Applying statistical equations is essential across all professional fields in Kenya, from healthcare management at county referral hospitals to financial analysis in SACCOs and operational planning in county government offices. Mastery of linear equations enables professionals to model relationships between variables, predict trends, and make informed decisions based on quantitative data. Understanding how to solve and graph linear equations builds a strong foundation for analyzing business problems and optimizing resources effectively.

1.1 Linear Equations; Solving and Graphs

Linear equations represent the simplest form of mathematical relationships where the highest power of the variable is one. In professional contexts, these equations model relationships such as cost versus production volume, revenue against sales, or resource allocation over time. This section covers how to solve linear equations and graph their solutions, providing essential tools for data interpretation and decision-making.

1.1.1 Understanding Linear Equations

A linear equation in one variable is an algebraic equation of the form \( ax + b = 0 \), where \(a\) and \(b\) are constants, and \(x\) is the variable. The solution to this equation is the value of \(x\) that satisfies the equality. Linear equations in two variables take the form \( y = mx + c \), where \(m\) is the slope and \(c\) is the y-intercept; these are essential for graphing straight lines on the Cartesian plane.

$$ ax + b = 0 $$

$$ y = mx + c $$

Worked Examples

Example 1: A retail shop's monthly fixed cost is Ksh 50,000, and the variable cost per unit is Ksh 200. The total cost \(C\) for producing \(x\) units is modeled by the linear equation \( C = 200x + 50000 \). Find the cost when producing 150 units.

Given: \( x = 150 \) units, \( C = 200x + 50000 \)

$$ C = 200(150) + 50000 $$

$$ C = 30000 + 50000 $$

$$ C = 80000 $$

Answer: Ksh 80,000

Example 2: A cooperative's revenue \(R\) in Ksh depends on the number of members \(x\) through the equation \( R = 350x + 12000 \). Find the revenue when the cooperative has 75 members.

Given: \( x = 75 \), \( R = 350x + 12000 \)

$$ R = 350(75) + 12000 $$

$$ R = 26250 + 12000 $$

$$ R = 38250 $$

Answer: Ksh 38,250

Example 3: A county hospital records that the number of patients \(P\) treated daily decreases linearly as the number of nurses on leave \(x\) increases. The relationship is \( P = 500 - 25x \). Find the number of patients treated when 8 nurses are on leave.

Given: \( x = 8 \), \( P = 500 - 25x \)

$$ P = 500 - 25(8) $$

$$ P = 500 - 200 $$

$$ P = 300 $$

Answer: 300 patients

Example 4: The monthly electricity bill \(E\) in a hotel is modeled by \( E = 1500 + 75x \), where \(x\) is the number of rooms occupied. If the hotel wants to keep the bill under Ksh 9,000, find the maximum number of rooms that can be occupied.

Given: \( E < 9000 \), \( E = 1500 + 75x \)

$$ 1500 + 75x < 9000 $$

$$ 75x < 9000 - 1500 $$

$$ 75x < 7500 $$

$$ x < \frac{7500}{75} $$

$$ x < 100 $$

Answer: Maximum 99 rooms

Example 5: A SACCO's profit \(P\) in Ksh is given by the equation \( P = -500x + 10000 \), where \(x\) is the number of months since a new policy was introduced. After how many months will the SACCO break even (profit zero)?

Given: \( P = 0 \), \( P = -500x + 10000 \)

$$ 0 = -500x + 10000 $$

$$ 500x = 10000 $$

$$ x = \frac{10000}{500} $$

$$ x = 20 $$

Answer: Break-even after 20 months

1.1.2 Solving Linear Equations in One Variable

Solving linear equations in one variable involves isolating the variable on one side of the equation to find its value. This process is fundamental in budgeting, cost analysis, and resource allocation in various sectors such as education and agriculture. The general steps include simplifying both sides, moving variable terms to one side, constants to the other, and dividing to solve for the variable.

$$ ax + b = 0 $$

Worked Examples

Example 1: A farmer has Ksh 20,000 to buy fertilizer costing Ksh 400 per bag. Let \(x\) be the number of bags. Find how many bags can be bought if all money is spent.

Given: Total money \(20,000\), cost per bag \(400\), equation: \(400x = 20000\)

$$ 400x = 20000 $$

$$ x = \frac{20000}{400} $$

$$ x = 50 $$

Answer: 50 bags

Example 2: A school has 800 students. The number of boys is \(x\), and the number of girls is \(800 - x\). If the ratio of boys to girls is 3:5, find the number of boys.

