ECO406 Assignment 2 Solution Fall 2020-VU-Digilearnerspoint

SEMESTER FALL 2020
MATHEMATICAL ECONOMICS(ECO406) 
ASSIGNMENT NO.2 
DUE DATE: 21 DECEMBER, 2020
TOTAL MARKS:15 

#ECO406, ECO406 Assignment 2 Solution Fall 2020-VU-Digilearnerspoint
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ECO406 Assignment 2 Solution Fall 2020-VU-Digilearnerspoint

Question No.1 (Mark5)

A linear function is one in which the power of an independent variable is equal to one. Given below is an example of a linear function showing the relationship between demand and price.

Derive the linear equation of demand function from the following table:

P

550

225

Q

100

150

Question No.2: (Mark:5)

Solve the following equations using the substitution method and find the values of the unknown.

   6x – 2y = 12              

– 8x + 4y = –16

Question No.3: (Mark5)

 Solve the following equation by the factorizing method and find the values of X.

2X2 -X-15=0


Solution#01

The general form of linear demand function or equation is 

Qd= a-bP A

By Plugin in values of given table in A we get

100 = a-b550 1

150 = a-b225 2

By subtracting 1 from 2

  150 = a-b225

-(100=a-b550)

50=b325

b = 50/325=0.15

Put value of b in equation 1

100 = a – 0.15*550

100 = a – 84

a=100+84=184

Thus linear demand function is 

Qd = 184-0.15P


Solution#02


   6x – 2y = 12                                                                                                      A

– 8x + 4y = –16                                                                                                  B

Divide equation A by 2 and B by 4, we get

  3x - y = 6                                                                                                            1

-2x + y = -4                                                                                                          2

By adding 1 and 2

 3x – y = 6

-2x +y = -4

X = 2

To find the value of y, substitute value of  x either in 1, we get

3*2 – y = 6

6 – y = 6

By subtracting 6 from both sides

6-6-y = 6-6

-y = 0

Multiplying both sides by minus 1, we get

-1*-y = 0*-1

Y = 0

Thus solution set is (2,0)

Verification

Put x=2 and y=0 in equation B

-8*2 + 4*0 = -16

-16 = -16

As both sides are same, hence proved


Solution#03

2X2 -X-15=0                                                                         A

2X2 -6X + 5x -15=0

2x(x-3) + 5 (x-3) = 0

(2x + 5)*(x – 3) = 0

2x + 5 = 0                                             x – 3 =0

Subtracting 5 from both side       Adding 3 on both side

2x + 5 -5 = 0 - 5                                  x – 3+3 =0+3

2x = -5                                                   x = 3

Dividing both side by 2

X = -5/2

Thus solution set is (-5/2, 3)

Verification

Put value of x=-5/2 in A

2(-5/2)^2 –(-5/2)-15=0

25/2 +5/2 – 15 = 0

(25+5-30)/2 =0

0/2=0

0=0

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