IBEW: Electrical Training Alliance Aptitude Test • Algebra & Functions

Solve Linear Equations — Solve single-variable linear equations.

Master the foundational skill of isolating a variable—essential for every electrical calculation on the job.

IBEW aptitude test emblem with an energized plug, circuit board, and trade wrench
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The ability to solve a linear equation is one of the oldest and most practical skills in mathematics. Long before anyone wrote equations with letters and symbols, ancient engineers, builders, and tradespeople faced a fundamental problem: if you know the total and some of the parts, how do you find what is missing? Whether an Egyptian surveyor was calculating the length of a canal or a Babylonian merchant was figuring out the price per unit of grain, the underlying logic was the same—work backward from what you know to find the unknown quantity. This same logic is precisely what you use on the job when you rearrange Ohm's Law to find current, or back-calculate the length of conduit needed for a run.

~1800 BCE • Babylonian Algebra

Babylonian scribes on clay tablets solved problems equivalent to linear and quadratic equations, using verbal recipes rather than symbols. They described steps like "add to both sides" in natural language.

~300 BCE • Greek Geometric Algebra

Euclid and Greek mathematicians reframed algebraic relationships as geometric constructions—lengths, areas, and ratios—laying the groundwork for formal proofs of equality.

~820 CE • Al-Khwarizmi's Al-Jabr

The Persian mathematician al-Khwarizmi published a systematic treatise on solving equations. The word algebra itself comes from "al-jabr," meaning "restoration" or "completion"—the act of moving a term to the other side.

1637 • Descartes & Symbolic Notation

René Descartes introduced the convention of using letters like x and y for unknowns and a, b, c for constants, creating the compact symbolic notation we use today.

Modern Era • Trades & Engineering Applications

Linear equations now form the backbone of electrical formulas (Ohm's Law, power equations), building codes, load calculations, and the IBEW aptitude test itself.

The central question that linear equations answer is deceptively simple: What value of the unknown makes both sides of the equation equal? Every technique you will learn in this lesson—combining like terms, using inverse operations, clearing fractions—is a tool for answering that question efficiently and accurately. Mastering these techniques is not optional for the IBEW Electrical Training Alliance Aptitude Test; it is the single most tested algebraic skill.

SECTION 2

Core Principles & Definitions

Before diving into solving procedures, you need a clear understanding of the building blocks. A linear equation in one variable is any equation that can be written in the form ax + b = c, where x is the unknown, and a, b, and c are known numbers (with a ≠ 0). The word "linear" means the variable appears only to the first power—no x², no √x, no 1/x. These equations graph as straight lines, hence the name, and they always have exactly one solution (assuming a ≠ 0).

1

Variable

A letter (usually x) representing the unknown quantity you need to find. Think of it as a blank space in a circuit diagram—you know the slot exists, you just need to determine what goes there.
2

Coefficient

The number multiplied by the variable. In 3x, the coefficient is 3. It tells you how many "copies" of x are in that term—like saying you have 3 identical resistors in series.
3

Constant

A fixed number with no variable attached (e.g., the 7 in 2x + 7 = 15). Constants are the known values in your equation—like a known voltage or fixed length.
4

Inverse Operations

Operations that undo each other: addition undoes subtraction, multiplication undoes division. These are your primary tools for isolating the variable on one side.
5

Balance Principle

Whatever operation you perform on one side of the equation, you must perform on the other side. An equation is like a balanced scale—move weight on one side and you must compensate on the other.
✦

KEY TAKEAWAY

Think of solving a linear equation like troubleshooting a circuit to find a single unknown component. You already know the total voltage and most of the resistances; solving the equation is just methodically eliminating what you already know until only the unknown remains. Every step must keep the "circuit" balanced—if you disconnect something from one side, you must account for it on the other.

SECTION 3

Visual Explanation — The Balance Model

The most powerful mental model for solving linear equations is the balance scale. The equals sign is the fulcrum; whatever is on the left side must weigh the same as what is on the right side. The diagram below illustrates how solving 2x + 3 = 11 works step by step: each operation you perform is applied to both sides simultaneously, keeping the scale level until x stands alone.