Given: \( \frac{x}{800-x} = \frac{3}{5} \)

$$ 5x = 3(800 - x) $$

$$ 5x = 2400 - 3x $$

$$ 5x + 3x = 2400 $$

$$ 8x = 2400 $$

$$ x = \frac{2400}{8} $$

$$ x = 300 $$

Answer: 300 boys

Example 3: A retail business sells pens at Ksh 15 each and notebooks at Ksh 25 each. If a customer buys a total of 10 items for Ksh 190, find how many pens were bought.

Given: Number of pens \(x\), notebooks \(10 - x\)

Equation: \( 15x + 25(10 - x) = 190 \)

$$ 15x + 250 - 25x = 190 $$

$$ 15x - 25x = 190 - 250 $$

$$ -10x = -60 $$

$$ x = \frac{-60}{-10} $$

$$ x = 6 $$

Answer: 6 pens

Example 4: A hotel has Ksh 50,000 budgeted for staff salaries. If the salary per staff is Ksh 4,000, find the maximum number of staff that can be paid.

Given: \( 4000x \leq 50000 \)

$$ 4000x \leq 50000 $$

$$ x \leq \frac{50000}{4000} $$

$$ x \leq 12.5 $$

Answer: Maximum 12 staff

Example 5: A cooperative's expenses \(E\) and income \(I\) satisfy \( E = 1.2I - 5000 \). If expenses are Ksh 70,000, find the income.

Given: \( E = 70000 \), \( E = 1.2I - 5000 \)

$$ 70000 = 1.2I - 5000 $$

$$ 1.2I = 70000 + 5000 $$

$$ 1.2I = 75000 $$

$$ I = \frac{75000}{1.2} $$

$$ I = 62500 $$

Answer: Ksh 62,500 income

1.1.3 Graphing Linear Equations in Two Variables

Graphing linear equations helps visualize relationships between two variables, such as cost versus quantity or time versus output. The general equation is \( y = mx + c \), where \(m\) is the slope indicating the rate of change, and \(c\) is the y-intercept where the line crosses the y-axis. Plotting points for values of \(x\) and \(y\) and joining them with a straight line represents the equation graphically.

$$ y = mx + c $$

Worked Examples

Example 1: Plot the equation \( y = 2x + 3 \) for values of \(x = 0, 1, 2\).

Calculate \(y\):

For \(x=0\), \( y = 2(0) + 3 = 3 \)

For \(x=1\), \( y = 2(1) + 3 = 5 \)

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

Points: (0,3), (1,5), (2,7)

Plot these points on coordinate axes and draw the line through them.

Example 2: A SACCO's monthly savings \(y\) in Ksh is modeled by \( y = 150x + 1000 \), where \(x\) is the number of months. Plot for \(x = 0, 2, 4\).

Calculate \(y\):

\(x=0\), \(y=150(0)+1000=1000\)

\(x=2\), \(y=150(2)+1000=1300\)

\(x=4\), \(y=150(4)+1000=1600\)

Plot points (0,1000), (2,1300), (4,1600) and join.

Example 3: The cost \(C\) in Ksh of producing \(x\) units is \( C = 300x + 20000 \). Plot for \(x = 0, 5, 10\).

Calculate \(C\):

\(x=0\), \(C=20000\)

\(x=5\), \(C=300(5)+20000=21500\)

\(x=10\), \(C=300(10)+20000=23000\)

Points: (0,20000), (5,21500), (10,23000)

Plot and draw the line.

Example 4: A hotel’s monthly profit \(P\) in Ksh is given by \( P = -500x + 10000 \), where \(x\) is the number of months since opening. Plot for \(x = 0, 10, 20\).

Calculate \(P\):

\(x=0\), \(P=10000\)

\(x=10\), \(P=-500(10)+10000=5000\)

\(x=20\), \(P=-500(20)+10000=0\)

Points: (0,10000), (10,5000), (20,0)

Plot and draw the line.

Example 5: The number of patients \(P\) treated in a clinic per day depends on the number of nurses \(x\) according to \( P = 600 - 30x \). Plot for \(x = 0, 5, 10\).

Calculate \(P\):

\(x=0\), \(P=600\)

\(x=5\), \(P=600 - 150 = 450\)

\(x=10\), \(P=600 - 300 = 300\)

Points: (0,600), (5,450), (10,300)

Plot and draw the line.

1.1.4 Solving Systems of Linear Equations

Systems of linear equations involve two or more linear equations with multiple variables. Solving these systems helps in business and management for resource allocation, budgeting, and optimization problems. Common methods include substitution, elimination, and graphical solutions.

