By the end of this chapter, you will be able to: - perform calculations involving indices correctly using the right mathematical concepts - represent linear equations accurately in mathematical form - use a scientific calculator properly to solve mathematical problems following the manufacturer’s manual - solve simultaneous equations correctly by applying the correct mathematical rules - solve simple algebraic equations accurately using the right concepts - form simple algebraic equations correctly based on mathematical principles
Mastering these algebra skills will help you solve real problems efficiently and confidently in your technical trade.
Linear equations are fundamental in solving problems across various sectors in Kenya, such as budgeting in county government offices, inventory management in retail businesses, or dosage calculations in county referral hospitals. Mastery of linear equations enables professionals to model relationships where one variable depends linearly on another, facilitating decision-making and problem-solving in everyday work scenarios.
A linear equation in one variable represents a straight-line relationship and can be written in the general form
$$ ax + b = 0 $$
where \(a\) and \(b\) are constants, and \(x\) is the variable to solve for.
Example 1: A hotel manager wants to find the number of rooms \(x\) that must be booked to break even if the fixed cost is Ksh 5000 and each room generates Ksh 400 in revenue. The break-even condition is \(400x - 5000 = 0\).
Given: \(a = 400\), \(b = -5000\)
$$ ax + b = 0 $$
$$ 400x - 5000 = 0 $$
$$ 400x = 5000 $$
$$ x = \frac{5000}{400} $$
Answer: 12.5 rooms (meaning at least 13 rooms must be booked)
Example 2: A SACCO offers a loan with a fixed processing fee of Ksh 1000 plus a variable interest of Ksh 0.05 per Ksh 100 borrowed. If the total fee charged is Ksh 2500, find the amount \(x\) borrowed.
Given: \(a = 0.05\), \(b = 1000\), total fee = 2500
$$ 0.05x + 1000 = 2500 $$
$$ 0.05x = 2500 - 1000 $$
$$ 0.05x = 1500 $$
$$ x = \frac{1500}{0.05} $$
Answer: Ksh 30,000 borrowed
Example 3: A county government office pays Ksh 1200 monthly for internet plus Ksh 200 per gigabyte of data used. The total monthly bill is Ksh 3200. Find the data usage \(x\) in gigabytes.
Given: \(a = 200\), \(b = 1200\), total bill = 3200
$$ 200x + 1200 = 3200 $$
$$ 200x = 3200 - 1200 $$
$$ 200x = 2000 $$
$$ x = \frac{2000}{200} $$
Answer: 10 gigabytes
Transposition involves moving terms from one side of the equation to the other by changing their signs to isolate the variable.
The general process for
$$ ax + b = c $$
is to subtract \(b\) from both sides and then divide by \(a\),
$$ ax = c, b $$
$$ x = \frac{c, b}{a} $$
Example 1: A retail business spends Ksh 1500 on fixed costs and Ksh 300 per item sold. If total revenue is Ksh 4500, find the number of items \(x\) sold.
Given: \(a = 300\), \(b = 1500\), \(c = 4500\)
$$ 300x + 1500 = 4500 $$
$$ 300x = 4500 - 1500 $$
$$ 300x = 3000 $$
$$ x = \frac{3000}{300} $$
Answer: 10 items
Example 2: A farmer sells eggs at Ksh 15 each and has fixed daily costs of Ksh 200. If the farmer wants to make Ksh 550 in total revenue, how many eggs \(x\) must be sold?
Given: \(a = 15\), \(b = 200\), \(c = 550\)
$$ 15x + 200 = 550 $$
$$ 15x = 550 - 200 $$
$$ 15x = 350 $$
$$ x = \frac{350}{15} $$
Answer: 23.33 eggs (at least 24 eggs)
Example 3: A TVET college charges Ksh 10,000 registration fee plus Ksh 5,000 tuition per semester. A student paid Ksh 25,000. Find the number of semesters \(x\) paid.
Given: \(a = 5000\), \(b = 10000\), \(c = 25000\)
$$ 5000x + 10000 = 25000 $$
$$ 5000x = 25000 - 10000 $$
$$ 5000x = 15000 $$
$$ x = \frac{15000}{5000} $$
Answer: 3 semesters
Linear equations frequently model cost and revenue relationships, enabling businesses and organizations to calculate break-even points, profits, and losses.
The general revenue or cost function can be expressed as
$$ R = px $$
and
$$ C = mx + b $$
where \(p\) is price per unit, \(m\) is variable cost per unit, and \(b\) is fixed cost.
Example 1: A cooperative sells maize at Ksh 80 per bag. The cost to produce each bag is Ksh 50, and fixed costs are Ksh 10,000. Find the number of bags \(x\) needed to break even.
