Business mathematics is essential for Accountancy professionals in Kenya, enabling them to analyse data, solve financial problems, and make informed decisions. Understanding how to carry out statistical equations, particularly linear equations, equips accountants to model relationships between financial variables, forecast trends, and optimize resource allocation. This chapter focuses on solving linear equations and graphing them, foundational skills that support advanced financial analysis and reporting.

1.1 Linear equations; solving and graphs

Linear equations describe relationships where the change between variables is constant, making them highly relevant in accounting contexts such as budgeting, cost analysis, and revenue forecasting. Mastery of solving these equations and interpreting their graphical representations enables accountants to visualize financial scenarios and solve practical problems efficiently.

1.1.1 Solving linear equations in one variable

Linear equations in one variable have the general form \( ax + b = 0 \), where \( a \) and \( b \) are constants and \( x \) is the variable. Solving such equations involves isolating \( x \) to find its value.

$$ ax + b = 0 $$

Worked Examples

Example 1: A small retail business finds that its monthly profit \( P \) in thousands of Ksh is related to units sold \( x \) by the equation \( 5x - 20 = 0 \). Find the number of units sold to break even.

Given: \( 5x - 20 = 0 \)

$$ 5x = 20 $$

$$ x = \frac{20}{5} $$

$$ x = 4 \text{ units} $$

Answer: 4 units

Example 2: A cooperative has expenses represented by \( 3x + 1500 = 0 \), where \( x \) is the number of days. Find the value of \( x \).

Given: \( 3x + 1500 = 0 \)

$$ 3x = -1500 $$

$$ x = \frac{-1500}{3} $$

$$ x = -500 \text{ days} $$

Answer: -500 days (no practical meaning; indicates no solution in real context)

Example 3: A bank's monthly loan recovery \( R \) follows \( 2x - 8 = 0 \), where \( x \) is in thousands of Ksh. Find \( x \).

Given: \( 2x - 8 = 0 \)

$$ 2x = 8 $$

$$ x = \frac{8}{2} $$

$$ x = 4 \text{ thousand Ksh} $$

Answer: 4 thousand Ksh

1.1.2 Solving linear equations in two variables

Linear equations in two variables have the form \( ax + by = c \), where \( a, b, c \) are constants and \( x, y \) are variables. Solutions are pairs \( (x, y) \) that satisfy the equation. Such equations model relationships like cost and quantity or price and demand.

$$ ax + by = c $$

Worked Examples

Example 1: A hotel charges Ksh 1200 per night for a single room and Ksh 2000 for a double room. If the total revenue \( R \) from \( x \) single rooms and \( y \) double rooms is Ksh 16000, write and solve the equation for \( x = 5 \).

Given: \( 1200x + 2000y = 16000 \), \( x=5 \)

$$ 1200(5) + 2000y = 16000 $$

$$ 6000 + 2000y = 16000 $$

$$ 2000y = 16000 - 6000 $$

$$ 2000y = 10000 $$

$$ y = \frac{10000}{2000} $$

$$ y = 5 $$

Answer: 5 double rooms

Example 2: A SACCO lends Ksh 8000 at 5% interest and Ksh 12000 at 7% interest. The total interest collected \( I \) is Ksh 1160. Express as an equation and find \( x \) when \( y=1 \) (number of loans at each rate).

Given: \( 8000 \times 0.05 x + 12000 \times 0.07 y = 1160 \), \( y=1 \)

$$ 400x + 840 = 1160 $$

$$ 400x = 1160 - 840 $$

$$ 400x = 320 $$

$$ x = \frac{320}{400} $$

$$ x = 0.8 \text{ loans} $$

Answer: 0.8 loans (partial loan; interpret as 1 loan with adjusted amount)

Example 3: A cooperative sells maize and beans. Maize sells at Ksh 150 per kg and beans at Ksh 200 per kg. Total sales \( S \) is Ksh 11,000. If 30 kg of beans were sold, find the quantity of maize sold.

Given: \( 150x + 200(30) = 11000 \)

$$ 150x + 6000 = 11000 $$

$$ 150x = 11000 - 6000 $$

$$ 150x = 5000 $$

$$ x = \frac{5000}{150} $$

$$ x = 33.33 \text{ kg} $$

Answer: 33.33 kg maize

1.1.3 Graphing linear equations

Graphing linear equations helps visualize relationships between variables. The graph of \( ax + by = c \) is a straight line. Plotting points or using intercepts simplifies drawing the graph.

$$ ax + by = c $$

Worked Examples

Example 1: Graph the equation \( 2x + y = 6 \) by finding intercepts.

