By the end of this chapter, you will be able to:
Mastering these skills will help you make smart, data-driven decisions that are essential for success in any business environment.
Compiled by: John Gitau
This chapter develops the essential mathematical tools for solving equations commonly encountered in human resource management analysis, such as workforce planning, salary budgeting, and performance metrics. Mastery of linear and quadratic equations enables HR professionals to quantify relationships among variables like employee numbers, training costs, and productivity indices. These skills support informed decision-making and accurate forecasting in HR operations within Kenyan organizations.
Linear equations model relationships where variables appear to the first power only, reflecting proportional changes common in HR scenarios such as determining total payroll costs or calculating average training hours. Solving these equations requires isolating the variable(s) to find their numerical values that satisfy the equation. When multiple variables are involved, simultaneous equations techniques are used to find values that satisfy all constraints simultaneously.
A single-variable linear equation has the form \( ax + b = 0 \), where \( a \) and \( b \) are constants, and \( x \) is the variable. The goal is to find the value of \( x \) that makes the equation true.
$$ ax + b = 0 $$
Example 1: A retail business pays a fixed monthly salary plus a commission. The total monthly payment is Ksh 50,000, and the fixed salary is Ksh 30,000. If \( x \) is the commission amount, solve for \( x \).
Given: \( a = 1 \), \( b = -20,000 \), equation: \( x + 30,000 = 50,000 \)
$$ x + 30,000 = 50,000 $$
$$ x = 50,000 - 30,000 $$
$$ x = 20,000 $$
Answer: Ksh 20,000
Example 2: An HR manager calculates overtime hours \( h \) where total hours worked are 45, regular hours are 40, and overtime hours satisfy \( h + 40 = 45 \). Find \( h \).
Given: \( a = 1 \), \( b = -5 \), equation: \( h + 40 = 45 \)
$$ h + 40 = 45 $$
$$ h = 45 - 40 $$
$$ h = 5 $$
Answer: 5 hours
Example 3: Training cost per employee is Ksh 3,000 and total training budget is Ksh 75,000. If \( n \) is the number of employees trained, solve for \( n \).
Given: \( a = 3,000 \), \( b = 0 \), equation: \( 3,000 n = 75,000 \)
$$ 3,000 n = 75,000 $$
$$ n = \frac{75,000}{3,000} $$
$$ n = 25 $$
Answer: 25 employees
Simultaneous linear equations involve two equations with two unknowns. They represent constraints such as budget limits and staffing requirements that must be satisfied together. The solution is the pair of values that satisfy both equations.
General form:
$$ a_1 x + b_1 y = c_1 $$
$$ a_2 x + b_2 y = c_2 $$
Example 1: A SACCO pays Ksh 1,000 per month for casual workers and Ksh 3,000 per month for permanent workers. The total monthly wage bill is Ksh 27,000 for 12 workers. Find the number of casual (\( x \)) and permanent (\( y \)) workers.
Given:
$$ x + y = 12 $$
$$ 1,000 x + 3,000 y = 27,000 $$
Step 1: Express \( y \) from first equation:
$$ y = 12 - x $$
Step 2: Substitute into second equation:
$$ 1,000 x + 3,000 (12 - x) = 27,000 $$
$$ 1,000 x + 36,000 - 3,000 x = 27,000 $$
$$ -2,000 x + 36,000 = 27,000 $$
$$ -2,000 x = 27,000 - 36,000 $$
$$ -2,000 x = -9,000 $$
$$ x = \frac{-9,000}{-2,000} $$
$$ x = 4.5 $$
Since number of workers must be whole, adjust to nearest integer:
\( x = 5 \) casual workers
Step 3: Find \( y \):
$$ y = 12 - 5 = 7 $$
Answer: 5 casual workers and 7 permanent workers
Example 2: A university HR department has two types of staff: academic and non-academic. Total staff is 50. Academic staff earn Ksh 40,000 monthly, non-academic earn Ksh 25,000. Total payroll is Ksh 1,700,000. Find the number of academic (\( x \)) and non-academic (\( y \)) staff.
