Algebra, Equations and Inequalities

Introduction

Business analytics frequently converts operational situations into algebraic models. A price, cost, sales volume, staffing level, inventory quantity or advertising budget can be represented by a variable. Relationships among these variables can then be manipulated and solved to support pricing, budgeting, allocation and break-even decisions.

Algebra is therefore more than symbolic calculation. It provides a language for describing assumptions, testing scenarios and identifying the conditions under which a business decision is feasible. A mathematically correct solution must also be interpreted in context: a negative number of products, a fractional employee or a price outside the market range may be algebraically possible but operationally meaningless.

Learning Objectives

By the end of this chapter, you should be able to:

1. Simplify and factor algebraic expressions.
2. Use exponents, ratios and percentages in business calculations.
3. Solve linear, quadratic, polynomial and rational equations.
4. Solve simultaneous equations using substitution, elimination and matrices.
5. Solve linear and quadratic inequalities and represent their solution sets.
6. Construct and interpret revenue, cost, profit and break-even models.
7. Evaluate the assumptions, limitations and ethical implications of algebraic business models.

1. Algebraic Expressions and Notation

An algebraic expression combines numbers, variables and operations. In 4x² - 3x + 7, x is the variable, 4 and −3 are coefficients, 7 is a constant, and the terms are 4x², -3x and 7. An expression does not contain an equals sign; an equation does.

A variable represents a quantity that may change. A constant is fixed within the stated problem. The coefficient describes the numerical multiplier of a variable. The degree of a polynomial is the highest exponent of its variable.

Order of operations

Calculations follow the order of operations: brackets, exponents, multiplication and division, then addition and subtraction. Multiplication and division are performed from left to right, as are addition and subtraction.

For example, if a product has a list price of £80 and a 15% discount, the sale price is:

The result is £68. The percentage must be expressed as a decimal before it is used in the formula.

Laws of exponents

For nonzero values where the expressions are defined:

A negative exponent means a reciprocal: a^-n = 1/aⁿ. These rules apply only when the bases are the same in multiplication or division. Thus x² + x³ cannot be simplified to x⁵, because the terms are being added rather than multiplied.

Like terms and distributive law

Like terms have identical variable parts. Therefore 3x + 5x = 8x, but 3x + 5x² cannot be combined. The distributive law is:

For example:

Factoring reverses distribution. The expression 6x + 9 becomes 3(2x + 3). Factoring is particularly useful for solving polynomial equations.

2. Linear Equations and Business Models

A linear equation has the general form:

where a, b and c are constants and x is the unknown. Solving means finding the value of x that makes the equation true. Operations must be applied equally to both sides.

Worked Example 1: pricing

A retailer’s variable cost is £18 per item and fixed costs are £4,800. Management wants a profit of £2,700 from selling 500 items. Let p be the selling price per item. The model is:

Since 500 × 18 = 9000,

The required price is £33 per item, assuming that 500 units can be sold at that price and that costs are accurately classified as fixed and variable.

Rearranging formulas

A formula may need to be rearranged to isolate a particular variable. If total cost is:

then:

provided v ≠ 0. Here C is total cost, F is fixed cost, v is variable cost per unit and q is quantity. Units are consistent: money minus money divided by money per unit gives units.

Break-even analysis

Revenue and cost are commonly modeled as:

where p is selling price per unit, F is fixed cost, v is variable cost per unit and q is quantity. Profit is:

Break-even occurs when profit is zero. If p > v, the break-even quantity is:

The denominator p-v is the contribution per unit. The formula is not valid when p = v, because each unit then contributes nothing toward fixed costs. If p < v, increasing sales increases losses rather than eliminating them.

Worked Example 2: break-even

Suppose F = £30,000, p = £50 and v = £32. Then:

Because products are discrete, at least 1,667 units must be sold to reach or exceed break-even. At 1,667 units:

The break-even revenue based on the continuous quantity is approximately 50 × 1666.67 = £83,333.50. The rounded operational target is 1,667 units.

