Marketing  ·  Level 5
Business Mathematics And Statistics
Chapter 1: Apply Statistical Equations
📚 7 Topics
What you will be able to do

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

  • determine linear equations correctly based on given objective functions.
  • find quadratic equations accurately according to specific objective functions.
  • formulate simultaneous equations that align with the objective functions.
  • compute break-even analysis following your organization's goals.
  • carry out differentiation and integration correctly to solve objective-related problems.
  • create total revenue, total cost, and profit equations based on organizational objectives.

These skills will help you make smart, data-driven decisions that are essential for success in any business environment.

Applying statistical equations often begins with mastering linear equations, a fundamental tool in business mathematics used to model relationships between variables. In Kenya’s diverse sectors such as banking, healthcare, education, and agriculture, solving linear equations allows professionals to analyze costs, revenues, resource allocation, and performance trends. Graphs of linear equations provide visual insights that support decision-making in organizations like county hospitals, universities, SACCOs, and retail businesses. This chapter focuses on solving linear equations and interpreting their graphs, equipping students across all fields with practical quantitative skills.

1.1 Linear equations; solving and graphs

Linear equations are algebraic expressions representing straight lines when graphed. They describe a constant rate of change between two variables, commonly written in the form \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. Mastering the solution of linear equations and their graphical representation is essential for analyzing trends, forecasting, and budgeting across professional fields in Kenya.

1.1.1 Solving linear equations in one variable

Solving linear equations with one variable means finding the value of \( x \) that makes the equation true. These equations take the form \( ax + b = 0 \), where \( a \) and \( b \) are constants and \( a eq 0 \).

The general approach involves isolating \( x \) on one side of the equation using inverse operations such as addition, subtraction, multiplication, or division.

$$ ax + b = 0 $$

$$ x = \frac{-b}{a} $$

Worked Examples

Example 1: A retail shop calculates that the monthly electricity bill \( E \) in Ksh depends on the number of units \( x \) consumed as \( 15x + 300 = E \). If the bill is Ksh 1200, find the units consumed.

Given: \( 15x + 300 = 1200 \) Ksh

$$ 15x + 300 = 1200 $$

$$ 15x = 1200 - 300 $$

$$ 15x = 900 $$

$$ x = \frac{900}{15} $$

Answer: 60 units

Example 2: A SACCO charges a fixed fee of Ksh 500 and a variable fee of Ksh 50 per transaction. The total fee \( F \) for \( x \) transactions is \( 50x + 500 = F \). If a member pays Ksh 2000, how many transactions did they make?

Given: \( 50x + 500 = 2000 \) Ksh

$$ 50x + 500 = 2000 $$

$$ 50x = 2000 - 500 $$

$$ 50x = 1500 $$

$$ x = \frac{1500}{50} $$

Answer: 30 transactions

Example 3: A farm produces vegetables whose total revenue \( R \) in Ksh is given by \( 120x - 200 = R \), where \( x \) is the kilograms sold. If the revenue is Ksh 2800, find the kilograms sold.

Given: \( 120x - 200 = 2800 \) Ksh

$$ 120x - 200 = 2800 $$

$$ 120x = 2800 + 200 $$

$$ 120x = 3000 $$

$$ x = \frac{3000}{120} $$

Answer: 25 kg

Example 4: The cost \( C \) of photocopying documents at a university is \( 10x + 250 = C \), where \( x \) is the number of pages. If the total cost is Ksh 850, find the number of pages copied.

Given: \( 10x + 250 = 850 \) Ksh

$$ 10x + 250 = 850 $$

$$ 10x = 850 - 250 $$

$$ 10x = 600 $$

$$ x = \frac{600}{10} $$

Answer: 60 pages

Example 5: A hotel charges a fixed booking fee of Ksh 1000 and Ksh 300 per night. The total charge \( T \) for \( x \) nights is \( 300x + 1000 = T \). If the total charge is Ksh 4900, find the number of nights booked.