Equations:
$$ x + y = 50 $$
$$ 40,000 x + 25,000 y = 1,700,000 $$
Step 1: Express \( y \):
$$ y = 50 - x $$
Step 2: Substitute:
$$ 40,000 x + 25,000 (50 - x) = 1,700,000 $$
$$ 40,000 x + 1,250,000 - 25,000 x = 1,700,000 $$
$$ 15,000 x + 1,250,000 = 1,700,000 $$
$$ 15,000 x = 450,000 $$
$$ x = \frac{450,000}{15,000} $$
$$ x = 30 $$
Step 3: Find \( y \):
$$ y = 50 - 30 = 20 $$
Answer: 30 academic staff and 20 non-academic staff
Example 3: A hotel employs two categories of cleaners: supervisors and assistants. Supervisors clean 3 rooms per hour, assistants 1 room per hour. Together they clean 24 rooms in 8 hours. There are 6 employees in total. Find the number of supervisors (\( x \)) and assistants (\( y \)).
Equations:
$$ x + y = 6 $$
$$ 3x \times 8 + 1y \times 8 = 24 $$
Simplify second equation:
$$ 24x + 8y = 24 $$
Divide whole equation by 8:
$$ 3x + y = 3 $$
Step 1: Express \( y \):
$$ y = 6 - x $$
Step 2: Substitute:
$$ 3x + (6 - x) = 3 $$
$$ 3x + 6 - x = 3 $$
$$ 2x + 6 = 3 $$
$$ 2x = 3 - 6 $$
$$ 2x = -3 $$
$$ x = \frac{-3}{2} $$
Negative number of supervisors is impossible. Check problem assumptions or data.
If the problem is realistic, re-check conditions.
Answer: No valid solution with given data; revise inputs.
Two main methods are used to solve simultaneous equations:
Substitution involves expressing one variable in terms of the other and substituting into the second equation.
Elimination involves adding or subtracting equations to eliminate one variable.
Example 1 (Substitution): In a cooperative, total members are 40. Male members are twice the female members. Find the number of males (\( x \)) and females (\( y \)).
Equations:
$$ x + y = 40 $$
$$ x = 2 y $$
Step 1: Substitute \( x \):
$$ 2 y + y = 40 $$
$$ 3 y = 40 $$
$$ y = \frac{40}{3} = 13.33 $$
Since number of members must be whole, approximate:
\( y = 13 \), \( x = 2 \times 13 = 26 \)
Check sum:
$$ 26 + 13 = 39 $$ (close to 40, slight adjustment needed)
Answer: Approximately 26 males and 13 females
Example 2 (Elimination): A county office has 10 vehicles: cars and vans. Total passenger capacity is 46. Cars seat 4, vans seat 6. Find number of cars (\( x \)) and vans (\( y \)).
Equations:
$$ x + y = 10 $$
$$ 4x + 6 y = 46 $$
Step 1: Multiply first equation by 4:
$$ 4x + 4 y = 40 $$
Step 2: Subtract from second equation:
$$ (4x + 6 y) - (4x + 4 y) = 46 - 40 $$
$$ 2 y = 6 $$
$$ y = 3 $$
Step 3: Find \( x \):
$$ x = 10 - 3 = 7 $$
Answer: 7 cars and 3 vans
Example 3 (Elimination): A bank employs tellers and customer service officers. Total employees are 25. Total monthly salary is Ksh 1,000,000. Tellers earn Ksh 30,000, officers Ksh 50,000. Find number of tellers (\( x \)) and officers (\( y \)).
Equations:
$$ x + y = 25 $$
$$ 30,000 x + 50,000 y = 1,000,000 $$
Step 1: Multiply first equation by 30,000:
$$ 30,000 x + 30,000 y = 750,000 $$
Step 2: Subtract from second equation:
$$ (30,000 x + 50,000 y) - (30,000 x + 30,000 y) = 1,000,000 - 750,000 $$
$$ 20,000 y = 250,000 $$
$$ y = \frac{250,000}{20,000} = 12.5 $$
Step 3: Find \( x \):
$$ x = 25 - 12.5 = 12.5 $$
Since number of employees must be whole, adjust to nearest integer:
\( x = 13 \), \( y = 12 \)
Answer: 13 tellers and 12 officers
Linear equations assist in budgeting for salaries, training, and benefits where costs vary with employee numbers or time. Understanding how to form and solve these equations enables accurate allocation of resources.
