TOPICS
Exercise - 3.4

Question-1 :-  Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x - y/3 = 3

Solution :-
(i) a. By Elimination Method :
  
  x + y = 5 .....(i) 
  2x – 3y = 4 ......(ii)

  For equal of any one variable(x or y), mutiplied by their coefficients following below:
  x + y = 5 ........X 2 
  2x – 3y = 4 ......X 1

  Now,
  2x + 2y = 10 ....(iii)
  2x - 3y = 4 .....(iv)

  By substracting eq (iv) and (iii)
  2x + 2y - 2x + 3y = 10 - 4
  5y = 6
   y = 6/5

  Put the value of y = 6/5 in Equation (i)
  x + 6/5 = 5
  x = 5 - 6/5
  x = 19/5

  x = 19/5, y = 6/5


b. By Substitution Method :

  x + y = 5 .....(i) 
  2x – 3y = 4 ......(ii)

  Let us consider the Equation (i)
  x + y = 5
  x = 5 - y ......(iii)

  Put the value of x = 5 - y in Equation (i), we get
  2(5 - y) - 3y = 4
  10 - 2y - 3y = 4
  -5y = 4 - 10
  -5y = -6
   y = 6/5
  
  Put the value of y = 6/5 in equation (i)
  x + 6/5 = 4
  x = 5 - 6/5
  x = 19/5, y = 6/5
(ii) a. By Elimination Method :
  
  3x + 4y = 10 .....(i) 

  2x – 2y = 2
  2(x - y) = 2 
   x - y = 1 ......(ii)

  For equal of any one variable(x or y), mutiplied by their coefficients following below:
  3x + 4y = 10 ........X 1
  x - y = 1 ...........X 3

  Now,
  3x + 4y = 10 ....(iii)
  3x - 3y = 3 .....(iv)

  By substracting eq (iv) and (iii)
  3x + 4y - 3x + 3y = 10 - 3
  7y = 7
   y = 7/7
   y = 1 

  Put the value of y = 1 in Equation (ii)
  x - 1 = 1
  x = 1 + 1
  x = 2

  x = 2, y = 1


b. By Substitution Method :

  3x + 4y = 10 .....(i) 

  2x – 2y = 2
  2(x - y) = 2 
   x - y = 1 ......(ii)

  Let us consider the Equation (ii)
  x - y = 1
  x = 1 + y ......(iii)

  Put the value of x = 1 + y in Equation (i), we get
  3(1 + y) + 4y = 10
  3 + 3y + 4y = 10
  7y = 10 - 3
  7y = 7
   y = 7/7
   y = 1
  
  Put the value of y = 6/5 in equation (iii)
  x = 1 + 1
  x = 2

  x = 2, y = 1
(iii) a. By Elimination Method :
  
  3x – 5y – 4 = 0 
  3x - 5y = 4 .....(i) 
  
  9x = 2y + 7 
  9x - 2y = 7 .....(ii)

  For equal of any one variable(x or y), mutiplied by their coefficients following below:
  3x - 5y = 4 ........X 9
  9x - 2y = 7 ........X 3

  Now,
  27x - 45y = 36 ....(iii)
  27x - 6y = 21 .....(iv)

  By substracting eq (iv) and (iii)
  27x - 45y - 27x + 6y = 36 - 21
  -39y = 15
   y = 15/(-39)
   y = -5/13

  Put the value of y = -5/13 in Equation (i)
  3x - 5 x (-5/13) = 4
  3x + 25/13 =  4
  3x = 4 - 25/13
  3x = 27/13
   x = 9/13

   x = 9/13, y = -5/13

b. By Substitution Method :

  3x – 5y – 4 = 0 
  3x - 5y = 4 .....(i) 
  
  9x = 2y + 7 
  9x - 2y = 7 .....(ii)

  Let us consider the Equation (i)
  3x - 5y = 4
  x = (4 + 5y)/3 ......(iii)

  Put the value of x = (4 + 5y)/3 in Equation (ii), we get
  9(4 + 5y)/3 - 2y = 7
  3(4 + 5y) - 2y = 7
  12 + 15y - 2y = 7 
  13y = 7 - 12
  13y = -5
  y = -5/13
  
  Put the value of y = -5/13 in equation (iii)
  x = [4 + 5 x (-5/13)]/3
  x = [4 - 25/13]/3
  x = 27/(13 x 3 )
  x = 9/13

  x = 9/13, y = -5/13
(iv) a. By Elimination Method :
  
  x/2 + 2y/3 = -1
  3x + 4y = -6 .....(i)
      
  x - y/3 = 3 
  3x - y = 9 ......(ii)

