Chapter 2: Apply statistical equations

Apply statistical equations are fundamental in diverse Kenyan professional settings such as building construction, road works, water supply, and manufacturing. Mastery of quadratic and simultaneous equations enables problem-solving in budgeting, resource allocation, data analysis, and operational planning. This chapter focuses on practical methods for solving quadratic equations and systems of simultaneous equations using substitution, elimination, and matrix techniques. These mathematical tools empower diploma-level students to apply quantitative reasoning effectively across multiple sectors.

2.1 Solving linear equations with one or more variables

A linear equation is an equation in which every variable is to the power one and no variable is multiplied by another variable. The equation 3x + 7 = 25 is linear. The equations x² = 9 and xy = 12 are not. A linear equation with one variable, such as 3x + 7 = 25, has one solution. A linear equation with two or more variables, such as 2x + 3y = 12, has many solutions, and more equations are needed to fix a single answer.

Purchasing officers, accounts clerks and project managers meet linear equations every day. They work out how many cartons a given sum buys, what price lies behind a discounted invoice, or how a budget can be split between two items.

Equations with one variable

To solve an equation is to find the value of the variable that makes both sides equal. The method is the balance method: whatever is done to one side is done to the other, so that the equation stays true.

1. Remove brackets by multiplying each term inside by the number outside.
2. Clear fractions by multiplying every term by a common multiple of the denominators.
3. Collect like terms on each side.
4. Move the variable terms to one side and the numbers to the other side, by adding or subtracting the same amount on both sides.
5. Divide both sides by the coefficient of the variable.
6. Check by putting the answer back into the original equation.

Worked examples with one variable

Example 1: Solve 3x + 7 = 25.

3x + 7 = 25
3x     = 25 - 7 = 18
x      = 18 ÷ 3  = 6
Check: 3 × 6 + 7 = 25

Example 2: Solve 5x - 4 = 2x + 17, where the variable is on both sides.

5x - 4 = 2x + 17
5x - 2x = 17 + 4
3x     = 21
x      = 7
Check: left side 5 × 7 - 4 = 31, right side 2 × 7 + 17 = 31

Example 3: Solve 4(2x - 3) = 3(x + 6) - 5, which has brackets.

8x - 12 = 3x + 18 - 5
8x - 12 = 3x + 13
8x - 3x = 13 + 12
5x      = 25
x       = 5
Check: left side 4 × (10 - 3) = 28, right side 3 × 11 - 5 = 28

Example 4: Solve (x + 3) ÷ 4 = (2x - 1) ÷ 3, which has fractions.

Multiply both sides by 12:   3(x + 3) = 4(2x - 1)
3x + 9 = 8x - 4
9 + 4  = 8x - 3x
13     = 5x
x      = 13 ÷ 5 = 2.6
Check: (2.6 + 3) ÷ 4 = 1.4 and (5.2 - 1) ÷ 3 = 1.4

Example 5: Solve x ÷ 2 + x ÷ 3 = 10.

Multiply every term by 6:   3x + 2x = 60
5x = 60
x  = 12
Check: 12 ÷ 2 + 12 ÷ 3 = 6 + 4 = 10

Equations with no solution or many solutions

Sometimes the variable disappears when both sides are simplified. If the result is a false statement, there is no solution. For 2x + 3 = 2x + 5, taking 2x from both sides gives 3 = 5, which is false, so no value of x works. If the result is a true statement, every value of x works. For 2x + 4 = 2(x + 2), removing the bracket gives 2x + 4 = 2x + 4, which is true for every x.

Forming and solving equations from business problems

A word problem is solved in four steps: let a letter stand for the unknown quantity, write the statement as an equation, solve it, and check the answer against the words of the problem. The answer is given with its unit.

Example 6: An office buys 5 identical chairs and a desk that costs Ksh 9,500. The bill is Ksh 34,500. Find the price of one chair.

