Procurement management professionals frequently encounter mathematical problems involving quadratic and simultaneous equations when analyzing cost functions, optimizing resource allocation, or forecasting demand and supply relationships. Mastery of these equations allows for accurate decision-making in contract pricing, budget planning, and supplier negotiations. This chapter develops skills in solving quadratic equations by formula and factorization methods, as well as solving simultaneous equations using substitution, elimination, and matrix methods. Each technique is illustrated with practical examples relevant to procurement scenarios in Kenya.

2.1 Solving linear equations with one or more variables

A linear equation is an equation in which each variable appears only to the first power: there are no squares, square roots or products of variables, and its graph is a straight line. Examples are 5x − 7 = 18 and 1,200a + 800b = 48,000. Solving a linear equation means finding the value of the unknown that makes both sides equal. In procurement, linear equations answer everyday questions such as how many units can be bought within a budget, how far a hired truck can travel for a given amount, or how a formula can be rearranged to find the missing figure.

Terms used

• Variable (unknown): A letter such as x, y or q that stands for a quantity whose value is to be found.
• Coefficient: The number multiplying a variable; in 2,400x the coefficient is 2,400.
• Constant: A fixed number in the equation, such as a delivery charge of 3,000.
• Expression: A combination of terms such as 2,400x + 3,000.
• Equation: A statement that two expressions are equal, joined by the sign =.
• Solution (root): The value of the variable that makes the equation true.

The balance rule

An equation is like a balanced scale: whatever is done to one side must be done to the other so that the two sides stay equal. The permitted operations are:

• Adding the same number to both sides.
• Subtracting the same number from both sides.
• Multiplying both sides by the same non-zero number.
• Dividing both sides by the same non-zero number.
• Expanding brackets and collecting like terms on either side.

Steps in solving a linear equation in one variable

1. Clear fractions by multiplying every term on both sides by the lowest common multiple (LCM) of the denominators.
2. Remove brackets by multiplying out.
3. Collect all terms containing the variable on one side and all constants on the other, changing the sign of any term that crosses the equals sign.
4. Simplify each side by combining like terms.
5. Divide both sides by the coefficient of the variable.
6. Check the answer by substituting it into the original equation.

Worked examples: one variable

Example 1: Solve 5x − 7 = 18.

5x − 7 = 18

5x = 18 + 7

5x = 25

x = 25 ÷ 5 = 5

Check: 5 × 5 − 7 = 25 − 7 = 18, which agrees.

Example 2 (brackets): Solve 3(2x − 4) = 2(x + 6).

6x − 12 = 2x + 12

6x − 2x = 12 + 12

4x = 24

x = 6

Check: left side 3(12 − 4) = 3 × 8 = 24; right side 2(6 + 6) = 24, which agrees.

Example 3 (fractions): Solve x ÷ 4 + x ÷ 6 = 15.

The LCM of 4 and 6 is 12. Multiply every term by 12:

3x + 2x = 180

5x = 180

x = 36

Check: 36 ÷ 4 + 36 ÷ 6 = 9 + 6 = 15, which agrees.

Forming equations from procurement problems

Most practical problems are given in words, so the first task is to turn the words into an equation.

1. Read the problem and identify the unknown quantity.
2. Choose a letter for the unknown and state what it stands for, with its unit.
3. Write each piece of information in terms of the letter.
4. Form the equation from the condition given, such as a total cost or a budget.
5. Solve the equation.
6. Answer the question in words, with units, and check that the answer makes sense.

Example 4: A procurement officer has a budget of Ksh 75,000 for printing paper. Each carton costs Ksh 2,400 and the supplier charges a fixed delivery fee of Ksh 3,000. How many cartons can be bought?

Let x be the number of cartons.

2,400x + 3,000 = 75,000

2,400x = 75,000 − 3,000 = 72,000

x = 72,000 ÷ 2,400 = 30

The officer can buy 30 cartons. Check: 30 × 2,400 + 3,000 = 72,000 + 3,000 = 75,000.

Example 5: A truck is hired at a fixed charge of Ksh 2,000 plus Ksh 150 per kilometre. The total charge for one trip was Ksh 9,500. How far did the truck travel?

Let d be the distance in kilometres.

C = 150d + 2,000

9,500 = 150d + 2,000

150d = 7,500

d = 7,500 ÷ 150 = 50

The truck travelled 50 km.

Linear equations with two or more variables

A linear equation in two variables, such as 1,200a + 800b = 48,000, has many solutions: every pair of values that satisfies it. It can be solved for one variable in terms of the other, which shows how a change in one affects the other. A single equation with two unknowns cannot give one unique answer; for that, two equations are needed and are solved together (simultaneous equations, 2.3).

Example 6: A store has Ksh 48,000 to spend on two items: safety helmets (a) at Ksh 1,200 each and reflective vests (b) at Ksh 800 each. Express b in terms of a and find the possible combinations.

1,200a + 800b = 48,000

800b = 48,000 − 1,200a

b = (48,000 − 1,200a) ÷ 800

b = 60 − 1.5a

Every extra helmet bought means 1.5 fewer vests, because 1,200 ÷ 800 = 1.5.

The straight-line form y = mx + c

A linear equation in two variables can be written as y = mx + c, where m is the gradient (the change in y for each one-unit increase in x) and c is the intercept (the value of y when x = 0). In Example 5, C = 150d + 2,000 has gradient 150 (the cost per extra kilometre) and intercept 2,000 (the fixed charge). Plotting such equations on a graph is the basis of cost and break-even graphs (2.4 and 2.6).

Changing the subject of a formula

Formulas with several variables are linear in the variable being found, so the same balance rule is used to make that variable the subject.

Example 7: The total cost of an order is T = pq + d, where p is the unit price, q the quantity and d the delivery charge. Make q the subject, and find q when T = 64,500, p = 1,250 and d = 2,000.

T = pq + d

T − d = pq

q = (T − d) ÷ p

q = (64,500 − 2,000) ÷ 1,250

q = 62,500 ÷ 1,250 = 50 units

Common errors to avoid

• Changing the sign wrongly: A term that moves across the equals sign changes its sign.
• Doing an operation to one side only: Every operation must be applied to the whole of both sides.
• Not multiplying every term when clearing fractions: Each term, including whole numbers, must be multiplied by the LCM.
• Dividing by part of a side: When dividing, divide every term on that side.
• Not checking: Substituting the answer back into the original equation catches most errors.

Chapter Summary

This chapter covered essential mathematical techniques for solving quadratic equations using the formula method and factorization, enabling the determination of roots in various civil engineering problems. It explored methods for solving simultaneous equations, including substitution, elimination, and matrix approaches, which are vital for handling multiple variable relationships in practical scenarios. The concept of breakeven analysis was introduced to identify the point where costs and revenues balance, supporting effective financial decision-making. Calculus fundamentals were addressed through differentiation and integration, providing tools for analyzing rates of change and accumulated quantities. Finally, the chapter demonstrated the formulation of total revenue, total cost, and profit equations, linking mathematical principles to business and economic evaluations in engineering projects. Together, these topics equip students with a robust foundation in applying statistical and mathematical equations to solve real-world technical and financial challenges in the field.