Study guide
5.1 Solving Single-Variable Linear Equations
ACCUPLACER Quantitative Reasoning, Algebra, and Statistics Exam
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1 Chapter 1: Rational Numbers & Operations
1.1 Rational Number Arithmetic & Signed Numbers 1.2 Fractions, Decimals, & Mixed Numbers Operations 1.3 Absolute Value, Number Line, & Estimation
2 Chapter 2: Ratios, Proportions & Rates
2.1 Ratios, Unit Rates, & Dimensional Analysis 2.2 Proportions & Direct/Inverse Variation 2.3 Percent Applications, Markup, Discount, & Simple Interest
3 Chapter 3: Exponents & Scientific Notation
3.1 Integer Exponents & Laws of Exponents 3.2 Scientific Notation & Operations
4 Chapter 4: Algebraic Expressions
4.1 Evaluating & Translating Algebraic Expressions 4.2 Simplifying, Combining Like Terms, & Factoring
5 Chapter 5: Linear Equations & Inequalities
5.1 Solving Single-Variable Linear Equations 5.2 Solving & Graphing Linear Inequalities 5.3 Systems of Linear Equations
6 Chapter 6: Linear Applications & Graphing
6.1 Slope & Rate of Change 6.2 Forms of Linear Equations & Graphing 6.3 Linear Modeling & Inequality Graphs
7 Chapter 7: Probability & Set Theory
7.1 Fundamental Probability Concepts 7.2 Set Theory, Operations, & Venn Diagrams
8 Chapter 8: Descriptive Statistics & Data Displays
8.1 Measures of Center: Mean, Median, Mode, & Weighted Average 8.2 Measures of Spread: Range, IQR, Variance, & Standard Deviation 8.3 Data Displays: Histograms, Box Plots, & Scatter Plots
9 Chapter 9: Geometry Concepts & Measurement
9.1 2D Geometry: Perimeter, Area, & Circles 9.2 3D Geometry: Surface Area & Volume of Solids 9.3 Triangles, Angles, & the Pythagorean Theorem
5.1 Solving Single-Variable Linear Equations
Key Takeaways Explain This Topic AI
A linear equation in one variable can be written in the standard form ax + b = c (where a ≠ 0), representing a first-degree equality where the variable has an exponent of 1.
The Fundamental Properties of Equality (Addition, Subtraction, Multiplication, Division) preserve balance across equations: any algebraic operation performed on one side must be identically applied to the other.
Solving multi-step equations follows a standard sequential algorithm: clear fractions by multiplying every term by the Least Common Denominator (LCD), clear decimals, apply the distributive property, combine like terms, isolate variable terms on one side, and isolate the variable.
Linear equations produce three distinct solution outcomes: Conditional equations (exactly one unique solution), Inconsistent equations (no solution, simplifying to false contradictions like 0 = 7), and Identities (infinitely many solutions, simplifying to universal truths like 5 = 5).
Applied word problems (consecutive integers, age comparisons, and geometric perimeter relationships) require defining a single variable, translating verbal constraints into algebraic expressions, and solving the resulting linear equation.
Last updated: August 2026
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: a x + b = c ax + b = c a x + b = c
where:
x x x is the unknown variable (with an implicit exponent of 1 1 1 ).
a a a , b b b , and c c c are real constants, with the strict condition that a ≠ 0 a \neq 0 a = 0 .
Because the exponent on the variable is strictly 1 1 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:
Property of Equality Formal Mathematical Definition Algebraic Interpretation
Addition Property If a = b a = b a = b , then a + c = b + c a + c = b + c a + c = b + c Adding the same real number to both sides preserves equality
Subtraction Property If a = b a = b a = b , then a − c = b − c a - c = b - c a − c = b − c Subtracting the same real number from both sides preserves equality
Multiplication Property If a = b a = b a = b , then a ⋅ c = b ⋅ c a \cdot c = b \cdot c a ⋅ c = b ⋅ c Multiplying both sides by any real number c c c preserves equality
Division Property If a = b a = b a = b and c ≠ 0 c \neq 0 c = 0 , then a c = b c \frac{a}{c} = \frac{b}{c} c a = c b Dividing both sides by any non-zero real number c c c preserves equality
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) : 3 x 4 − 5 6 = x 3 + 1 2 \frac{3x}{4} - \frac{5}{6} = \frac{x}{3} + \frac{1}{2} 4 3 x − 6 5 = 3 x + 2 1
Identify denominators: 4 , 6 , 3 , 2 4, 6, 3, 2 4 , 6 , 3 , 2 . The LCD ( 4 , 6 , 3 , 2 ) = 12 \text{LCD}(4, 6, 3, 2) = 12 LCD ( 4 , 6 , 3 , 2 ) = 12 .
