TOPICS
Exercise - 3.3

Question-1 :-  Solve the following pair of linear equations by the substitution method.
(i) x + y = 14, x – y = 4
(ii) s – t = 3, s/3 + t/2 = 6
(iii) 3x – y = 3, 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
(v) √2x + √3y = 0, √3x - √8y = 0
(vi) 3x/2 - 5y/3 = -2, x/3 + y/2 = 13/6

Solution :-
(i) x + y = 14 .....(i) 
  x – y = 4 ......(ii)

  Let us consider the Equation (ii)
  x – y = 4
  x = 4 + y ......(iii)

  Put the value of x = 4 + y in Equation (i), we get
  4 + y + y = 14
  2y = 14 - 4
  2y = 10
   y = 10/2
   y = 5
  
  Put the value of y = 5 in equation (i)
  x + 5 = 14
  x = 14 - 5
  x = 9, y = 5
(ii) s – t = 3 .....(i) 

  s/3 + t/2 = 6 
  (2s + 3t)/6 = 6
  2s + 3t = 36 ......(ii)
 
  Let us consider the Equation (i)
  s – t = 3 
  s = t + 3 ......(iii)

  Put the value of s = t + 3 in Equation (ii), we get
  2(t + 3) + 3t = 36
  2t + 6 + 3t = 36
  5t = 36 - 6
  5t = 30
   t = 30/5
   t = 6
  
  Put the value of t = 6 in equation (iii)
  s = 6 + 3
  s = 9, t = 6
  x = 14 - 5
  x = 9, y = 5
(iii) 3x – y = 3 .....(i) 
  9x – 3y = 9 ......(ii)

  Let us consider the Equation (i)
  3x – y = 3
   y = 3x - 3 ......(iii)

  Put the value of y = 3x - 3 in Equation (ii), we get
  9x – 3(3x - 3) = 9
  9x - 9x + 9 = 9
   9 = 9
  This is always true.
  Therefore, possible solution is x = 1, y = 0.
(iv) 0.2x + 0.3y = 1.3  
  2x + 3y = 13 .....(i)
      
  0.4x + 0.5y = 2.3 
  4x + 5y = 23  ......(ii)

  Let us consider the Equation (ii)
  4x + 5y = 23
  x = (23 - 5y)/4 ......(iii)

  Put the value of x = (23 - 5y)/4  in Equation (i), we get
  2(23 - 5y)/4 + 3y = 13
  (23 - 5y)/2 + 3y = 13
  23 - 5y + 6y = 26
  y = 26 - 23
  y = 3
  
  Put the value of y = 3 in equation (iii)
  x = (23 - 5 x 3)/4
  x = (23 - 15)/4
  x = 8/4
  x = 2

  x = 2, y = 3
(v) √2x + √3y = 0  .....(i) 
  √3x - √8y = 0 ......(ii)

  Let us consider the Equation (ii)
  √3x - √8y = 0
  x = √8y/√3 ......(iii)

  Put the value of x = √8y/√3 in Equation (i), we get
  √2 x √8y/√3 + √3y = 0
  √16y + 3y = 0
  y(√16 + 3) = 0
  y = 0 , x = 0
(vi) 3x/2 - 5y/3 = -2  .....(i) 
  x/3 + y/2 = 13/6 ......(ii)

  Let us consider the Equation (ii)
  x/3 + y/2 = 13/6
  x/3 = 13/6 - y/2
  x = 3(13 - 3y)/6 
  x = (13 - 3y)/2  ......(iii)

  Put the value of x = (13 - 3y)/2 in Equation (i), we get
  4 + y + y = 14
  2y = 14 - 4
  2y = 10
   y = 10/2
   y = 5
  
  Put the value of y = 5 in equation (i)
  x + 5 = 14
  x = 14 - 5
  x = 9, y = 5
    

Question-2 :-  Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution :-
  2x + 3y = 11  .......(i)

  2x – 4y = – 24
   x - 2y = -12  ......(ii)
  
  Let us consider the Equation (ii)
  x - 2y = -12
  x = -12 + 2y ......(iii)

  Put the value of x = -12 + 2y in eq (i)
  2(-12 + 2y) + 3y = 11
  -24 + 4y + 3y = 11
  7y = 11 + 24
  7y = 35
   y = 35/7
   y = 5

  Put the value of y = 5 in eq (iii)
  x = -12 + 2 x 5
  x = -12 + 10
  x = -2

  x = -2, y = 5

  Put the value of x and y  in y = mx + 3
  5 = -2m + 3
  -2m = 5 - 3
  -2m = 2
    m = -1
    

