Apply statistical equations is a critical skill for banking and finance professionals in Kenya, enabling precise data analysis, risk assessment, and decision-making. Mastery of quadratic and simultaneous equations supports modeling of financial scenarios such as loan repayments, investment returns, and portfolio optimization. This chapter focuses on solving these equations efficiently using algebraic methods, ensuring practical application in banking reports, financial forecasting, and operational planning.

1.1 Solving linear equations with one or more variables

A linear equation is a statement that two expressions are equal, in which each variable appears only to the power one and is never multiplied by another variable. Solving it means finding the value of the unknown that makes both sides equal. Linear equations are the basic tool of business mathematics: they are used to find prices, quantities, costs, loan terms and break-even points. An equation may contain one unknown, or several unknowns that are related to each other. Example: a farm in Nakuru that knows its total feed bill and the price per bag uses a linear equation to find how many bags were bought.

1.1.1 Meaning and form of a linear equation

A linear equation in one variable can be written in the form ax + b = c, where x is the unknown and a, b and c are known numbers with a not equal to 0. A linear equation in two variables takes the form ax + by = c. Its graph is a straight line, which is why it is called linear. Example: 3x + 7 = 22 is a linear equation in one variable, while 2x + 3y = 12 is a linear equation in two variables.

Linear and non-linear equations compared

Key terms

• Variable (unknown): the letter, such as x or y, that stands for the value to be found.
• Coefficient: the number multiplying a variable, such as 3 in 3x.
• Constant: a number on its own, such as 7 in 3x + 7.
• Solution (root): the value of the variable that makes the equation true.
• Equivalent equations: equations that have the same solution, such as 2x = 10 and x = 5.

1.1.2 Rules for solving linear equations

An equation is like a balanced scale. Whatever is done to one side must also be done to the other, so that the balance is kept. The aim is to isolate the unknown on one side, with all other terms moved to the opposite side. Example: to solve x + 8 = 20, subtract 8 from both sides to leave x = 12.

The balance rules

• Add the same number to both sides: this removes a negative constant, for example x − 5 = 9 becomes x = 14.
• Subtract the same number from both sides: this removes a positive constant, for example x + 8 = 20 becomes x = 12.
• Multiply both sides by the same non-zero number: this removes a fraction, for example x ÷ 4 = 6 becomes x = 24.
• Divide both sides by the same non-zero number: this removes a coefficient, for example 5x = 35 becomes x = 7.
• Collect like terms: terms with the unknown are gathered on one side and constants on the other.

Inverse operations

1.1.3 Solving linear equations in one variable

The general procedure works for every linear equation with one unknown. Example: a hardware shop in Thika buys nails at a fixed delivery charge plus a price per kilogram, and the owner needs to find the number of kilograms bought.

1. Remove brackets: multiply out any brackets, taking care with signs.
2. Clear fractions: multiply every term by the lowest common denominator.
3. Collect unknown terms on one side: add or subtract terms so that all terms containing the unknown are together.
4. Collect constants on the other side: move the numbers to the opposite side.
5. Simplify each side: combine like terms.
6. Divide by the coefficient: this gives the value of the unknown.
7. Check the answer: substitute it back into the original equation.

Worked example 1: two-step equation

Equation: 3x + 7 = 22

Subtract 7 from both sides: 3x = 22 − 7 = 15

Divide both sides by 3: x = 15 ÷ 3 = 5

Result: x = 5

Check: 3 × 5 + 7 = 15 + 7 = 22, which equals the right side.

Worked example 2: negative coefficient

Equation: 10 − 2x = 4

Subtract 10 from both sides: −2x = 4 − 10 = −6

Divide both sides by −2: x = −6 ÷ (−2) = 3

Result: x = 3

Check: 10 − 2 × 3 = 10 − 6 = 4, which equals the right side.

1.1.4 Equations with variables on both sides

Many equations have the unknown on both sides. The unknown terms are first brought together on one side. It is simpler to move the smaller coefficient so that the coefficient of the unknown stays positive. Example: a school compares two suppliers of exercise books whose costs both depend on the number of dozens bought.

Worked example 3

Equation: 5x − 12 = 3x + 8

Subtract 3x from both sides: 2x − 12 = 8

Add 12 to both sides: 2x = 20

Divide both sides by 2: x = 10

Result: x = 10

Check: left side 5 × 10 − 12 = 38, right side 3 × 10 + 8 = 38.

1.1.5 Equations with brackets

Brackets are removed by multiplying every term inside by the number outside. Signs must be handled carefully, because a negative number outside changes the sign of every term inside. Example: a contractor in Kisumu pays a fixed amount plus an hourly rate to each of several workers.