3. Simultaneous Linear Equations

A simultaneous system contains two or more equations whose unknowns must satisfy all equations at once. A two-variable system can be written:

A unique solution exists when:

If the determinant is zero, the equations either describe parallel lines with no solution or the same line with infinitely many solutions.

Elimination method

A company sells basic and premium subscriptions. In one month it sells 240 subscriptions and receives £5,760. Basic subscriptions cost £18 and premium subscriptions cost £30. Let b and p denote the quantities:

Multiply the first equation by 18:

Subtract from the second equation:

Therefore b = 240 - 120 = 120. The company sold 120 of each type. Verification gives 18(120) + 30(120) = 5760.

Matrix representation

The system may be represented as AX = B, where:

For larger systems, spreadsheet solvers, linear algebra software or optimization systems are useful. However, software output should be checked against the original equations and business constraints.

4. Quadratic Equations

A quadratic equation has the form:

It may be solved by factoring, completing the square or using the quadratic formula:

The discriminant is Δ = b² - 4ac. If Δ > 0, there are two distinct real solutions; if Δ = 0, there is one repeated real solution; if Δ < 0, there are no real solutions.

Worked Example 3: profit target

A simplified profit model is:

To find quantities producing zero profit:

Multiplying by −50 gives:

Using the quadratic formula:

Thus the roots are approximately 268.34 and 931.66 units. In a continuous model, profit is positive between these values. If q is an integer, quantities from 269 through 931 produce positive modeled profit, subject to capacity and demand constraints.

The quadratic model illustrates two break-even points, which can occur when high sales volumes cause overtime, congestion or other nonlinear costs. A simple linear break-even model would not capture this effect.

5. Polynomial Equations

A polynomial is an expression such as:

where n is a non-negative integer and the coefficients are constants. Polynomial equations can sometimes be solved by factoring. The zero-product property states that if:

then A = 0 or B = 0.

For example:

Factor out x:

Therefore x = 0, 2 or 3. In an application where x represents production volume, x = 0 may be operationally relevant, while negative roots—if present—would normally be rejected.

When exact factoring is unavailable, numerical methods such as Newton’s method or software solvers may be used. Newton’s iteration is:

It requires a differentiable function and a suitable initial value. It may fail to converge or may converge to a root that is not relevant to the business problem. Numerical answers should be reported with appropriate precision, not spurious certainty.

6. Rational Equations

A rational equation contains fractions involving variables. Before solving, identify restrictions that make any denominator zero. These restricted values are excluded even if algebraic cancellation later occurs.

Consider:

The restrictions are q ≠ 2 and q ≠ 0. Multiplying by the common denominator 2q(q−2) gives:

Factoring:

The candidates are q = 2/3 and q = 2. Since q = 2 violates the original denominator restriction, the only valid solution is q = 2/3. In a quantity model, even this solution might be rejected if quantities must be whole units.

7. Inequalities and Feasible Decisions

An inequality uses <, >, ≤ or ≥. Solving an inequality identifies a set of values rather than one value. Adding or subtracting the same quantity preserves the inequality direction. Multiplying or dividing by a negative quantity reverses it.

For example:

Dividing by −4 reverses the sign:

If a business must spend no more than £12,000 on materials at £24 per unit, the constraint is 24q ≤ 12000, so q ≤ 500. If q must be a non-negative integer, the feasible quantities are 0 through 500.

Compound inequalities

A capacity requirement may be written:

This means q is at least 200 and at most 600. In allocation problems, several inequalities combine to form a feasible region. For example:

The first two inequalities represent resource limits, and the last two impose non-negativity. Such constraints form the foundation of linear programming, although optimization itself is beyond the scope of this chapter.

Quadratic inequalities

To solve (q-2)(q-5) ≥ 0, identify critical points 2 and 5. The product is non-negative outside the roots, so:

A sign chart or number line helps prevent errors. Endpoints are included because the inequality is ≥.

8. Applied Case: Capacity, Pricing and Break-Even Planning

Background

The following is an illustrative case. A small analytics consultancy offers standard and advanced data reports. The standard report sells for £400 and costs £160 in analyst time and software usage. The advanced report sells for £700 and costs £280. Monthly fixed costs are £18,000. The firm has 120 analyst-hours available; each standard report uses 1 hour and each advanced report uses 2 hours.