Given two linear equations:

$$ a_1x + b_1y = c_1 $$

$$ a_2x + b_2y = c_2 $$

Worked Examples

Example 1 (Substitution): A hotel sells two types of rooms: standard at Ksh 3000 and deluxe at Ksh 5000. On a certain day, 50 rooms were booked, earning Ksh 200,000. Find how many of each type were booked.

Let \(x\) = number of standard rooms, \(y\) = deluxe rooms

Equations:

$$ x + y = 50 $$

$$ 3000x + 5000y = 200000 $$

From first: \( y = 50 - x \)

Substitute into second:

$$ 3000x + 5000(50 - x) = 200000 $$

$$ 3000x + 250000 - 5000x = 200000 $$

$$ -2000x + 250000 = 200000 $$

$$ -2000x = 200000 - 250000 $$

$$ -2000x = -50000 $$

$$ x = \frac{-50000}{-2000} $$

$$ x = 25 $$

Then \( y = 50 - 25 = 25 \)

Answer: 25 standard rooms, 25 deluxe rooms

Example 2 (Elimination): A cooperative sells maize and beans. Maize costs Ksh 80/kg, beans Ksh 100/kg. A customer buys 15 kg total, paying Ksh 1300. Find the weight of each.

Let \(x\) = maize kg, \(y\) = beans kg

Equations:

$$ x + y = 15 $$

$$ 80x + 100y = 1300 $$

Multiply first by 80:

$$ 80x + 80y = 1200 $$

Subtract from second:

$$ (80x + 100y) - (80x + 80y) = 1300 - 1200 $$

$$ 20y = 100 $$

$$ y = \frac{100}{20} = 5 $$

Then \( x = 15 - 5 = 10 \)

Answer: 10 kg maize, 5 kg beans

Example 3 (Graphical): A hospital has two wards, A and B. Ward A has 30 beds, ward B has 40 beds. Total patients are 60. Total cost per patient is Ksh 5000 in ward A and Ksh 7000 in ward B. Total cost is Ksh 370,000. Find number of patients in each ward.

Let \(x\) = patients in ward A, \(y\) in ward B

Equations:

$$ x + y = 60 $$

$$ 5000x + 7000y = 370000 $$

From first: \( y = 60 - x \)

Substitute:

$$ 5000x + 7000(60 - x) = 370000 $$

$$ 5000x + 420000 - 7000x = 370000 $$

$$ -2000x + 420000 = 370000 $$

$$ -2000x = -50000 $$

$$ x = 25 $$

Then \( y = 60 - 25 = 35 \)

Graph points: (25, 35) satisfy both equations.

Answer: 25 patients in ward A, 35 in ward B

Example 4 (Substitution): A retail store sells pens and pencils. Total items sold are 120. Pens cost Ksh 10, pencils Ksh 5. Total revenue Ksh 950. Find quantity sold of each.

Let \(x\) = pens, \(y\) = pencils

Equations:

$$ x + y = 120 $$

$$ 10x + 5y = 950 $$

From first: \( y = 120 - x \)

Substitute:

$$ 10x + 5(120 - x) = 950 $$

$$ 10x + 600 - 5x = 950 $$

$$ 5x + 600 = 950 $$

$$ 5x = 350 $$

$$ x = 70 $$

Then \( y = 120 - 70 = 50 \)

Answer: 70 pens, 50 pencils

Example 5 (Elimination): A farm produces maize and beans. Total yield is 300 bags. Maize bags sell at Ksh 2500 each, beans at Ksh 3500 each. Total revenue Ksh 780,000. Find number of bags of each.

Let \(x\) = maize bags, \(y\) = beans bags

Equations:

$$ x + y = 300 $$

$$ 2500x + 3500y = 780000 $$

Multiply first by 2500:

$$ 2500x + 2500y = 750000 $$

Subtract from second:

$$ (2500x + 3500y) - (2500x + 2500y) = 780000 - 750000 $$

$$ 1000y = 30000 $$

$$ y = \frac{30000}{1000} = 30 $$

Then \( x = 300 - 30 = 270 \)

Answer: 270 maize bags, 30 beans bags

Chapter Summary

This chapter covered the methods of solving linear equations and representing them graphically to analyze relationships between variables. It then explored quadratic equations, including their solutions and graph shapes, which are useful for modeling curved data trends. The chapter introduced differentiation as a tool for finding rates of change and slopes of curves, followed by integration for calculating areas under curves and accumulated quantities. Techniques for solving simultaneous equations were explained to handle multiple unknowns in business problems. Break-even analysis was presented to determine the point where total costs equal total revenue, aiding decision-making on profitability. The chapter concluded with the formulation of total revenue, total cost, and profit equations, emphasizing the importance of identifying and correcting errors in these calculations to ensure accurate business assessments.