Multiply every individual term by 12 12 12 : 12 ⋅ ( 3 x 4 ) − 12 ⋅ ( 5 6 ) = 12 ⋅ ( x 3 ) + 12 ⋅ ( 1 2 ) 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) 12 ⋅ ( 4 3 x ) − 12 ⋅ ( 6 5 ) = 12 ⋅ ( 3 x ) + 12 ⋅ ( 2 1 ) 3 ( 3 x ) − 2 ( 5 ) = 4 ( x ) + 6 ( 1 ) 3(3x) - 2(5) = 4(x) + 6(1) 3 ( 3 x ) − 2 ( 5 ) = 4 ( x ) + 6 ( 1 ) 9 x − 10 = 4 x + 6 9x - 10 = 4x + 6 9 x − 10 = 4 x + 6
Isolate the variable: 9 x − 4 x = 6 + 10 ⟹ 5 x = 16 ⟹ x = 16 5 9x - 4x = 6 + 10 \implies 5x = 16 \implies x = \frac{16}{5} 9 x − 4 x = 6 + 10 ⟹ 5 x = 16 ⟹ x = 5 16
2. Clearing Decimals
If an equation contains decimals, multiply every term by 10 k 10^k 1 0 k , where k k k is the maximum number of decimal places appearing in any term: 0.25 x + 1.4 = 0.05 x − 0.2 0.25x + 1.4 = 0.05x - 0.2 0.25 x + 1.4 = 0.05 x − 0.2
Multiply every term by 100 100 100 (since 0.25 0.25 0.25 and 0.05 0.05 0.05 have two decimal places): 25 x + 140 = 5 x − 20 25x + 140 = 5x - 20 25 x + 140 = 5 x − 20 20 x = − 160 ⟹ x = − 8 20x = -160 \implies x = -8 20 x = − 160 ⟹ 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:
Classification Algebraic Outcome Solution Set Geometric Meaning
Conditional Equation x = k x = k x = k (where k k k is a constant) Exactly one unique solution { k } \{k\} { k } A single point on the real number line
Inconsistent Equation (Contradiction) False statement (e.g., 0 = 7 0 = 7 0 = 7 or − 4 = 9 -4 = 9 − 4 = 9 ) No solution (Empty set ∅ \emptyset ∅ ) Parallel lines that never intersect
Identity True statement (e.g., 0 = 0 0 = 0 0 = 0 or 5 = 5 5 = 5 5 = 5 ) Infinitely many real solutions (All real numbers R \mathbb{R} R ) Coinciding, identical lines
Analytical Examples of Solution Types
Conditional : 3 ( x − 2 ) = 2 x + 1 ⟹ 3 x − 6 = 2 x + 1 ⟹ x = 7 3(x - 2) = 2x + 1 \implies 3x - 6 = 2x + 1 \implies x = 7 3 ( x − 2 ) = 2 x + 1 ⟹ 3 x − 6 = 2 x + 1 ⟹ x = 7 . Exactly one solution.
Inconsistent : 2 ( 3 x + 1 ) = 6 x − 5 ⟹ 6 x + 2 = 6 x − 5 ⟹ 2 = − 5 2(3x + 1) = 6x - 5 \implies 6x + 2 = 6x - 5 \implies 2 = -5 2 ( 3 x + 1 ) = 6 x − 5 ⟹ 6 x + 2 = 6 x − 5 ⟹ 2 = − 5 (False) . No real number x x x can satisfy this equation; solution is ∅ \emptyset ∅ .
Identity : 4 ( 2 x − 3 ) + 7 = 8 x − 5 ⟹ 8 x − 12 + 7 = 8 x − 5 ⟹ 8 x − 5 = 8 x − 5 ⟹ 0 = 0 4(2x - 3) + 7 = 8x - 5 \implies 8x - 12 + 7 = 8x - 5 \implies 8x - 5 = 8x - 5 \implies 0 = 0 4 ( 2 x − 3 ) + 7 = 8 x − 5 ⟹ 8 x − 12 + 7 = 8 x − 5 ⟹ 8 x − 5 = 8 x − 5 ⟹ 0 = 0 (True) . Every real number is a valid solution; solution is R = ( − ∞ , ∞ ) \mathbb{R} = (-\infty, \infty) R = ( − ∞ , ∞ ) .