Given: \( 300x + 1000 = 4900 \) Ksh

$$ 300x + 1000 = 4900 $$

$$ 300x = 4900 - 1000 $$

$$ 300x = 3900 $$

$$ x = \frac{3900}{300} $$

Answer: 13 nights

1.1.2 Solving linear equations in two variables

Linear equations with two variables express relationships where the solution is a pair \((x, y)\) satisfying the equation. The standard form is:

$$ ax + by = c $$

where \( a \), \( b \), and \( c \) are constants, and \( a \) and \( b \) are not both zero. Solutions are infinite and lie on a straight line when graphed.

To find specific solutions, one variable can be expressed in terms of the other.

$$ y = \frac{c, ax}{b} $$

Worked Examples

Example 1: A county hospital sells two types of medical kits: basic kits at Ksh 1500 each and advanced kits at Ksh 3000 each. If the total revenue from selling \( x \) basic and \( y \) advanced kits is Ksh 75,000, find the equation representing this.

Given: \( 1500x + 3000y = 75000 \)

Express \( y \) in terms of \( x \):

$$ 3000y = 75000 - 1500x $$

$$ y = \frac{75000 - 1500x}{3000} $$

$$ y = 25 - 0.5x $$

Example 2: A retail store sells pens at Ksh 20 each and notebooks at Ksh 50 each. If a customer buys \( x \) pens and \( y \) notebooks for Ksh 900, write the linear equation and express \( y \) in terms of \( x \).

Given: \( 20x + 50y = 900 \)

$$ 50y = 900 - 20x $$

$$ y = \frac{900 - 20x}{50} $$

$$ y = 18 - 0.4x $$

Example 3: A cooperative sells two types of maize flour bags: small bags at Ksh 800 and large bags at Ksh 1500. If \( x \) small bags and \( y \) large bags were sold to generate Ksh 21,000, write the equation and express \( y \) in terms of \( x \).

Given: \( 800x + 1500y = 21000 \)

$$ 1500y = 21000 - 800x $$

$$ y = \frac{21000 - 800x}{1500} $$

$$ y = 14 - \frac{16}{30}x $$

Example 4: A university bookstore sells novels at Ksh 120 and textbooks at Ksh 350. If the total sales from \( x \) novels and \( y \) textbooks is Ksh 11,700, express \( y \) as a function of \( x \).

Given: \( 120x + 350y = 11700 \)

$$ 350y = 11700 - 120x $$

$$ y = \frac{11700 - 120x}{350} $$

$$ y = 33.43 - 0.343x $$

Example 5: A farm sells eggs and milk. Eggs cost Ksh 10 each and milk Ksh 70 per litre. If the total revenue from selling \( x \) eggs and \( y \) litres of milk is Ksh 2800, find the linear equation and express \( y \) in terms of \( x \).

Given: \( 10x + 70y = 2800 \)

$$ 70y = 2800 - 10x $$

$$ y = \frac{2800 - 10x}{70} $$

$$ y = 40 - 0.143x $$

1.1.3 Graphing linear equations

Graphing linear equations visually represents the relationship between variables, showing how one variable changes with respect to the other. The graph of a linear equation in two variables is a straight line, defined by its slope \( m \) and y-intercept \( c \) in the form:

$$ y = mx + c $$

where \( m = -\frac{a}{b} \) and \( c = \frac{c}{b} \) from the standard form \( ax + by = c \).

Plotting two points that satisfy the equation and joining them produces the line.

Worked Examples

Example 1: Graph the equation \( y = 2x + 1 \) representing a hotel’s revenue where \( y \) is revenue in Ksh and \( x \) is the number of rooms booked.

Calculate points:

For \( x=0 \), \( y = 2(0) + 1 = 1 \)

For \( x=3 \), \( y = 2(3) + 1 = 7 \)

Plot points (0,1) and (3,7) and draw the line through them.

Example 2: Graph \( y = -x + 4 \) representing a pharmacy’s stock levels where \( y \) is remaining stock and \( x \) is days.

Calculate points:

For \( x=0 \), \( y = -0 + 4 = 4 \)

For \( x=4 \), \( y = -4 + 4 = 0 \)

Plot points (0,4) and (4,0) and draw the line.

Example 3: Graph the equation \( 3x + 2y = 12 \) from a retail business’s sales data.