Problem

Management wants to determine a sales mix that covers fixed costs and understand whether a proposed mix is feasible.

Let s be standard reports and a be advanced reports. The capacity constraint is:

The contribution per standard report is 400 - 160 = £240, and the contribution per advanced report is 700 - 280 = £420. Profit is therefore:

Analysis of a proposed mix

Suppose the firm plans 60 standard and 25 advanced reports. Capacity usage is:

so the plan is feasible. Profit is:

The proposal generates an estimated monthly profit of £6,900.

The break-even condition for any mix is:

For a standard-only plan, at least 18000/240 = 75 reports are needed, using 75 hours. For an advanced-only plan, at least 18000/420 ≈ 42.86, or 43 reports, are needed, using 86 hours. These calculations show why product mix matters: advanced reports yield more contribution per report, but also consume more capacity.

Outcome and learning points

The algebraic model identifies feasible and infeasible plans before resources are committed. It also makes assumptions visible: prices are fixed, demand is sufficient, costs are linear and analyst hours are interchangeable. In practice, management should test alternative scenarios, include demand limits and consider quality risks from excessive workload.

The central lesson is that equations calculate consequences; they do not independently choose the best strategy. A decision requires economic, operational and ethical judgment in addition to algebra.

9. Technology, Governance and Limitations

Spreadsheets, computer algebra systems and business intelligence platforms can simplify expressions, solve equations and perform sensitivity analysis. These tools are valuable when models contain many variables, but automation does not guarantee validity. A spreadsheet can return a precise answer from an incorrect formula.

Best practice includes:

• define each variable and unit before calculation;
• record assumptions and restrictions;
• distinguish estimated inputs from observed data;
• verify results by substitution into the original equation;
• apply business constraints such as non-negativity, capacity and integer quantities;
• conduct sensitivity analysis on prices, costs and demand;
• retain an audit trail of formulas and revisions.

Algebraic models may also create governance concerns. A pricing rule can disadvantage vulnerable customers if it uses inappropriate segmentation. An allocation model may reproduce historical discrimination if its input data reflect unequal access. Analysts should document model scope, test outcomes across relevant groups and ensure that automated recommendations remain subject to human review and organizational policy.

Linear models are especially limited when economies of scale, tiered prices, capacity bottlenecks or uncertain demand exist. Quadratic or piecewise models may represent some of these effects, but greater complexity does not automatically mean greater accuracy. Model choice should be justified by the decision context and evidence.

Chapter Summary

Algebra represents business quantities and relationships using variables, constants and operations. Simplification, distribution, factoring and exponent rules support reliable manipulation of expressions. Linear equations model relationships such as total cost, revenue and target profit. Simultaneous equations identify unknown quantities when several conditions apply at once.

Quadratic equations can produce two, one or no real solutions, with the discriminant determining the number of real roots. Polynomial equations are often solved through factoring, while rational equations require careful attention to denominator restrictions. Inequalities describe feasible ranges and are essential for capacity, budget and resource constraints.

The linear break-even model is based on contribution per unit and is valid when price exceeds variable cost and the relevant cost relationships are approximately linear. Business interpretation is as important as symbolic manipulation: solutions must be checked for units, restrictions, integer requirements, feasibility and ethical consequences. Digital tools improve speed, but verification, documentation and professional judgment remain necessary.

Key Terms

Variable: A symbol representing a quantity that may change.

Constant: A fixed value within a specified model.

Coefficient: A numerical factor multiplying a variable.

Expression: A combination of numbers, variables and operations without an equals sign.

Polynomial: An expression whose variable exponents are non-negative integers.

Linear equation: An equation in which the highest power of the variable is one.

Quadratic equation: An equation of the form ax² + bx + c = 0, with a ≠ 0.

Discriminant: The quantity b² - 4ac used to classify quadratic roots.

Rational equation: An equation containing variable expressions in denominators.

Factor: An expression multiplied by another expression to produce a product.

Simultaneous equations: Equations solved together for common unknowns.

