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Example 1
Consider the following system of linear equations
3 x  5 y  13
7 x  11 y  15
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
3x  5 y  13  7
21x  35 y  91
Subtracting gives 68 y  136 .

7 x  11y  15  3
21x  33 y  45
Thus y  2 . Substituting into the first equation we get 3x  5(2)  13
 3x  10  13  3x  3  x  1.Thus solution is x  1 y  2 .
If this is correct solution it should satisfy 7 x  11y  15 and it does since
7(1)  11(2)  7  22  15 ..
Geometric solution
7
  3 x   13 

  
 5   5
6
 7 x   15 
   
 11   11 
3
5
4
2
1
 3 1.667
 0.333 1 2.3333.667 5
x
To find where straight line 3 x  5 y  13 cuts y-axis( x=0)
we put x  0  5 y  13  y  13 / 5  2.6 .
To find where straight line 3 x  5 y  13 cuts x-axis( y=0)
we put y  0  3x  13  x  13 / 3  4.33
Draw a straight between the points (0,2.6) and ( 4.33,0) to give the graph of the straight
line 3 x  5 y  13 as shown above.
To find where straight line 7 x  11y  15 cuts y-axis( x=0)
we put x  0  11y  15  y  15 / 11  1.36 .
To find where straight line 7 x  11y  15 cuts x-axis( y=0)
we put y  0  7 x  15  x  15 / 7  2.14
Draw a straight between the points (0,1.36) and (2.14,0) to give the graph of the straight
line 7 x  11y  15 as shown above.
The lines 3 x  5 y  13 and 7 x  11y  15 intersect at the point x  1, y  2 which is the
solution of the system of equations
3 x  5 y  13
7 x  11 y  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 P  100  0.5Q .
(a) Find the slope and intercepts of P  100  0.5Q .(b) Plot P  100  0.5Q for
0  Q  220 (c) What is the quantity demanded when(i) P  5 ? (ii) P  20 ?
(d) Find an expression for the demand function in the form Q  g (P) .
Solution (a) slope=-0.5, intercepts 100 and 200 (b) Plot given below
(c) P  100  0.5Q  5  100  0.5Q  0.5Q  95  Q  190 .
P  100  0.5Q  20  100  0.5Q  0.5Q  80  Q  160 .
(d) P  100  0.5Q  0.5Q  100  P  Q  100 /(1 / 2)  1 /(1 / 2) P  Q  200  2 P
150
100
100 0.5Q
50
0
0
50
100
Q
150
200
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
108
96
84
72
10 0.5Q 60
48
36
24
12
0
0
12
24
36
48
60
72
84
96
108
120
Q
Example The supply function is given by P  10  0.5Q .
Find(a) the slope and intercepts (b) plot P  10  0.5Q for 0  Q  120
(c) What is the price of the quantity Q  70
Solution (a) slope= +0.5 and intercepts vertical  10 and horizontal  20 (c)
Q=70  P  10  0.5(70)  45 .
Equilibrium in goods
Goods market equilibrium occurs when the quantity demanded Q d by consumers and the
quantity supplied Qs by the producers of a good is equal. Equivalently, market
equilibrium occurs when the price that a consumer is willing to pay Pd is equal to the
price that a producer is willing to accept Ps . The equilibrium condition then is expressed
as Qd  Qs and Pd  Ps .
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
Pd  100  0.5Qd -----demand function
Ps  10  0.5Qs -----supply function
Calculate the equilibrium price and quantity graphically. Confirm your answer
algebraically.
Solution
Graphical solution is given below.
120
100
80
100 0.5Q
10 0.5Q
60
40
20
0
0
36.667
73.333
110
146.667
183.333
Q
Equilibrium point is approximately Q=90,P=55.
Algebraically Pd  Ps  100  0.5Q  10  0.5Q  Q  90
P  0.5Q  100
OR
Subtracting 2P  110  P  55
P  0.5Q  10
55  0.5Q  100  0.5Q  100  55  45  Q  90 .
220
Tutorial
Example
Consider the following system of linear equations
2x  3y  5
 x  4y  3
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
2x  3 y  5 1

2x  3y  5
Adding we get 11y  11  y  1
 x  4y  3 2
 2x  8 y  6
Substituting into first equation we get 2 x  3(1)  5  2 x  2  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
7
  2 x   5 

  
 3   3
6
 1 x   3 
   
 4   4
3
5
4
2
1
 3 2  1 0 1
2 3
4
5
x
To find where straight line 2 x  3 y  5 cuts y-axis( x=0)
we put x  0  3 y  5  y  5 / 3  1.67 .
To find where straight line 2 x  3 y  5 cuts x-axis( y=0)
we put y  0  2 x  5  x  5 / 2  2.5
Draw a straight between the points (0,1.67) and (2.5,0) to give the graph of the straight
line 2 x  3 y  5 as shown above.
To find where straight line  x  4 y  3 cuts y-axis( x=0)
we put x  0  4 y  3  y  3 / 4  0.75 .
To find where straight line  x  4 y  3 cuts x-axis( y=0)
we put y  0   x  3  x  3
Draw a straight line between the points (0,0.75) and (3,0) to give the graph of the
straight line  x  4 y  3 as shown above.
The lines 2 x  3 y  5 and  x  4 y  3 intersect at the point x  1, y  1 which is the
solution of the system of equations
2x  3y  5
 x  4y  3
and confirms the algebraic solution.
Example 2
Consider the following system of equations
11x  17 y  45
13 x  5 y  3
Solve algebraically. Confirm your answer by solving geometrically.
Answer
Algebraic solution
11x  17 y  45  13
143 x  221 y  585
Subtracting gives 276 y  552 .

13x  5 y  3  11
143 x  55 y  33
Thus y  2 . Substituting into the first equation we get 11x  17(2)  45
 11x  34  45  11x  11  x1  1.Thus solution is x  1 y  2 .
If this is correct solution it should satisfy 13 x  5 y  3 and it does since
13(1)  5(2)  13  10  3 .
Geometrical Solution
solution of 11x+17y=45,13x-5y=3
5
  11  x  45 

  
 17   17 
 13  x 3 
   
 5   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
Pd  300  0.5Qd -----demand function
Ps  90  0.7Qs -----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
200
250
Q
Algebraic solution is 300-0.5Q=90+0.7Q  300  90  1.2Q  Q 
Substituting P=300-0.5(175)=300-87.5=212.5.
Hence solution is P=212.5 and Q=175.
210
 175
1.2
300