Given: price \(p = 80\), variable cost \(m = 50\), fixed cost \(b = 10,000\)
Break-even when revenue equals cost:
$$ px = mx + b $$
$$ 80x = 50x + 10000 $$
$$ 80x - 50x = 10000 $$
$$ 30x = 10000 $$
$$ x = \frac{10000}{30} $$
Answer: 333.33 bags (at least 334 bags)
Example 2: A county hospital charges Ksh 1500 per patient visit. The hospital's fixed monthly costs are Ksh 2,000,000, and variable cost per patient is Ksh 900. Find the number of patients \(x\) to break even.
Given: \(p = 1500\), \(m = 900\), \(b = 2,000,000\)
$$ 1500x = 900x + 2000000 $$
$$ 1500x - 900x = 2000000 $$
$$ 600x = 2000000 $$
$$ x = \frac{2000000}{600} $$
Answer: 3333.33 patients (at least 3334 patients)
Example 3: A retail store buys goods at Ksh 120 each and sells them at Ksh 180. Fixed monthly rent is Ksh 15,000. Find the number of goods \(x\) to sell to make a profit of Ksh 30,000.
Profit formula:
$$ \text{Profit} = \text{Revenue} - \text{Cost} $$
$$ \text{Profit} = px - (mx + b) $$
Given profit = 30000
$$ 30000 = 180x - (120x + 15000) $$
$$ 30000 = 180x - 120x - 15000 $$
$$ 30000 + 15000 = 60x $$
$$ 45000 = 60x $$
$$ x = \frac{45000}{60} $$
Answer: 750 goods
Sometimes linear equations contain fractions or decimals, common in measurements or financial calculations. The key is to eliminate fractions by multiplying through by the least common denominator (LCD) or carefully handle decimals.
Example 1: A retail shop sells fruit at Ksh 3.75 per kilogram. The total cost for \(x\) kilograms is Ksh 45. Find \(x\).
Given:
$$ 3.75x = 45 $$
$$ x = \frac{45}{3.75} $$
Calculate:
$$ x = 12 $$
Answer: 12 kilograms
Example 2: A farmer uses \(\frac{2}{3}\) of a bag of fertilizer per hectare. If the farmer has used 8 bags, find the number of hectares \(x\) farmed.
Given:
$$ \frac{2}{3} x = 8 $$
Multiply both sides by 3:
$$ 2x = 8 \times 3 $$
$$ 2x = 24 $$
Divide both sides by 2:
$$ x = \frac{24}{2} $$
$$ x = 12 $$
Answer: 12 hectares
Example 3: A hotel charges Ksh 1200 per night plus Ksh 0.25 per litre of water consumed. If a guest pays Ksh 1350, find the litres of water \(x\) used.
Given:
$$ 1200 + 0.25x = 1350 $$
$$ 0.25x = 1350 - 1200 $$
$$ 0.25x = 150 $$
$$ x = \frac{150}{0.25} $$
$$ x = 600 $$
Answer: 600 litres
Example 4: A SACCO charges a monthly fee of Ksh 500 and interest of \(\frac{1}{20}\) of the loan amount \(x\). If the total charge is Ksh 4500, find the loan amount.
Given:
$$ 500 + \frac{1}{20}x = 4500 $$
Subtract 500:
$$ \frac{1}{20}x = 4500 - 500 $$
$$ \frac{1}{20}x = 4000 $$
Multiply both sides by 20:
$$ x = 4000 \times 20 $$
$$ x = 80000 $$
Answer: Ksh 80,000
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Create a free accountThis chapter covered essential concepts in basic algebra starting with indices and logarithms, focusing on their properties and how to manipulate them for solving mathematical expressions. It then explored linear equations, demonstrating methods to find unknown variables in straightforward algebraic statements. The use of a calculator was introduced as a practical tool to efficiently solve algebraic problems, enhancing accuracy and speed. Techniques for solving simultaneous equations were presented, enabling the determination of multiple unknowns through systems of equations. The chapter also addressed simple algebraic equations, emphasizing the steps to isolate variables and find solutions. Finally, the various forms of simple algebraic equations were discussed, helping students recognize and work with different equation structures commonly encountered in civil engineering calculations.
The monthly electricity consumption \(E\) (in kWh) at a county referral hospital is modelled by the formula \(E = 2^x\), where \(x\) is the number of days since the hospital upgraded to energy-saving bulbs. If on day 5, find the electricity consumption. (4 marks)
Solve for \(x\) in the equation \(3x - 7 = 2x + 8\). (4 marks)
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