Example 1: An HR department plans to train \( n \) employees. Training cost per employee is Ksh 4,000. Budget is Ksh 120,000. Find maximum \( n \).
Given:
$$ 4,000 n = 120,000 $$
$$ n = \frac{120,000}{4,000} $$
$$ n = 30 $$
Answer: 30 employees
Example 2: A company pays Ksh 20,000 fixed salary plus Ksh 1,000 commission per sale. If total monthly salary is Ksh 50,000, find number of sales \( s \).
Given:
$$ 20,000 + 1,000 s = 50,000 $$
$$ 1,000 s = 50,000 - 20,000 $$
$$ 1,000 s = 30,000 $$
$$ s = \frac{30,000}{1,000} $$
$$ s = 30 $$
Answer: 30 sales
Example 3: A firm employs full-time and part-time employees. Full-time earn Ksh 50,000, part-time Ksh 25,000. Total payroll is Ksh 1,250,000 for 30 employees. Find number of full-time (\( x \)) and part-time (\( y \)) employees.
Equations:
$$ x + y = 30 $$
$$ 50,000 x + 25,000 y = 1,250,000 $$
Step 1: Express \( y \):
$$ y = 30 - x $$
Step 2: Substitute:
$$ 50,000 x + 25,000 (30 - x) = 1,250,000 $$
$$ 50,000 x + 750,000 - 25,000 x = 1,250,000 $$
$$ 25,000 x + 750,000 = 1,250,000 $$
$$ 25,000 x = 500,000 $$
$$ x = \frac{500,000}{25,000} $$
$$ x = 20 $$
Step 3: Find \( y \):
$$ y = 30 - 20 = 10 $$
Answer: 20 full-time and 10 part-time employees
A retail store has 15 employees: sales and cashiers. Sales earn Ksh 35,000, cashiers Ksh 25,000 monthly. Total payroll is Ksh 460,000. Find the number of sales employees and cashiers. (5 marks)
An HR officer calculates overtime hours where total hours worked are 48, regular hours 40. If \( h \) is overtime hours, solve \( h + 40 = 48 \). (3 marks)
A training budget of Ksh 150,000 covers training costs at Ksh 5,000 per employee. Find number of employees trained. (3 marks)
A cooperative has 50 members, with male members twice the female members. Find the number of male and female members. (5 marks)
A hotel employs 8 cleaners: supervisors and assistants. Supervisors clean 5 rooms/hour, assistants 2 rooms/hour. Together they clean 32 rooms in 4 hours. Find number of supervisors and assistants. (6 marks)
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Create a free accountThis chapter covered the application of statistical equations starting with solving linear equations involving one or more variables, which form the basis for many business calculations. It then explored quadratic equations, demonstrating how to solve them using both the formula method and factorization to find values that satisfy these polynomial expressions. The study of simultaneous equations followed, where solutions were obtained through substitution, elimination, and the matrix method, providing multiple approaches for handling systems of equations. Breakeven analysis was introduced as a practical tool to determine the point at which total costs equal total revenue, crucial for business decision-making. The chapter also delved into calculus, focusing on differentiation and integration as methods to analyze rates of change and accumulated quantities, respectively. Finally, it discussed the formulation of total revenue, total cost, and profit equations, tying together the mathematical concepts to model and optimize business performance effectively.
A Human Resource officer needs to solve the linear equation \(3x - 7 = 11\) to find the number of training sessions (\(x\)) required. Find \(x\). (2 marks)
The turnover rate \(x\) in a company satisfies the quadratic equation \(x^2 - 5x + 6 = 0\). Calculate the possible values of \(x\) using the formula method. (3 marks)
A human resource manager at a Nairobi-based bank wants to calculate the number of staff to train if the total training budget \(B\) in Ksh 500,000 is shared equally among \(x\) employees. Formulate and solve the linear equation if each employee receives Ksh 25,000. (4 marks)
In a recruitment drive, the number of applicants \(x\) satisfies the quadratic equation \(x^2 - 11x + 24 = 0\). Solve for \(x\) using the formula method. (4 marks)
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