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
- Verbal Phrase Algebraic Translation
- " k k k more than a number", "a number increased by k k k " x + k x + k x + k
- " k k k less than a number", "a number decreased by k k k " x − k x - k x − k (Note: not k − x k - x k − x )
- "The difference of a number and k k k " x − k x - k x − k
- "The product of k k k and a number", " k k k times a number" k x kx k x
- "The quotient of a number and k k k " x k \frac{x}{k} k x
- " k k k times the sum of a number and p p p " k ( x + p ) k(x + p) k ( x + p )
- "Is", "equals", "yields", "amounts to" = = =
- Step-by-Step Worked Examples
- Worked Example 1: Multi-Step Equation with Rational Fractions & Distribution
Solve the equation for x x x : 2 3 ( 2 x − 1 ) − 1 4 ( x + 3 ) = 5 12 x + 2 \frac{2}{3}(2x - 1) - \frac{1}{4}(x + 3) = \frac{5}{12}x + 2 3 2 ( 2 x − 1 ) − 4 1 ( x + 3 ) = 12 5 x + 2
Solution Plan & Execution :
Identify the LCD of all denominators ( 3 , 4 , 12 3, 4, 12 3 , 4 , 12 ): LCD = 12 \text{LCD} = 12 LCD = 12 .
Multiply both sides of the entire equation by 12 12 12 : 12 ⋅ [ 2 3 ( 2 x − 1 ) ] − 12 ⋅ [ 1 4 ( x + 3 ) ] = 12 ⋅ ( 5 12 x ) + 12 ⋅ ( 2 ) 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) 12 ⋅ [ 3 2 ( 2 x − 1 ) ] − 12 ⋅ [ 4 1 ( x + 3 ) ] = 12 ⋅ ( 12 5 x ) + 12 ⋅ ( 2 ) 4 ⋅ 2 ( 2 x − 1 ) − 3 ⋅ 1 ( x + 3 ) = 5 x + 24 4 \cdot 2(2x - 1) - 3 \cdot 1(x + 3) = 5x + 24 4 ⋅ 2 ( 2 x − 1 ) − 3 ⋅ 1 ( x + 3 ) = 5 x + 24 8 ( 2 x − 1 ) − 3 ( x + 3 ) = 5 x + 24 8(2x - 1) - 3(x + 3) = 5x + 24 8 ( 2 x − 1 ) − 3 ( x + 3 ) = 5 x + 24
Apply the Distributive Property : 16 x − 8 − 3 x − 9 = 5 x + 24 16x - 8 - 3x - 9 = 5x + 24 16 x − 8 − 3 x − 9 = 5 x + 24
Combine like terms on the left side : ( 16 x − 3 x ) + ( − 8 − 9 ) = 5 x + 24 (16x - 3x) + (-8 - 9) = 5x + 24 ( 16 x − 3 x ) + ( − 8 − 9 ) = 5 x + 24 13 x − 17 = 5 x + 24 13x - 17 = 5x + 24 13 x − 17 = 5 x + 24
Isolate the variable terms and constant terms : 13 x − 5 x = 24 + 17 13x - 5x = 24 + 17 13 x − 5 x = 24 + 17 8 x = 41 8x = 41 8 x = 41 x = 41 8 x = \frac{41}{8} x = 8 41
Worked Example 2: Consecutive Integer Problem
The sum of three consecutive odd integers is 177 177 177 . What is the value of the largest integer?
Solution Plan & Execution :
Define the variables :
Let the first (smallest) odd integer be n n n .
Consecutive odd integers differ by 2 2 2 : second integer = n + 2 = n + 2 = n + 2 , third (largest) integer = n + 4 = n + 4 = n + 4 .
Set up the linear equation : n + ( n + 2 ) + ( n + 4 ) = 177 n + (n + 2) + (n + 4) = 177 n + ( n + 2 ) + ( n + 4 ) = 177
Combine like terms and solve : 3 n + 6 = 177 3n + 6 = 177 3 n + 6 = 177 3 n = 177 − 6 = 171 3n = 177 - 6 = 171 3 n = 177 − 6 = 171 n = 171 3 = 57 n = \frac{171}{3} = 57 n = 3 171 = 57
Find the largest integer : Largest integer = n + 4 = 57 + 4 = 61 \text{Largest integer} = n + 4 = 57 + 4 = 61 Largest integer = n + 4 = 57 + 4 = 61
- (Check: 57 + 59 + 61 = 177 57 + 59 + 61 = 177 57 + 59 + 61 = 177 . Correct.)