Inequality: A comparison using <, >, ≤ or ≥.

Feasible region: The set of values satisfying all stated constraints.

Fixed cost: A cost treated as independent of output within a relevant range.

Variable cost: A cost that changes with output under an assumed cost model.

Contribution per unit: Selling price minus variable cost per unit.

Break-even point: The output or sales level at which profit equals zero.

Sensitivity analysis: Examination of how results change when assumptions change.

Review Questions

Short Answer Questions

1. Distinguish between an algebraic expression and an equation.
2. State the condition required for the standard linear break-even formula to be meaningful.
3. What happens to an inequality when both sides are divided by a negative number?
4. Why must denominator restrictions be identified before solving a rational equation?
5. What does a zero determinant imply for a two-equation linear system?

Long Answer Questions

1. Explain how revenue, cost and profit equations can support a pricing decision. Include assumptions and limitations.
2. Compare factoring, completing the square and the quadratic formula as methods for solving quadratic equations.
3. Explain how simultaneous equations can be used to determine the sales quantities of two products from total units and total revenue.
4. Discuss the role of inequalities in representing budgets, capacity and non-negativity constraints.
5. Explain why verification and model governance are necessary when equations are solved using spreadsheets or automated systems.

Analytical / Application Questions

1. A firm has fixed costs of £24,000, a selling price of £80 and a variable cost of £50 per unit. Calculate the break-even quantity and explain why the result must be rounded upward for whole units.
2. A company sells two services. Service A contributes £120 per sale and Service B contributes £200 per sale. Its fixed costs are £10,000. Formulate the break-even equation for sales quantities x and y, then test whether x = 40 and y = 30 covers fixed costs.
3. Solve the inequality (q-4)(q-9) < 0 and interpret the result when q represents a continuous production level.

Multiple Choice Questions with Answers

Question 1

Which expression is equivalent to 4(2x - 3) + 5x?

A. 13x - 12

B. 8x - 7

C. 13x + 12

D. 8x - 12

Correct Answer: A. 13x - 12

Question 2

A product has fixed costs of £20,000, a price of £50 and variable cost of £30. What is the break-even quantity?

A. 400 units

B. 667 units

C. 1,000 units

D. 2,000 units

Correct Answer: C. 1,000 units

Question 3

When solving an inequality, what happens when both sides are divided by a negative number?

A. The inequality sign reverses

B. The inequality sign remains unchanged

C. The variable must be squared

D. The inequality becomes an equation

Correct Answer: A. The inequality sign reverses

Question 4

What is the discriminant of 2x² + 5x - 3 = 0?

A. 1

B. 25

C. 49

D. −1

Correct Answer: C. 49

Question 5

The system x+y=100 and 2x+y=140 has which solution?

A. x=20, y=80

B. x=40, y=60

C. x=60, y=40

D. x=80, y=20

Correct Answer: B. x=40, y=60

Question 6

Which value must be excluded from the equation 1/(q-3)=2?

A. q=0

B. q=1

C. q=2

D. q=3

Correct Answer: D. q=3

Question 7

The solutions of (x-2)(x-5) ≥ 0 are:

A. 2 ≤ x ≤ 5

B. x ≤ 2 or x ≥ 5

C. x < 2 or x > 5

D. x ≥ 2 or x ≤ 5

Correct Answer: B. x ≤ 2 or x ≥ 5

Question 8

If a firm’s selling price equals its variable cost per unit, the standard break-even formula has:

A. A negative denominator

B. A denominator of zero

C. A fixed quantity of one

D. A guaranteed positive profit

Correct Answer: B. A denominator of zero

Question 9

In the equation C=F+vq, what does v represent?

A. Total revenue

B. Fixed cost

C. Variable cost per unit

D. Production capacity

Correct Answer: C. Variable cost per unit

Question 10

A proposed plan uses 80 hours when the available capacity is 75 hours. Which conclusion is justified?

A. The plan is feasible because hours are continuous

B. The plan violates the capacity constraint

C. The plan must be profitable

D. The plan has zero variable cost

Correct Answer: B. The plan violates the capacity constraint