Convert to slope-intercept form:

$$ 2y = 12 - 3x $$

$$ y = 6 - 1.5x $$

Calculate points:

For \( x=0 \), \( y=6 \)

For \( x=4 \), \( y=6 - 1.5(4) = 6 - 6 = 0 \)

Plot (0,6) and (4,0) and draw the line.

Example 4: Graph \( y = \frac{1}{2}x - 3 \) representing a county government’s budget deficit reduction over months.

Calculate points:

For \( x=0 \), \( y = -3 \)

For \( x=6 \), \( y = 3 - 3 = 0 \)

Plot (0,-3) and (6,0) and draw the line.

Example 5: Graph \( 4x, y = 8 \) representing milk sales in litres \( y \) versus days \( x \) at a dairy cooperative.

Convert to slope-intercept form:

$$ -y = 8 - 4x $$

$$ y = 4x - 8 $$

Calculate points:

For \( x=2 \), \( y = 4(2) - 8 = 0 \)

For \( x=4 \), \( y = 4(4) - 8 = 8 \)

Plot (2,0) and (4,8) and draw the line.

1.1.4 Applications of linear equations in business contexts

Linear equations model real-world relationships such as cost, revenue, and profit, enabling professionals to make informed decisions. For example, banks use linear equations to determine loan repayment schedules, while farms calculate production costs. Graphing these equations helps visualize trends and forecast outcomes.

Worked Examples

Example 1: A SACCO charges Ksh 2000 fixed membership fee and Ksh 150 monthly contribution per member. Write the linear equation for total payment \( T \) after \( x \) months and calculate total payment after 12 months.

Given:

$$ T = 150x + 2000 $$

Substitute \( x = 12 \):

$$ T = 150(12) + 2000 $$

$$ T = 1800 + 2000 $$

$$ T = 3800 \text{ Ksh} $$

Answer: Ksh 3800

Example 2: A retail business sells T-shirts at Ksh 500 each with a fixed monthly rent of Ksh 30,000. Write the equation for total monthly cost \( C \) if \( x \) shirts are bought and calculate cost for 50 shirts.

Given:

$$ C = 500x + 30000 $$

Substitute \( x = 50 \):

$$ C = 500(50) + 30000 $$

$$ C = 25000 + 30000 $$

$$ C = 55000 \text{ Ksh} $$

Answer: Ksh 55,000

Example 3: A county hospital estimates patient costs as Ksh 800 per day plus a fixed admission fee of Ksh 1500. Write the cost equation and calculate for a 7-day stay.

Given:

$$ C = 800x + 1500 $$

Substitute \( x = 7 \):

$$ C = 800(7) + 1500 $$

$$ C = 5600 + 1500 $$

$$ C = 7100 \text{ Ksh} $$

Answer: Ksh 7,100

Example 4: A university’s printing service charges Ksh 10 per page plus a fixed maintenance fee of Ksh 1200 monthly. Calculate the total cost for printing 350 pages.

Given:

$$ C = 10x + 1200 $$

Substitute \( x = 350 \):

$$ C = 10(350) + 1200 $$

$$ C = 3500 + 1200 $$

$$ C = 4700 \text{ Ksh} $$

Answer: Ksh 4,700

Example 5: A farm’s fertilizer cost is Ksh 2500 fixed plus Ksh 150 per bag. Write the cost equation and calculate cost for 20 bags.

Given:

$$ C = 150x + 2500 $$

Substitute \( x = 20 \):

$$ C = 150(20) + 2500 $$

$$ C = 3000 + 2500 $$

$$ C = 5500 \text{ Ksh} $$

Answer: Ksh 5,500

Practice Questions

  1. A hotel charges Ksh 3500 per night plus a fixed booking fee of Ksh 1500. Write the equation for total charge \( T \) for \( x \) nights and calculate the total charge for 5 nights. (5 marks)

  2. A school sells textbooks at Ksh 1200 each and exercise books at Ksh 350 each. If a student buys \( x \) textbooks and \( y \) exercise books for Ksh 7550, write the linear equation and express \( y \) in terms of \( x \). (5 marks)