Let c be the price of one chair, in Ksh.
5c + 9,500 = 34,500
5c         = 34,500 - 9,500 = 25,000
c          = 25,000 ÷ 5     = 5,000
Check: 5 × 5,000 + 9,500 = 34,500

A chair costs Ksh 5,000.

Example 7: A supplier charges a delivery fee of Ksh 1,200 plus Ksh 85 for each carton. A buyer pays Ksh 11,400 in all. How many cartons were delivered?

Let n be the number of cartons.
1,200 + 85n = 11,400
85n         = 11,400 - 1,200 = 10,200
n           = 10,200 ÷ 85    = 120
Check: 85 × 120 + 1,200 = 10,200 + 1,200 = 11,400

The buyer received 120 cartons.

Example 8: A buyer pays Ksh 8,550 for goods after a trade discount of 5% on the list price. Find the list price.

Let L be the list price, in Ksh.
L - 0.05L = 8,550
0.95L     = 8,550
L         = 8,550 ÷ 0.95 = 9,000
Check: 9,000 × 0.95 = 8,550

The list price is Ksh 9,000.

Example 9: Two clerks share Ksh 18,000 of overtime pay. The first clerk receives Ksh 2,000 more than the second. How much does each receive?

Let x be the second clerk's share, in Ksh. The first clerk's share is x + 2,000.
x + (x + 2,000) = 18,000
2x              = 16,000
x               = 8,000

The second clerk receives Ksh 8,000 and the first receives Ksh 10,000. Check: 8,000 + 10,000 = 18,000.

Example 10: A casual worker is paid a fixed sum of Ksh 600 plus Ksh 120 for each hour worked. The total pay is Ksh 2,160. How many hours were worked?

Let h be the hours worked.
600 + 120h = 2,160
120h       = 1,560
h          = 13
Check: 600 + 120 × 13 = 600 + 1,560 = 2,160

The worker worked 13 hours.

Making one variable the subject

A formula with several letters is also a linear equation when each letter is to the power one. To solve for one letter, the same balance method is used, treating the other letters as numbers. The total bill in Example 7 is T = 1,200 + 85n. To find n from T, subtract the delivery fee and then divide.

T = F + p × n
T - F = p × n
n = (T - F) ÷ p

Putting in T = 11,400, F = 1,200 and p = 85 gives n = (11,400 - 1,200) ÷ 85 = 10,200 ÷ 85 = 120, the same answer as before.

Equations with two or more variables

A single linear equation with two variables has many pairs of values that satisfy it. Take 2x + 3y = 12. If x = 0, then 3y = 12 and y = 4. If x = 3, then 3y = 6 and y = 2. If x = 6, then y = 0. All the pairs lie on one straight line when plotted on a graph. Writing one variable in terms of the other gives the line's rule, here y = (12 - 2x) ÷ 3.

Example 11: An office has Ksh 4,000 to spend on packs of markers at Ksh 200 each and files at Ksh 100 each. Let m be the number of packs of markers and f the number of files.

200m + 100f = 4,000
f = 40 - 2m

Each choice of m gives the number of files that uses the whole budget.

All five choices spend exactly Ksh 4,000, so the single equation does not tell the office which one to pick. A second condition is needed, for example that the office wants twice as many files as packs of markers. With two independent equations and two unknowns there is one solution, found by the methods under Simultaneous equations. An equation with three variables, such as x + y + z = 60, needs three independent equations to fix one solution.

Chapter Summary

This chapter focused on applying various mathematical techniques essential for solving engineering and business problems. It began with methods for solving quadratic equations, exploring both the formula method and factorization to find roots efficiently. The chapter then covered simultaneous equations, demonstrating how substitution, elimination, and matrix methods can be used to solve systems of linear equations. Breakeven analysis was introduced to determine the point where total costs equal total revenue, a critical concept in financial decision-making. The calculus section addressed fundamental operations, including differentiation to find rates of change and integration to calculate areas under curves. Finally, the chapter concluded with formulating total revenue, total cost, and profit equations, providing a framework for analysing business performance mathematically. These topics collectively equip students with practical tools for quantitative analysis in technical and economic contexts.