Example 1
Consider the following system of linear equations
3 x
7 x


5 y
11 y


13

15
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
3
7 x x


5 y
11 y


13

7

15

3

7 ( 1 )

11 ( 2 )

7

22
21 x

15 ..

35 y

33 y


91

45
Subtracting gives 68 y

136 .
Thus y

2 . Substituting into the first equation we get 3 x

5 ( 2 )

13

3 x

10

13

3 x

3
 x

1 .Thus solution is
If this is correct solution it should satisfy 7 x

11 y
 x

1 y

2 .

15 and it does since

21 x
Geometric solution

 
5
3 x





13
5




7
11 x





15
11


5
4
3
2
1
7
6

3
 
0.333
1 2.3333.667 5 x
To find where straight line we put x

0

5 y

13

To find where straight line we put y

0

3 x

13

3 x

5 y

13 cuts y-axis( x=0) y

13 /
3 x x


5 y
13 / 3
5


13
2 .
6 . cuts x-axis( y=0)

4 .
33
Draw a straight between the points line 3 x

5 y

13 as shown above.
( 0 , 2 .
6 ) and ( 4 .
33 , 0 ) to give the graph of the straight
To find where straight line 7 x

11 y
 
15 cuts y-axis( x=0) we put x

0
 
11 y
 
15
To find where straight line 7
 x y


11 y
15

/ 11

15

1 .
36 . cuts x-axis( y=0) we put y

0

7 x
 
15
 x
 
15 / 7
 
2 .
14
Draw a straight between the points line 7 x

11 y
 
15
( 0 , 1 .
36 )
as shown above.
and (

2 .
14 , 0 ) to give the graph of the straight

The lines 3 x

5 y

13 and 7 x

11 y
 
15 intersect at the point x

1 , y

2 which is the solution of the system of equations
3 x
7 x

5 y

11 y

13
 
15 and confirms the algebraic solution.

Laws of Supply and Demand,
Applications of the straight line equation
Demand and supply decisions by consumers, firms and governments determine the level of economic activity within an economy. As these decisions play a vital role in business and consumer activity, it is important to mathematically model and analyse them. This can be done by modelling the simple laws of economics, namely the demand and supply laws , by linear equations as a first approximate model.
There are several variables that influence the demand for a certain good or service X .
These may be expressed by the general demand function
P
 f ( Q , Y , T , A , O ) where
Q is the quantity demanded for good X
P is the price of good X
Y is the income of the consumer
T is the fashion or taste of the consumer
O is other factors if there are any.
The simplest model is
P
 f ( Q ) where the price depends mainly on the quantity(or other factors remain fixed or negligible).This is called the law of demand in economics and the function f the demand function.
The demand function P
 f ( Q ) can be modelled by the general simple linear equation
P
 a
 bQ where a

0 and b

0 .
A plot of P
 a
 bQ with a

100 and b
 
0 .
5 is shown below.
Example The demand function is given by
(a) Find the slope and intercepts of
0

Q

220
P

P
100


100

0 .
5 Q
0 .
5 Q .
.(b) Plot
(d) Find an expression for the demand function in the form
P
Q


100
 g ( P ) .
0 .
5 Q
(c) What is the quantity demanded when(i) P

5 ? (ii) P

20 ? for
Solution (a) slope=-0.5, intercepts 100 and 200 (b) Plot given below
(c) P

100

0 .
5 Q

5

100

0 .
5 Q

0 .
5 Q

95

Q

190 .
(d)
P
P

100

0 .
5 Q

100

0 .
5 Q


20

0 .
5 Q
100


100
0 .
5 Q

P


Q
0 .
5 Q


100
80
/( 1 /

2 )
Q

160

1 /( 1 /
.
2 ) P

Q

200

2 P

150
100
50
0
0 50 100 150 200
Q

Supply Function The law of supply is a basic law in economics and is given by the linear equation
P
 c
 dQ where c

0 and d

0 and of course P and Q are the price and quantity of a good X .
A plot of P
 c
 dQ where c

10 and d

0 .
5 is given below
Example
120
24
12
0
0
60
48
36
108
96
84
72
12 24 36 48 60
Q
72 84 96 108 120
Example The supply function is given by
Find(a) the slope and intercepts (b) plot P
P


10
10


0 .
0 .
5
5 Q
Q . for 0

Q

120
(c) What is the price of the quantity Q

70
Solution (a) slope= +0.5 and intercepts vertical
Q=70

P

10

0 .
5 ( 70 )

45 .