  By substracting eq (iv) and (iii)
  3x + 4y - 3x + y = -6 - 9
  5y = -15
   y = -15/5
   y = -3

  Put the value of y = -3 in Equation (ii)
  3x - (-3) = 9
  3x = 9 - 3
  x = 6/3
  x = 2

  x = 2, y = -3


b. By Substitution Method :

  x/2 + 2y/3 = -1
  3x + 4y = -6 .....(i)
      
  x - y/3 = 3 
  3x - y = 9 ......(ii)

  Let us consider the Equation (ii)
  3x - y = 9
  3x = 9 + y
  x = (9 + y)/3 ......(iii)

  Put the value of x = (9 + y)/3 in Equation (i), we get
  3(9 + y)/3 + 4y = -6
  9 + y + 4y = -6
  5y = -6 - 9
  5y = -15
   y = -15/5
   y = -3
  
  Put the value of y = -3 in equation (iii)
  x = [9 + (-3)]/3
  x = 6/3
  x = 2
   
  x = 2, y = -3
    

Question-2 :-  Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution :-
(i) Let the fraction be x/y.
  According to question : 

  (x + 1)/(y - 1) = 1
   x + 1 = y - 1
   x - y = -2 ....(i)

  x/(y + 1) = 1/2 
  2x = y + 1
  2x - y = 1 ....(ii)

  By substracting eq (i) and (ii)
  x - y - 2x + y = -2 - 1
  -x = -3
   x = 3

  Put the value of x = 3 in Equation (i)
  3 - y = -2
  -y = -2 - 3
  -y = -5
   y = 5

  x = 3, y = 5
  Hence, The fraction is 3/5.
(ii) Let the present age of Nuri and Sonu are be the x and y respectively.
  According to Question : 

  (x - 5) = 3(y - 5)
  x - 5 = 3y - 15
  x - 3y = -15 + 5 
  x - 3y = -10 .....(i)

  (x + 10) = 2(y + 10)
  x + 10 = 2y + 20
  x - 2y = 20 - 10
  x - 2y = 10 .....(ii)

  By substracting eq (i) and (ii)
  x - 3y - x + 2y = -10 - 10
  -y = -20
   y = 20

  Put the value of y = 20 in Equation (i)
  x - 3 x 20 = -10
  x = -10 + 60
  x = 50

  x = 50, y = 20
  Hence, the Present age of Nuri and Sonu are 50 and 20 years respectively.
(iii) Let the unit digitand tens digits of the numberbe x and y respectively. Then, The number = 10y + x
  Reversing digit = 10x + y
  According to Question : 

  x + y = 9 ....(i)
  
  9(10y + x) = 2(10x + y)
  90y + 9x = 20x + 2y
  9x - 20x + 90y - 2y = 0
  -11x + 88y = 0 
  11(-x + 8y) = 0 
  -x + 8y = 0 .....(ii)


  By adding eq (i) and (ii)
  x + y - x + 8y = 9 + 0
  9y = 9
   y = 9/9
   y = 1

  Put the value of y = 1 in Equation (i)
  x + 1 = 9
  x = 9 - 1
  x = 8

  x = 8, y = 1
  Hence, the tens number digit = 10 x 1 + 8 = 18 and 
  Reverse digit = 10 x 8 + 1 = 81.
(iv) Let the number of ₹ 50 Notes x and number of ₹ 100 Notes y.
  According to question : 
  x + y = 25 ....(i)
  
  50x + 100y = 2000
  50(x + 2y) = 2000
  x + 2y = 40 ....(ii)
  
  By substracting eq (i) and (ii)
  x + y - x - 2y = 25 - 40
  -y = -15
   y = 15

  Put the value of y = 15 in Equation (i)
  x + 15 = 25
  x = 25 - 15
  x = 10

  x = 10, y = 15
  Hence, Meena has 10 Notes of ₹ 50 and 15 Notes of ₹ 100.
(v) Let the fixed charge is x and each day charge is y.
  According to question :
    
  x + 4y = 27 .....(i)
  x + 2y = 21 .....(ii) 

  By Substracting eq (i) and (ii)
  x + 4y - x - 2y = 27 - 21
  2y = 6
   y = 6/2
   y = 3

  Put the value of y = 3 in eq (i)
  x + 4 x 3 = 27
  x = 27 - 12
  x = 15

  x = 15, y = 3
  Hence, Fixed charge = ₹ 15 and each day charge = ₹ 3
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