Solving 2x + 3 = 11 Using the Balance ModelSTEP 0: Original Equation2x+311=STEP 1: Subtract 3 from both sides2x8=−3 removed from both sidesSTEP 2: Divide both sides by 2x4=✓ x = 4
Each row shows the equation as a balanced scale. In Step 1, the constant (+3) is removed from both sides. In Step 2, the coefficient (2) is divided out from both sides. The result: x = 4.

Notice that at every stage the scale remains balanced. This is the inviolable rule of equation solving: perform the same operation on both sides. If you subtract 3 from the left but forget to subtract 3 from the right, the scale tips and you get a wrong answer. This is the most common source of errors on the aptitude test—always double-check that your operation was applied to the entire right-hand side, not just one term.

SECTION 4

Mathematical Framework — Solving Procedures

Every single-variable linear equation, no matter how complex it initially appears, can be reduced to the standard form ax + b = c, and then solved in two moves. The general solution procedure relies on two inverse operations applied in the correct order. Understanding the underlying formulas ensures you can handle any variation the test throws at you.

STANDARD FORM

ax+b=cax + b = c

where a = coefficient of the variable (a ≠ 0), b = constant added to the variable term, c = value on the other side of the equation.

STEP 1 — ISOLATE THE VARIABLE TERM

ax=c−bax = c − b

Subtract b from both sides (or add if b is negative). This removes the constant from the variable's side, leaving only the variable term.

STEP 2 — ISOLATE THE VARIABLE

x=(c−b)/ax = (c − b) / a

Divide both sides by a (the coefficient). This isolates x completely, giving you the solution.

When the equation is more complex—variables on both sides, parentheses, or fractions—additional preliminary steps are required before you reach standard form. The general algorithm expands to the following sequence.

  1. Clear fractions or decimals by multiplying every term on both sides by the least common denominator (LCD).
  2. Distribute to eliminate parentheses using the distributive property: a(b + c) = ab + ac.
  3. Combine like terms on each side of the equation separately.
  4. Move variable terms to one side and constants to the other using addition or subtraction.
  5. Divide both sides by the coefficient to solve for x.
  6. Check by substituting the answer back into the original equation.
💡

Test Tip

On the IBEW aptitude test, the most time-efficient approach is to always simplify each side fully before moving terms across the equals sign. This minimizes sign errors, which are the number-one reason candidates miss algebra questions.

SECTION 5

Detailed Breakdown — Types of Linear Equations

Not all linear equations look the same on the test. Recognizing the type of equation you are dealing with lets you choose the most efficient approach. The diagram below categorizes the four main types you will encounter, ranging from basic one-step equations to multi-step equations with fractions and variables on both sides.

Types of Single-Variable Linear EquationsTYPE 1: One-Stepx + 5 = 123x = 21One inverse operation needed.Add/subtract OR multiply/divide.Difficulty: ●○○○TYPE 2: Two-Step2x + 7 = 15x/4 − 3 = 5Two inverse operations in sequence.Undo addition/subtraction first, then coefficient.Difficulty: ●●○○TYPE 3: Multi-Step / Distribution3(x − 2) + 4 = 165x + 3 = 2x + 18Requires distributing, combining liketerms, and/or moving variables across.Difficulty: ●●●○TYPE 4: Fractions & Decimals(x/3) + (x/6) = 50.5x − 1.2 = 3.8Clear fractions by multiplying by LCD;clear decimals by multiplying by 10ⁿ.Difficulty: ●●●●
Four types of linear equations organized by complexity. Type 1 (one-step) and Type 2 (two-step) form the foundation. Types 3 and 4 build on those skills by adding distribution, variable collection, and fraction clearing.
Equation TypeExampleKey First MoveTest Frequency
One-Stepx − 9 = 14Add 9 to both sidesCommon (warm-up)
Two-Step4x + 7 = 31Subtract 7, then divide by 4Very common
Multi-Step2(x + 3) = x + 10Distribute, then collect variablesCommon
Fractions/Decimalsx/2 + x/5 = 7Multiply by LCD (10)Occasional

Linear equation types and their recommended first moves

SECTION 6

Worked Example — Multi-Step Equation

Let us work through a multi-step equation of the type you will encounter on the IBEW aptitude test. This example combines distribution, combining like terms, and variable collection.

