}$

---

Worked Example 4: Age Comparison Word Problem

David is currently $28\text{ years}$ older than his nephew Liam. In $6\text{ years}$, David's age will be three times Liam's age at that time. How old is Liam now?

Solution Plan & Execution:

1. Set up an age table:

   • Liam's current age: $x$
   • David's current age: $x + 28$
   • Liam's age in 6 years: $x + 6$
   • David's age in 6 years: $(x + 28) + 6 = x + 34$

2. Translate the future age relation: $x + 34 = 3(x + 6)$

3. Solve for $x$: $x + 34 = 3x + 18$ $34 - 18 = 3x - x$ $16 = 2x \implies x = 8$

Liam is currently $8\text{ years old}$ (David is $8 + 28 = 36$). In 6 years, Liam will be 14 and David will be 42, with $42 = 3 \times 14$.

---

Common Pitfalls & ACCUPLACER Exam Traps

1. Incomplete LCD Multiplication: When clearing denominators, students often forget to multiply standalone constant terms by the LCD. In $\frac{x}{3} + 4 = 5$, multiplying by $3$ gives $x + 12 = 15$, NOT $x + 4 = 15$.
2. Negative Sign Distribution Across Fractions: In expressions like $-\frac{x - 3}{4}$, the negative sign applies to the ENTIRE numerator: $-(x - 3) = -x + 3$, not $-x - 3$.
3. Reversing Subtraction Order: "$8$ less than a number" must be written as $x - 8$. Writing $8 - x$ reverses the subtraction and yields an incorrect negative result.
4. Consecutive Even vs Odd Integer Setup: Consecutive integers are $n, n+1, n+2$. Consecutive even integers are $n, n+2, n+4$. Crucially, consecutive odd integers ALSO use $n, n+2, n+4$ (because odd numbers are spaced 2 units apart).
5. Confusing $x = 0$ with No Solution: $x = 0$ is a valid, single, well-defined solution (a conditional equation). Only a contradictory equality with no variable (such as $0 = 5$) means "no solution."