Given: \( 2x + y = 6 \)

• For \( x=0 \):

$$ 2(0) + y = 6 $$

$$ y = 6 $$

• For \( y=0 \):

$$ 2x + 0 = 6 $$

$$ x = \frac{6}{2} = 3 $$

Points: (0,6), (3,0)

Answer: Plot points (0,6) and (3,0) and draw the line

Example 2: Graph \( 3x - 2y = 12 \) by intercepts.

Given: \( 3x - 2y = 12 \)

• For \( x=0 \):

$$ 3(0) - 2y = 12 $$

$$ -2y = 12 $$

$$ y = \frac{-12}{2} = -6 $$

• For \( y=0 \):

$$ 3x - 0 = 12 $$

$$ x = \frac{12}{3} = 4 $$

Points: (0,-6), (4,0)

Answer: Plot points (0,-6) and (4,0) and draw the line

Example 3: Graph \( x - y = 2 \).

Given: \( x - y = 2 \)

• For \( x=0 \):

$$ 0 - y = 2 $$

$$ y = -2 $$

• For \( y=0 \):

$$ x - 0 = 2 $$

$$ x = 2 $$

Points: (0,-2), (2,0)

Answer: Plot points (0,-2) and (2,0) and draw the line

1.1.4 Solving simultaneous linear equations by substitution and elimination

Simultaneous linear equations involve finding values of variables satisfying two or more linear equations. Substitution and elimination methods are essential tools for Accountancy professionals solving financial systems.

Substitution method

1. Solve one equation for one variable.
2. Substitute this expression into the other equation.
3. Solve for the remaining variable.
4. Substitute back to find the other variable.

Elimination method

1. Multiply equations to align coefficients.
2. Add or subtract equations to eliminate a variable.
3. Solve for the remaining variable.
4. Substitute to find the other variable.

Worked Examples

Example 1 (Substitution): Solve

\( x + y = 10 \)

\( 2x - y = 3 \)

Step 1: From \( x + y = 10 \), express \( y = 10 - x \)

Step 2: Substitute into second equation:

$$ 2x - (10 - x) = 3 $$

$$ 2x - 10 + x = 3 $$

$$ 3x - 10 = 3 $$

$$ 3x = 13 $$

$$ x = \frac{13}{3} = 4.33 $$

Step 3: Substitute \( x \) into \( y = 10 - x \):

$$ y = 10 - 4.33 = 5.67 $$

Answer: \( x = 4.33 \), \( y = 5.67 \)

Example 2 (Elimination): Solve

\( 3x + 2y = 16 \)

\( 5x - 2y = 4 \)

Step 1: Add equations to eliminate \( y \):

$$ (3x + 2y) + (5x - 2y) = 16 + 4 $$

$$ 8x = 20 $$

$$ x = \frac{20}{8} = 2.5 $$

Step 2: Substitute \( x \) into first equation:

$$ 3(2.5) + 2y = 16 $$

$$ 7.5 + 2y = 16 $$

$$ 2y = 16 - 7.5 = 8.5 $$

$$ y = \frac{8.5}{2} = 4.25 $$

Answer: \( x = 2.5 \), \( y = 4.25 \)

Example 3 (Substitution): Solve

\( 4x - y = 11 \)

\( x + 2y = 8 \)

Step 1: From second equation, \( x = 8 - 2y \)

Step 2: Substitute into first:

$$ 4(8 - 2y) - y = 11 $$

$$ 32 - 8y, y = 11 $$

$$ 32 - 9y = 11 $$

$$ -9y = 11 - 32 = -21 $$

$$ y = \frac{-21}{-9} = 2.33 $$

Step 3: Substitute \( y \):

$$ x = 8 - 2(2.33) = 8 - 4.66 = 3.34 $$

Answer: \( x = 3.34 \), \( y = 2.33 \)

Chapter Summary

This chapter covered the fundamental methods for solving and graphing linear equations, providing the basis for understanding relationships between variables. It then progressed to quadratic equations, detailing techniques for solving them and interpreting their parabolic graphs. The concept of differentiation was introduced to analyze rates of change, which is essential in optimizing business functions. Methods for solving simultaneous equations were explained to handle systems with multiple variables occurring together. The chapter also explored break-even analysis, demonstrating how to write equations that identify the point where total costs equal total revenue. Finally, it examined total revenue, total cost, and profit equations, including the application of error analysis to improve the accuracy and reliability of business decision-making. Throughout, practical examples emphasized the use of these mathematical tools in real-world business contexts.