Worked example 4

Equation: 4(x + 3) = 2(x + 11)

Expand: 4x + 12 = 2x + 22

Subtract 2x from both sides: 2x + 12 = 22

Subtract 12 from both sides: 2x = 10

Divide both sides by 2: x = 5

Result: x = 5

Check: left side 4 × (5 + 3) = 32, right side 2 × (5 + 11) = 32.

Worked example 5: negative sign outside the bracket

Equation: 7 − 2(x − 1) = 9

Expand: 7 − 2x + 2 = 9

Collect constants: 9 − 2x = 9

Subtract 9 from both sides: −2x = 0

Divide by −2: x = 0

Result: x = 0

1.1.6 Equations with fractions

To remove fractions, multiply every term on both sides by the lowest common denominator (LCD). The equation then has whole numbers only and is solved in the usual way. Example: a SACCO shares a surplus among members according to fractions of the total.

Worked example 6: fractions added

Equation: x ÷ 3 + x ÷ 4 = 7

The LCD of 3 and 4 is 12. Multiply every term by 12: 12 × x ÷ 3 + 12 × x ÷ 4 = 12 × 7

Simplify: 4x + 3x = 84

Combine like terms: 7x = 84

Divide by 7: x = 12

Result: x = 12

Check: 12 ÷ 3 + 12 ÷ 4 = 4 + 3 = 7.

Worked example 7: fraction on each side

Equation: (2x + 1) ÷ 3 = (x + 4) ÷ 2

Cross-multiply: 2 × (2x + 1) = 3 × (x + 4)

Expand: 4x + 2 = 3x + 12

Subtract 3x from both sides: x + 2 = 12

Subtract 2 from both sides: x = 10

Result: x = 10

Check: left side (2 × 10 + 1) ÷ 3 = 7, right side (10 + 4) ÷ 2 = 7.

1.1.7 Linear equations with more than one variable

When an equation has two or more unknowns, one equation alone cannot give a single number for each unknown. Instead it gives a relationship between them, and there are many pairs of values that satisfy it. The usual task is to express one variable in terms of the others, or to find the value of one variable when the others are known. Example: a bakery in Eldoret mixes two flours, and the cost equation 2x + 3y = 12 links the quantities x and y.

Solving for one variable in terms of the other

Equation: 2x + 3y = 12

Subtract 2x from both sides: 3y = 12 − 2x

Divide both sides by 3: y = (12 − 2x) ÷ 3

Result: y is expressed in terms of x.

Finding pairs of values

Check for x = 3 and y = 2: 2 × 3 + 3 × 2 = 6 + 6 = 12.

Graph of a linear equation in two variables

• Straight line: every pair of values (x, y) that satisfies the equation is a point on a straight line.
• Intercepts: the line meets the x axis where y = 0 and the y axis where x = 0, so the points (6, 0) and (0, 4) above are enough to draw it.
• Meaning: points on the line are all the combinations that satisfy the relationship, such as combinations of two inputs that cost exactly Ksh 12,000.

Equations with a known value for the other variables

When values of all variables except one are given, the equation becomes an ordinary equation in one variable. Example: in 2x + 3y = 12, if y = 2, then 2x + 6 = 12, so 2x = 6 and x = 3.

Link to simultaneous equations

Two unknowns can be found exactly only when two independent equations are given. Those methods, namely substitution, elimination and matrices, are treated in 1.3 and are not repeated here.

1.1.8 Equations with letters as constants

Some equations use letters for known quantities, called parameters, so that one rule works for many cases. Example: ax + b = c, where a, b and c stand for given numbers.

Equation: ax + b = c

Subtract b from both sides: ax = c − b

Divide both sides by a: x = (c − b) ÷ a

Formula: x = (c − b) ÷ a

Substitution: with a = 4, b = 5 and c = 29, x = (29 − 5) ÷ 4

Working: 29 − 5 = 24, and 24 ÷ 4 = 6

Result: x = 6

1.1.9 Changing the subject of a formula

A formula is an equation relating several quantities. Changing the subject means rearranging it so that a different variable stands alone on one side. The same balance rules are used. Example: a bank officer in Mombasa rearranges the simple interest formula to find the amount lent.

Worked example 8: simple interest

Formula: I = P × R × T ÷ 100

Multiply both sides by 100: 100 × I = P × R × T

Divide both sides by R × T: P = 100 × I ÷ (R × T)

Substitution: I = Ksh 9,000, R = 6 per cent, T = 3 years, so P = 100 × 9,000 ÷ (6 × 3)

Working: 100 × 9,000 = 900,000, and 6 × 3 = 18, then 900,000 ÷ 18 = 50,000

Result: the principal is Ksh 50,000.