- Worked Example 3: Geometric Perimeter Relation
The length of a rectangular sports field is 8 meters 8\text{ meters} 8 meters less than three times its width. If the total perimeter of the field is 112 meters 112\text{ meters} 112 meters , what are the dimensions of the field and its total area?
Solution Plan & Execution :
Define variables :
Let width = w = w = w .
Length = l = 3 w − 8 = l = 3w - 8 = l = 3 w − 8 .
Use the perimeter formula P = 2 l + 2 w P = 2l + 2w P = 2 l + 2 w : 2 ( 3 w − 8 ) + 2 w = 112 2(3w - 8) + 2w = 112 2 ( 3 w − 8 ) + 2 w = 112
Distribute, combine like terms, and solve for w w w : 6 w − 16 + 2 w = 112 6w - 16 + 2w = 112 6 w − 16 + 2 w = 112 8 w − 16 = 112 8w - 16 = 112 8 w − 16 = 112 8 w = 128 ⟹ w = 16 meters 8w = 128 \implies w = 16\text{ meters} 8 w = 128 ⟹ w = 16 meters
Compute the length and area : l = 3 ( 16 ) − 8 = 48 − 8 = 40 meters l = 3(16) - 8 = 48 - 8 = 40\text{ meters} l = 3 ( 16 ) − 8 = 48 − 8 = 40 meters Area = l ⋅ w = 40 × 16 = 640 square meters \text{Area} = l \cdot w = 40 \times 16 = 640\text{ square meters} Area = l ⋅ w = 40 × 16 = 640 square meters
Worked Example 4: Age Comparison Word Problem
David is currently 28 years 28\text{ years} 28 years older than his nephew Liam. In 6 years 6\text{ years} 6 years , David's age will be three times Liam's age at that time. How old is Liam now?
Solution Plan & Execution :
Set up an age table :
- Liam's current age: x x x
- David's current age: x + 28 x + 28 x + 28
- Liam's age in 6 years: x + 6 x + 6 x + 6
- David's age in 6 years: ( x + 28 ) + 6 = x + 34 (x + 28) + 6 = x + 34 ( x + 28 ) + 6 = x + 34
Translate the future age relation : x + 34 = 3 ( x + 6 ) x + 34 = 3(x + 6) x + 34 = 3 ( x + 6 )
Solve for x x x : x + 34 = 3 x + 18 x + 34 = 3x + 18 x + 34 = 3 x + 18 34 − 18 = 3 x − x 34 - 18 = 3x - x 34 − 18 = 3 x − x 16 = 2 x ⟹ x = 8 16 = 2x \implies x = 8 16 = 2 x ⟹ x = 8
Liam is currently 8 years old 8\text{ years old} 8 years old (David is 8 + 28 = 36 8 + 28 = 36 8 + 28 = 36 ). In 6 years, Liam will be 14 and David will be 42, with 42 = 3 × 14 42 = 3 \times 14 42 = 3 × 14 .
Common Pitfalls & ACCUPLACER Exam Traps
Incomplete LCD Multiplication : When clearing denominators, students often forget to multiply standalone constant terms by the LCD. In x 3 + 4 = 5 \frac{x}{3} + 4 = 5 3 x + 4 = 5 , multiplying by 3 3 3 gives x + 12 = 15 x + 12 = 15 x + 12 = 15 , NOT x + 4 = 15 x + 4 = 15 x + 4 = 15 .
Negative Sign Distribution Across Fractions : In expressions like − x − 3 4 -\frac{x - 3}{4} − 4 x − 3 , the negative sign applies to the ENTIRE numerator: − ( x − 3 ) = − x + 3 -(x - 3) = -x + 3 − ( x − 3 ) = − x + 3 , not − x − 3 -x - 3 − x − 3 .
Reversing Subtraction Order : " 8 8 8 less than a number" must be written as x − 8 x - 8 x − 8 . Writing 8 − x 8 - x 8 − x reverses the subtraction and yields an incorrect negative result.
Consecutive Even vs Odd Integer Setup : Consecutive integers are n , n + 1 , n + 2 n, n+1, n+2 n , n + 1 , n + 2 . Consecutive even integers are n , n + 2 , n + 4 n, n+2, n+4 n , n + 2 , n + 4 . Crucially, consecutive odd integers ALSO use n , n + 2 , n + 4 n, n+2, n+4 n , n + 2 , n + 4 (because odd numbers are spaced 2 units apart).
Confusing x = 0 x = 0 x = 0 with No Solution : x = 0 x = 0 x = 0 is a valid, single, well-defined solution (a conditional equation). Only a contradictory equality with no variable (such as 0 = 5 0 = 5 0 = 5 ) means "no solution."