  3. A retail store’s monthly electricity bill is given by \( 25x + 400 \), where \( x \) is the number of units consumed. If the bill is Ksh 1150, find the number of units consumed. (5 marks)

  4. Graph the equation \( y = -3x + 12 \) representing monthly profit in thousands of Ksh \( y \) and number of products sold \( x \). Calculate and plot two points. (5 marks)

  5. A cooperative sells maize flour bags at Ksh 900 each and charges a fixed transport fee of Ksh 3000. Write the cost equation and calculate the total cost for 15 bags. (5 marks)

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🔒1.2 Quadratic equations; solving and graphs

Quadratic equations frequently arise in various Kenyan professional contexts such as calculating areas for agricultural plots, optimizing costs in retail businesses, or determining dosage curves in healthcare. Understanding how to solve and graph quadratic equ…

🔒1.3 Differentiation

Differentiation is a fundamental mathematical tool used in business mathematics and statistics to analyze how quantities change. In Kenyan professional contexts such as finance, healthcare management, agriculture, and retail, differentiation helps in optimizin…

🔒1.4 Integration

Integration is a fundamental tool in business mathematics and statistics used for calculating areas under curves, total accumulated values, and solving problems involving continuous data. In Kenyan professional contexts such as county referral hospitals, banks…

🔒1.5 Simultaneous equations; solving

Simultaneous equations arise in many Kenyan professional fields when two or more conditions must be satisfied at the same time. For example, a county government office budgeting for staff salaries and office supplies may need to balance total expenditure with…

🔒1.6 Break-even analysis

Break-even analysis is a fundamental financial tool used by businesses and organizations across Kenya to determine the point at which total revenues equal total costs, resulting in neither profit nor loss. This analysis is crucial for decision-making in sector…

🔒1.7 Total revenue, total cost and profit equations; application of errors

In Kenyan businesses across sectors such as retail shops in Nairobi, county referral hospitals, and agricultural cooperatives in Central Kenya, understanding the relationship between total revenue, total cost, and profit is essential for effective financial ma…

Chapter Summary

This chapter covered key mathematical tools essential for business analysis, beginning with linear equations where methods of solving and graphing were explored to model relationships between variables. Quadratic equations were then examined, including techniques for finding solutions and interpreting their graphs to understand business scenarios involving nonlinear trends. Differentiation was introduced as a method to determine rates of change, useful in optimizing business functions such as cost and revenue. Integration followed, providing a way to calculate accumulated quantities, which is important in areas like total cost estimation over time. The chapter also addressed simultaneous equations, demonstrating how to solve systems of equations that arise when multiple variables interact in business problems. Break-even analysis was explained to identify the point where total revenue equals total cost, enabling informed decision-making about profitability. Finally, the relationships between total revenue, total cost, and profit were formulated, highlighting the application of error analysis to improve accuracy in business calculations.

Self-Assessment

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Written Assessment

  1. A retail business records its monthly sales \(S\) (in Ksh thousands) as a function of advertising expenses \(x\) (in Ksh thousands) by the linear equation \(S = 5x + 20\). Calculate the sales when the advertising expenses are Ksh 15,000. (2 marks)

  2. A county government office’s expense \(E\) (in Ksh thousands) on a project follows the quadratic equation \(E = 3t^2 - 4t + 10\), where \(t\) is time in months. Find the expense at \(t = 5\) months. (3 marks)

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Chapter Examination Questions

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SECTION A (40 Marks) - Answer ALL Questions

  1. A retail business has a linear cost function given by \(C = 1500 + 25x\), where \(x\) is the number of units produced. Find the cost when 100 units are produced. (4 marks)
  2. Solve the quadratic equation \(2x^2 - 5x - 3 = 0\) and interpret the roots in a business context. (4 marks)
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Am I competent?

At the start of this chapter we promised you would be able to:

  • determine linear equations correctly based on given objective functions.
  • find quadratic equations accurately according to specific objective functions.
  • formulate simultaneous equations that align with the objective functions.
  • compute break-even analysis following your organization's goals.
  • carry out differentiation and integration correctly to solve objective-related problems.
  • create total revenue, total cost, and profit equations based on organizational objectives.

Tick each one you can genuinely do.

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