MATHEMATICS IN EXCEL
LECTURE 2
Graphical representations in Business Analysis.
Part 2. Graphical representation of systems of equations
Use Excel to solve the problems
1 Break-even Analysis
The cost of producing x boxes of chocolates is given by the equation C  2.8x  600 and each box
sells for 4€.
a) Find the break-even point graphically;
b) If it is known that at least 450 units will be sold, show graphically what should be the price charged
for each item to guarantee no loss?
2 Supply-Demand Management
Imagine that we want to find the market equilibrium point for the plasma TV sets. The task is: At a
price (p) of 2400€ the supply (s) of the plasma TV sets is 120 units while their demand (d) is 560
units. If the price is raised to 2700€ per unit the supply and demand will be 160 units and 380 units,
respectively. The following supply and demand equations have been derived on the base of this data:
s
2
p  200,
15
d 
3
p  2000
5
Graph the obtained equations and determine the market equilibrium point graphically. Find the
revenue at the equilibrium point.
We can use Excel to graph these equations and determine the market equilibrium point from the table.
Solve the following problems by traditional methods:
1. (Supply) At a price of 2.5€ per unit, a firm will supply 8000 shirts per month; at 4€ per unit, the
firm will supply 14,000 shirts per month. Determine the supply equation, assuming it to be linear.
2. (Market Equilibrium) The following supply and demand equations have been proposed after some
observations of the process of producing and selling (weekly) cars:
s  2 p  10,
d
8000
 370 ,
p
where p is the price per unit of thousands of Euros and s and d are number of units supplied and
demanded per month.
a) Find the equilibrium point (algebraically or by using Excel).
b) Determine the total revenue received by the manufacturer at the equilibrium point.
Mathematical Tools
Methods for solving systems of linear equations in two variables
Substitution method.
The following steps have to be performed:
1) Solve one of the equations with respect to one of the variables;
2) Substitute this variable by the expression obtained in (1) in the remaining equation;
3) Solve the equation in one variable obtained in (2).
4) Find the value of the remaining variable by back substitution into the expression obtained in 1).
5) Check the solution.
Addition method.
According to this method :
1) Make the coefficients of one of the unknown variables in both equations exactly the same
magnitude and opposite in sign;
1
2) Add the two equations to eliminate one of the variables;
3) Solve the obtained equation in one variable;
4) Substitute the obtained variable in one of the given equations and solve this equation in respect to
the other variable;
5) Check the solution.
Example 1.1. Solve the following system:
4 x  7 y  3
3x  5 y  49
(i)
(ii )
Solution: Applying the substitution method, we solve (i) for x : 4 x  7 y  3
Substituting this value of x into (ii), we obtain
or
x
1
7 y  3
4
3
(7 y  3)  5 y  49 . We multiply both sides by 4
4
and solve for y :
3(7 y  3)  20 y  196  21y  9  20 y  196  41 y  196  9  205  y 
Putting y = 5 in (i) we get x 
205
 5.
41
1
(35  3)  8 .Thus the solution is x = 8 and y = 5 or the ordered pair (8,5).
4
x y 5
(i)
Example 1.2. Solve the system by using the addition method:
3x  2 y  1
(ii )
Solution: It is appropriate to multiply both sides of (i) by 2 and add the obtained equation to (ii):
x  y  5  2 x  2 y  10  (2 x  2 y )  (3x  2 y )  10  1  11  5 x  0. y  11  5 x  11 or x 
After replacing the obtained value of x in (11) we get
11
11 25  11 14
4
 y  5 y  5 

2
5
5
5
5
5
.
The solution is x  2
1
4
, y2
5
5
.
Graph of the Solution Set
To graph the solution set of system 1) we have to graph the solution sets of the two equations. These
are the straight lines l1 : a1 x  b1 y  c1 , l 2 : a 2 x  b2 y  c 2 .
Then the following three cases are possible:
a) l 1 intersects l 2 in one point - the solution set is only one point in the plane;
b) l 1 is parallel to l 2 - the solution set is empty set -  ;
c) l 1 = l 2 , i.e. the two lines coincide - infinitely many points belong to the solution set, namely all the
points on the line.
Exercises
I. Solve the following systems of linear equations:
1.
5.
3x  2 y  1
x y0
40 s  50t  10
100s  200t  30
2.
6.
x y2
x  4y  7
120 p  20q  10
120 p  15q  35
2x  y  2
3.
x  3y  1
7.
2
x  2y  2
2x  4 y  1
4.
2 C  3 R  1
22 C  13Ry  12
8.
s  2t  2
2s  4t  4
11
1
2
5
5