10 and horizontal
 
20 (c)

Equilibrium in goods
Goods market equilibrium occurs when the quantity demanded Q by consumers and the d quantity supplied Q s by the producers of a good is equal. Equivalently, market equilibrium occurs when the price that a consumer is willing to pay P d is equal to the price that a producer is willing to accept P . The equilibrium condition then is expressed s as Q d

Q s and P d

P s
.
In general this means that we have to solve two simultaneous linear equations
P
 a
 bQ and P
 c
 dQ .
Example The demand and supply functions for a good are given by
P d

100

0 .
5 Q d
-----demand function
P s

10

0 .
5 Q s
-----supply function
Calculate the equilibrium price and quantity graphically. Confirm your answer algebraically.
Solution
Graphical solution is given below.
120
100
80
60
40
20
0
0 36.667
73.333
110
Q
146.667
183.333
Equilibrium point is approximately Q=90,P=55.
Algebraically P d

P s

100

0 .
5 Q

10

0 .
5 Q
OR
P
P


0 .
5 Q
0 .
5 Q


100
10
Subtracting 2 P

110

P

55

0 .
5 Q

100

0 .
5 Q

100

55

45

Q


55
90 .
Q

90
220

Tutorial
Example
Consider the following system of linear equations
2
 x x


3
4 y y


5
3
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
2 x
 x

3 y

4 y


5

1
3

2

2 x


3 y

2 x

8 y
5

6
Adding we get 11 y
Substituting into first equation we get 2 x

3 ( 1 )

5


11

2 x

2 y


1 x

1
Solution is x

1 , y

1 .If this is correct solution second equation should be satisfied. which it is -1+4(1)=3.
Geometric solution

 
3
2 x





5
3




1
4 x





3
4


7
6
5
4
3
2
1

3

2

1 0 1 2 3 4 5 x
To find where straight line we put x

0

3 y

5
 y
2 x

To find where straight line we put y

0

2 x

5
 x
2 x


3 y

5 cuts y-axis( x=0)
5 / 3

3 y


1 .
67
5
. cuts x-axis( y=0)
5 / 2

2 .
5
Draw a straight between the points line 2 x

3 y

5 as shown above.
( 0 , 1 .
67 ) and ( 2 .
5 , 0 ) to give the graph of the straight
To find where straight line we put x

0

4 y

3
 y

 x

3 /
4 y

4

3 cuts y-axis( x=0)
0 .
75 .

To find where straight line we put y

0
  x

3
 x

 x


3
4 y

3 cuts x-axis( y=0)
Draw a straight line between the points straight line
The lines 2 x
 x


3 y
( 0 , 0 .
75 ) and (

3 , 0 )
4 y


5
3 as shown above. and
 x

4 y

3 intersect at the point to give the graph of the x

1 , y

1 which is the solution of the system of equations
2 x

3 y

5
 x

4 y

3 and confirms the algebraic solution.

Example 2
Consider the following system of equations
11 x
13 x


17
5 y y


3
45
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
11 x
13 x


Thus
17
5 y y
 y


2
45

3

11
13

143 x
143 x


221 y
55 y


585
Subtracting gives
33
. Substituting into the first equation we get 11 x
276 y


17 ( 2 )

552
45
.

11 x

34

45

11 x

11
 x
1

1 .Thus solution is
If this is correct solution it should satisfy 13 x

5 y

3 x

1 y

2 . and it does since
13 ( 1 )

5 ( 2 )

13

10

3 .
Geometrical Solution solution of 11x+17y=45,13x-5y=3
5

11
 
17 x

45
17
13
 
5 x

3
5
3
1

1

3

5

5

3

1 1 3 5 x
Solution is where the two lines meet namely x=1,y=2 confirming the algebraic solution.

Example
The demand and supply functions for a good are given by
P d

300

0 .
5 Q d
-----demand function
P s

90

0 .
7 Q s
-----supply function
By plotting these functions on the same graph calculate the equilibrium price
and quantity graphically. Confirm your answer algebraically.
Graph of PD=300-0.6Q,PS=90-0.7Q.Solution P=212.560,Q=175
300
250
200
( 300 0.5Q
)
( 90 0.7Q
)
150
100
50
0
0 50 100 150
Q
200 250
Algebraic solution is 300-0.5Q=90+0.7Q

300

90

1 .
2 Q

Q

210

175
1 .
2
Substituting P=300-0.5(175)=300-87.5=212.5.
Hence solution is P=212.5 and Q=175.
300