Question-3 :-  Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution :-
(i) Let x and y arebe the two numbers.
  According to question :

  x - y = 26  .......(i)
  x = 3y .........(ii)

  Put the value of x = 3y in eq (i)
  3y - y = 26
  2y = 26
  y = 26/2
  y = 13
 
  Put the value of y = 13 in eq (ii)
  x = 3 x 13
  x = 39

  x = 13, y = 39
  Hence, the numbers are 13 and 39.
(ii) Let the largest angle x and smallest angle be the y.
  We know that the sum of supplementary agles is 180°.
 
  x + y = 180° ......(i)
  x - y = 18° ......(ii)

  Let us consider eq (ii)
  x - y = 18°
  x = 18° + y  ......(iii)
  
  Put the value of x = 18° + y in equation (i)
  18° + y + y = 180°
  2y = 180° - 18°
  2y = 162°
   y = 81°

  Put the value of y = 81° in eq (iii)
  x = 18° + 81°
  x = 99°

  x = 99°, y = 81°
  Hence, largest value is 99° and smallest value is 81°.
(iii) Let cost of one bat and one ball are be the x and y respectively.
  According to question :
        
  7x + 6y = 3800 ......(i)
  3x + 5y = 1750 ......(ii) 

  Let us consider of eq (ii)
  3x + 5y = 1750
  3x = 1750 - 5y
   x = (1750 - 5y)/3 ......(iii)

  Put the value of x = (1750 - 5y)/3 in eq (i)
  7(1750 - 5y)/3 + 6y = 3800
  12250 - 35y + 18y = 3800 x 3
  -17y = 11400 - 12250
  -17y = -850
     y = 50

  Put the value of y = 50 in eq (iii)
  x = (1750 - 5 x 50)/3
  x = (1750 - 250)/3
  x = 1500/3
  x = 500

  x = 500, y = 50
  Hence, the cost of bat is ₹ 500 and cost of ball is ₹ 50. 
(iv) Let the fixed charge be the x and per km charge be the y.
  According to question :
  
  x + 10y = 105 ........(i)
  x + 15y = 155 ........(ii)
        
  Let us consider the eq (ii)
  x + 15y = 155
  x = 155 - 15y ......(iii)
        
  Put the value of x = 155 - 15y in eq (i)
  155 - 15y + 10y = 105
  -5y = 105 - 155
  -5y = -50
    y = 10

  Put the value of y = 10 in eq (iii)
  x = 155 - 15 x 10
  x = 155 - 150
  x = 5

  x = 5, y = 10
  Hence, Fixed charge is ₹ 5 and per km charge is ₹ 10.
  Charge for 25 km = x + 25y = 5 + 25 x 10 = 5 + 250 = ₹ 255
  
       
(v) Let the fraction is x/y.
  According to question :
        
  (x + 2)/(y + 2) = 9/11
  11x + 22 = 9y + 18
  11x - 9y = 18 - 22
  11x - 9y = -4 ......(i)
        
  (x + 3)/(y + 3) = 5/6
  6x + 18 = 5y + 15
  6x - 5y = 15 - 18
  6x - 5y = -3 .......(ii) 

  Let us consider the eq (ii)
  6x - 5y = -3
  6x = -3 + 5y
   x = (-3 + 5y)/6 ......(iii)
 
  Put the value of x = (-3 + 5y)/6 in eq (i)
  11(-3 + 5y)/6 - 9y = -4
  -33 + 55y - 54y = -24
  y = -24 + 33
  y = 9

  Put the value of y = 9  in eq (iii)
  x = (-3 + 5 x 9)/6
  x = (-3 + 45)/6
  x = 42/6  
  x = 7

  x = 7, y = 9
  Hence, the fraction is 7/9.
(vi) Let the age of Jacob is x and age of his son is y.
  According to question :

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

  (x - 5) = 7(y - 5)
  x - 5 = 7y - 35
  x - 7y = -35 + 5
  x - 7y = -30 .......(ii)

  Let us consider the eq (i)
  x - 3y = 10
  x = 10 + 3y .....(iii)

  Put the value of x = 10 + 3y in eq (ii)
  10 + 3y - 7y = -30
  -4y = -30 - 10
  -4y = -40
    y = 10

  Put the value of y = 10 in eq (iii)
  x = 10 + 3 x 10
  x = 10 + 30
  x = 40

  x = 40, y = 10
  Hence, the age og Jacob is 40 years and the age of his son is 10 years.
     
CLASSES

Connect with us:

Copyright © 2015-16 by a1classes.

www.000webhost.com