Simple Linear Equations

Equation:

This is a mathematical expression showing that the views being expressed are equal

e.g. \( \scriptsize x \: + \: 2 = 7 \)

This example is an equation with an unknown variable.

The above equation can be re-written as x = 7 – 2, therefore x = 5

Example 1.1.1:

Solve Z + 7 = 2x + 3    where x = 6

Solution

Z + 7 = 2(6) + 3

Z + 7 = 12 + 3

Z + 7 = 15

Z = 15 – 7

Z = 8

Example 1.1.2:

Z + 7 = ax + c, where a = 3, x = 10, c = 3

Solution

Substitute these values into the equation

Z + 7 = 3(10) + 3

Z + 7 = 30 + 3

Z + 7 = 33

Z = 33 – 7

 Z = 26

Simple Linear Equation:

This is an equation of the first degree. The graph of a simple linear equation is always a straight line. 

Example 1.1.3:

A market Demand curve for yams is given by D = 7x – 14
find x where D = 63   

Solution:

D = 7x – 14 

Substitute the value of D into the equation.

63 = 7x – 14

63 + 14 = 7x

7x = 63 + 14

7x = 77

x = \( \frac {77}{7} \)

x = 11

Example 1.1.4:

Given the equation \(\scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize P\)
where Q = quantity demanded and P = price

(a) Find the quantity demanded when price is (i) ₦30 (ii) ₦60
(b) Find the price when the quantity is (i) 0 (ii) 60

(a) Find the quantity demanded when price is (i) ₦30 (ii) ₦60

Solution:

(i) ₦30

Substitute 30 for P

Given the equation \(\scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize P \\ \scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize (30) \\ \scriptsize Q = 60 \: – \: 10 \\ \scriptsize Q = 50 \)

(ii) ₦60

Substitute 60 for P

Given the equation \(\scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize P \\ \scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize (60) \\ \scriptsize Q = 60 \: – \: 20 \\ \scriptsize Q = 40 \)

(b) Find the price when the quantity is (i) 0 (ii) 60

Solution:

(i) Q = 0

Substitute Q for 0

Given the equation \(\scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize P \\ \scriptsize 0 = 60 \: – \: \normalsize \frac{1}{3}\scriptsize (P) \\ \scriptsize 0 = 60 \: – \: \normalsize \frac{P}{3} \\ \scriptsize – 60 = \: – \normalsize \frac{P}{3} \\ \frac{P}{3} \scriptsize = 60 \\ \scriptsize P = 60 \: \times \: 3 \\ \scriptsize P = 180 \)

(ii) Q = 60

Substitute Q for 60

Given the equation \(\scriptsize Q = 60 \: – \: \normalsize \frac{1}{3}\scriptsize P \\ \scriptsize 60 = 60 \: – \: \normalsize \frac{1}{3}\scriptsize (P) \\ \scriptsize 60 = 60 \: – \: \normalsize \frac{P}{3} \\ \frac{P}{3} \scriptsize = 60 \: -\: 60 \\ \frac{P}{3} \scriptsize = 0 \\ \scriptsize P = 0 \)