Worked example 9: total cost

Formula: TC = F + v × Q, where F is fixed cost, v is cost per unit and Q is quantity

Subtract F from both sides: TC − F = v × Q

Divide both sides by v: Q = (TC − F) ÷ v

Substitution: TC = Ksh 190,000, F = Ksh 40,000, v = Ksh 250, so Q = (190,000 − 40,000) ÷ 250

Working: 190,000 − 40,000 = 150,000, and 150,000 ÷ 250 = 600

Result: the quantity produced is 600 units.

1.1.10 Forming and solving equations from word problems

Many practical problems are given in words. They are solved by turning the words into an equation, solving it and stating the answer in the original context. Example: a taxi firm in Nyeri charges a fixed fee plus a rate per kilometre.

1. Read the problem carefully: identify what is asked.
2. Define the unknown: write, for example, "let k be the number of kilometres".
3. Translate the words into an equation: link the known and unknown quantities.
4. Solve the equation: use the balance rules.
5. Check the answer in the problem: make sure it is sensible, for example not negative where a quantity cannot be negative.
6. State the answer with units: write a sentence in the context of the problem.

Worked example 10: taxi fare

A taxi charges Ksh 150 plus Ksh 60 for each kilometre. A passenger pays Ksh 570. Find the distance.

Let k be the number of kilometres.

Equation: 150 + 60k = 570

Subtract 150: 60k = 420

Divide by 60: k = 420 ÷ 60 = 7

Result: the journey was 7 km.

Worked example 11: savings

A SACCO member opened an account with Ksh 2,000 and saves Ksh 500 every month. Find the number of months needed to reach Ksh 14,000.

Let m be the number of months.

Equation: 2,000 + 500m = 14,000

Subtract 2,000: 500m = 12,000

Divide by 500: m = 12,000 ÷ 500 = 24

Result: it takes 24 months.

Worked example 12: two suppliers

A school compares two suppliers of desks. Supplier A charges Ksh 2,000 delivery plus Ksh 3,500 per desk. Supplier B charges Ksh 8,000 delivery plus Ksh 3,000 per desk. Find the number of desks at which both cost the same.

Let n be the number of desks.

Equation: 2,000 + 3,500n = 8,000 + 3,000n

Subtract 3,000n: 2,000 + 500n = 8,000

Subtract 2,000: 500n = 6,000

Divide by 500: n = 12

Result: for 12 desks both suppliers cost the same, Ksh 44,000. Check: 2,000 + 3,500 × 12 = 44,000, and 8,000 + 3,000 × 12 = 44,000.

1.1.11 Checking solutions

Every solution should be checked by substituting it back into the original equation, not into a later step, so that mistakes made on the way are found. Example: for 5x − 12 = 3x + 8, the value x = 10 gives 38 on both sides, so it is correct.

• Substitute into the original equation: replace the unknown with the answer.
• Evaluate both sides separately: work each side to a single number.
• Compare: the two sides must be equal.
• Re-check the working if they differ: look for sign errors and arithmetic slips.

1.1.12 Common errors in solving linear equations

Most mistakes come from careless handling of signs and from not applying an operation to both sides. Example: a student who adds 7 to one side only gets a wrong answer.

• Changing only one side: every operation must be done to both sides.
• Sign errors when moving terms: a term changes sign when it is moved to the other side by addition or subtraction.
• Forgetting to multiply every term in a bracket: 3(x + 4) is 3x + 12, not 3x + 4.
• Not multiplying every term when clearing fractions: constants must also be multiplied by the LCD.
• Dividing by the wrong number: divide by the coefficient of the unknown.
• Skipping the check: an error then goes unnoticed.

Chapter Summary

This chapter covered methods for solving quadratic equations, including the formula method and factorization, both essential for finding roots efficiently. It then explored solutions to simultaneous equations through substitution, elimination, and matrix methods, providing multiple approaches to handle linear systems. Breakeven analysis was computed to determine the point where total costs equal total revenue, a key concept in business decision-making. The chapter introduced calculus basics, focusing on differentiation and integration, which are vital tools for analyzing rates of change and areas under curves. Finally, the formulation of total revenue, total cost, and profit equations was examined, linking mathematical expressions directly to business performance evaluation. Together, these topics equip students with practical mathematical techniques for solving real-world business and economic problems.