Solve: 3(2x − 4) + 5 = 4x + 9

  1. 1

    Step 1 — Distribute

    Apply the distributive property to the left side. Multiply 3 by each term inside the parentheses: 3 × 2x = 6x and 3 × (−4) = −12. The equation becomes:
    6x − 12 + 5 = 4x + 9
  2. 2

    Step 2 — Combine Like Terms

    On the left side, combine the constants: −12 + 5 = −7. The right side has no like terms to combine.
    6x − 7 = 4x + 9
  3. 3

    Step 3 — Collect Variable Terms

    Subtract 4x from both sides to move all variable terms to the left. 6x − 4x = 2x on the left; 4x − 4x = 0 on the right.
    2x − 7 = 9
  4. 4

    Step 4 — Isolate the Variable Term

    Add 7 to both sides to move the constant away from the variable. −7 + 7 = 0 on the left; 9 + 7 = 16 on the right.
    2x = 16
  5. 5

    Step 5 — Solve for x

    Divide both sides by 2 (the coefficient of x). 2x ÷ 2 = x; 16 ÷ 2 = 8.
    x = 8
  6. 6

    Step 6 — Check

    Substitute x = 8 back into the original equation: 3(2(8) − 4) + 5 = 3(16 − 4) + 5 = 3(12) + 5 = 36 + 5 = 41. Right side: 4(8) + 9 = 32 + 9 = 41. Both sides equal 41. ✓
    41 = 41 ✓ Solution confirmed
ℹ

Why Checking Matters

On the aptitude test, you will not have time to check every problem by substitution. However, when you finish a problem and the answer "feels" off—a negative number when you expected positive, or a large number for a simple setup—take 15 seconds to plug it back in. This habit catches sign errors that cost points.

SECTION 7

Common Pitfalls & Strategies

Understanding the correct procedure is half the battle; the other half is avoiding the mistakes that even experienced test-takers make under time pressure. The table below identifies the most frequent errors on linear equation problems and pairs each with a concrete strategy to prevent it.

Common PitfallWhat Goes WrongPrevention Strategy
Sign errors in distributionForgetting to distribute the negative: −2(x − 3) written as −2x − 6 instead of −2x + 6.Negative × negative = positive. Rewrite the parentheses with explicit signs before multiplying.
Only applying to one termAdding a number to the left side but only adding it to the first term on the right, not the whole right side.Draw a vertical line through the equals sign; apply the operation to everything on each side.
Combining unlike termsAdding 3x + 5 to get "8x" — mixing variable and constant terms.Only terms with the same variable part can be combined. 3x and 5 are different "units"—like mixing amps and volts.
Division order of operationsDividing only the variable term by the coefficient instead of the entire other side.Make sure the variable term is the ONLY term on its side before dividing. Isolate first, then divide.
Forgetting to clear fractionsAttempting to solve fraction equations term-by-term instead of clearing all denominators first.Multiply every term on both sides by the LCD before doing anything else. This eliminates fractions entirely.

Top 5 pitfalls and their prevention strategies for linear equation problems

✦

KEY TAKEAWAY

Think of solving an equation like stripping wire: you remove layers in a specific order—outer jacket first, then inner insulation—until you expose the conductor (the variable). If you skip steps or cut in the wrong order, you risk damaging the wire or leaving insulation on. Similarly, distribute before combining like terms, move constants before dividing by the coefficient, and always check your work.

SECTION 8

Connection to Advanced Equations & Trades Math

Single-variable linear equations are the gateway to more complex algebraic problem solving. Once you have mastered the technique of isolating a variable through inverse operations, you can apply the exact same logic to rearranging multi-variable formulas—something you will do constantly in the electrical trades. Ohm's Law (V = IR), the power formula (P = IV), and voltage drop calculations (Vd = 2 × K × I × D / CM) are all linear equations in disguise. Solving for any one variable requires the same skills practiced in this lesson.

ConceptLinear Equations (This Lesson)Advanced Application
Number of variablesOne (x)Multiple (V, I, R, P, etc.) — but you solve for one at a time
Solution methodInverse operations to isolate xSame inverse operations, just with different letters
Equation formax + b = cV = IR → I = V/R (same algebraic step)
SystemsSingle equationSystems of equations (two unknowns, two equations)
GraphingSolution is a point on the number lineSolution is a point in the coordinate plane (two variables)

Progression from single-variable equations to trades applications

On the IBEW aptitude test, the algebra section tests whether you can reliably isolate a variable, handle sign changes, and work with fractions and decimals. These are the exact skills that apprenticeship instructors build upon when you begin calculating wire sizes, conduit fill, and load balancing. Invest in these fundamentals now and every formula you encounter later will feel like a variation of something you already know.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why you must perform the same operation on both sides of an equation. What would happen to the solution if you subtracted 5 from the left side of 2x + 5 = 13 but not from the right?
Show answer
An equation states that two expressions are equal—they represent the same value. If you subtract 5 only from the left side, you make the left side smaller while the right side stays the same, breaking the equality. The equation would become 2x = 13, which yields x = 6.5—incorrect. The correct procedure subtracts 5 from both sides: 2x = 8, so x = 4. You can verify: 2(4) + 5 = 13 ✓. The balance principle ensures the equality relationship is preserved through every step.

Question 2: BASIC CALCULATION

Solve for x: 5x − 9 = 26
Show answer
Add 9 to both sides: 5x = 35. Divide both sides by 5: x = 7. Check: 5(7) − 9 = 35 − 9 = 26 ✓. This is a standard two-step equation. The key first move is undoing the subtraction (add 9), then undoing the multiplication (divide by 5).

Question 3: INTERMEDIATE

Solve for x: 4(x + 3) − 2x = 22
Show answer
First, distribute the 4: 4x + 12 − 2x = 22. Combine like terms on the left: 2x + 12 = 22. Subtract 12 from both sides: 2x = 10. Divide both sides by 2: x = 5. Check: 4(5 + 3) − 2(5) = 4(8) − 10 = 32 − 10 = 22 ✓. The distribution step is crucial; forgetting to multiply 4 × 3 = 12 is a common error.

Question 4: APPLIED

An electrician has a total of 120 feet of wire. She has already used some wire for a first run and twice that amount for a second run. After both runs, she has 30 feet left. Set up and solve a linear equation to find how many feet were used in the first run.
Show answer
Let x = feet of wire used in the first run. The second run uses 2x feet. Total used plus remaining equals the original amount: x + 2x + 30 = 120. Combine like terms: 3x + 30 = 120. Subtract 30: 3x = 90. Divide by 3: x = 30. The first run used 30 feet and the second run used 60 feet. Check: 30 + 60 + 30 = 120 ✓. Setting up the equation from a word problem is the hardest part; identify the unknown, express all quantities in terms of that unknown, and write the equation before solving.

Question 5: CRITICAL THINKING

Solve for x: (2x + 1)/3 − (x − 2)/4 = 2. Show all steps and explain your strategy for clearing the fractions.
Show answer
The denominators are 3 and 4, so the LCD is 12. Multiply every term by 12: 12 × (2x + 1)/3 − 12 × (x − 2)/4 = 12 × 2, which simplifies to 4(2x + 1) − 3(x − 2) = 24. Distribute: 8x + 4 − 3x + 6 = 24. Note the critical sign: −3 × (−2) = +6. Combine like terms: 5x + 10 = 24. Subtract 10: 5x = 14. Divide by 5: x = 14/5 = 2.8. Check: (2(2.8) + 1)/3 − (2.8 − 2)/4 = (6.6)/3 − (0.8)/4 = 2.2 − 0.2 = 2 ✓. The strategy of multiplying by the LCD converts a fraction equation into a straightforward multi-step equation. The most dangerous step is distributing the negative sign into (x − 2).

SUMMARY

Lesson Summary

Summary

Solving single-variable linear equations is the most fundamental algebraic skill tested on the IBEW Electrical Training Alliance Aptitude Test. Every equation of the form ax + b = c is solved using the same two-move strategy: first, undo the addition or subtraction to isolate the variable term, then divide by the coefficient to isolate the variable itself. The balance principle — performing the same operation on both sides — is the unbreakable rule that preserves equality throughout the process.

For more complex equations, remember the expanded procedure: clear fractions by multiplying by the LCD, distribute to eliminate parentheses, combine like terms on each side, collect variable terms on one side and constants on the other, then divide by the coefficient. Always watch for sign errors — especially during distribution with negatives — and verify your answer by substituting it back into the original equation. These exact skills transfer directly to rearranging Ohm's Law, power formulas, and conduit calculations throughout your electrical career.

Varsity Tutors • IBEW: Electrical Training Alliance Aptitude Test • Solve Linear Equations — Solve single-variable linear equations.