TOPICS
Exercise - 3.5

Question-1 :-  Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0
(ii) 2x + y = 5 and 3x + 2y = 8
(iii) 3x – 5y = 20 and 6x – 10y = 40
(iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solution :-
(i) x – 3y – 3 = 0  .......(i)
  3x – 9y – 2 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 1, b₁ = -3, c₁ = -3
  a₂ = 3, b₂ = -9, c₂ = -2
        
  a₁/a₂ = 1/3, b₁/b₂ = -3/(-9), c₁/c₂ = -3/(-2)
  a₁/a₂ = 1/3, b₁/b₂ = 1/3, c₁/c₂ = 3/2
  a₁/a₂ = 1/3, b₁/b₂ = -3/(-9), c₁/c₂ = -3/(-2)

  Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are parallel to each other and its have only no possible solution.
  So, It is inconsitent.
(ii) 2x + y = 5  
  2x + y - 5 = 0  .......(i)

  3x + 2y = 8 
  3x + 2y - 8 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 2, b₁ = 1, c₁ = -5
  a₂ = 3, b₂ = 2, c₂ = -8
        
  a₁/a₂ = 2/3, b₁/b₂ = 1/2, c₁/c₂ = -5/(-8)
  a₁/a₂ = 2/3, b₁/b₂ = 1/2, c₁/c₂ = 5/8

  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersect to each other at a point and its have unique solution.
  So, It is consitent.

  By Cross Multilication Method : 

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(-8 + 10) = y/(-15 + 16) = 1/(4 - 3)
  x/2 = y/1 = 1/1
  Now,
  x/2 = 1/1; y/1 = 1/1
  x = 2; y = 1
(iii) 3x – 5y = 20 
  3x - 5y - 20 = 0  .......(i)
 
  6x – 10y = 40  
  6x - 10y - 40 = 0 ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 3, b₁ = -5, c₁ = -20
  a₂ = 6, b₂ = -10, c₂ = -40
        
  a₁/a₂ = 3/6, b₁/b₂ = -5/(-10), c₁/c₂ = -20/(-40)
  a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 1/2

  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
  Now, these linear equations are coincident and its have many solution.
  So, It is consitent.
(iv) x – 3y – 7 = 0   .......(i)
  3x – 3y – 15 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 1, b₁ = -3, c₁ = -7
  a₂ = 3, b₂ = -3, c₂ = -15
        
  a₁/a₂ = 1/3, b₁/b₂ = -3/(-3), c₁/c₂ = -7/(-15)
  a₁/a₂ = 1/3, b₁/b₂ = 1/1, c₁/c₂ = 7/15

  Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
  Now, these linear equations are intersect at a point and its have only unique solution.
  So, It is consitent.

  By Cross Multilication Method : 

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(45 - 21) = y/(-21 + 15) = 1/(-3 + 9)
  x/24 = y/(-6) = 1/6
  Now,
  x/24 = 1/6; y/(-6) = 1/6
  x = 4; y = -1

Question-2 :-  (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7 and (a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1 and (2k – 1) x + (k – 1) y = 2k + 1

Solution :-
(i) 2x + 3y = 7 
  2x + 3y - 7 = 0  .......(i)

  (a – b) x + (a + b) y = 3a + b – 2  
  (a – b) x + (a + b) y - (3a + b - 2) = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 2, b₁ = 3, c₁ = -7
  a₂ = a - b, b₂ = a + b, c₂ = -(3a + b - 2)
        
  a₁/a₂ = 2/(a - b), b₁/b₂ = 3/(a + b), c₁/c₂ = 7/(3a + b - 2)
  For infinitely many solutions,
  Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂

  Now, a₁/a₂ = c₁/c₂
  2/(a - b) = 7/(3a + b - 2)
  6a + 2b - 4 = 7a - 7b 
  7a - 6a - 7b - 2b = -4
  a - 9b = -4 ......(iii)

  Again, b₁/b₂ = c₁/c₂
  3/(a + b) = 7/(3a + b - 2)
  9a + 3b - 6 = 7a + 7b
  9a - 7a + 3b - 7b = 6
  2a - 4b = 6
  2(a - 2b) = 6
  a - 2b = 3 .......(iv)

  By substracting eq (iii) and  eq (iv)
  a - 9b - a + 2b = -4 - 3
  -7b = -7
    b = 7/7
    b = 1

  Put the value of b = 1 in eq (iv)
  a - 2 x 1 = 3
  a = 3 + 2
  a = 5

  Hence, a = 5 and b = 1 are those values which the given equations give infinitely many solution.
(ii) 3x + y = 1 
  3x + y - 1 = 0  .......(i)

  (2k – 1) x + (k – 1) y = 2k + 1 
  (2k – 1) x + (k - 1) y - (2k + 1) = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 3, b₁ = 1, c₁ = -1
  a₂ = 2k - 1, b₂ = k - 1, c₂ = -(2k + 1)
        
  a₁/a₂ = 3/(2k - 1), b₁/b₂ = 1/(k - 1), c₁/c₂ = 1/(2k + 1)
  For no any solutions,
  Therefore, a₁/a₂ = b₁/b₂ ≠  c₁/c₂

  Now, a₁/a₂ = b₁/b₂
  3/(2k - 1) = 1/(k - 1)
  3k - 3 = 2k - 1
  3k - 2k = -1 + 3
  k = 2

  Hence, k = 2 which the given equations has no solution.

Question-3 :-  Solve the following pair of linear equations by the substitution and cross-multiplication methods :
8x + 5y = 9 and 3x + 2y = 4

Solution :-
a. By using Substitution method :
  8x + 5y = 9  .......(i)
  3x + 2y = 4  ......(ii)

  Let us consider the eq (ii)
  3x + 2y = 4
  3x = 4 - 2y
   x = (4 - 2y)/3 .....(iii)

  Put the value of x = (4 - 2y)/3 in eq (i)
  8(4 - 2y)/3 + 5y = 9
  32 - 16y + 15y = 27
  -y = 27 - 32
  -y = -5
   y = 5 

  Put the value of y = 5 in eq (iii)
  x = (4 - 2 x 5)/3
  x = (4 - 10)/3
  x = -6/3
  x = -2

  x = -2: y = 5

b. By using of Cross Multipication method:
  8x + 5y = 9  
  8x + 5y - 9 = 0  .......(i)

  3x + 2y = 4 
  3x + 2y - 4 = 0  ......(ii)
  
  a₁ = 8, b₁ = 5, c₁ = -9
  a₂ = 3, b₂ = 2, c₂ = -4

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(-20 + 18) = y/(-27 + 32) = 1/(16 - 15)
  x/(-2) = y/5 = 1/1
  Now,
  x/(-2) = 1/1; y/5 = 1/1
  x = -2; y = 5

Question-4 :-  Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution :-
(i) Let x be the fixed charge and y be the per day charge of food.
  According to question :
  
  x + 20y = 1000
  x + 20y – 1000 = 0  .......(i)

  x + 26y = 1180  
  x + 26y - 1180 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 1, b₁ = 20, c₁ = -1000
  a₂ = 1, b₂ = 26, c₂ = -1180
        
  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(-23600 + 26000) = y/(-1000 + 1180) = 1/(26 - 20)
  x/2400 = y/180 = 1/6
  Now,
  x/2400 = 1/6; y/180 = 1/6
  x = 400; y = 30

  Hence, the fixed charge of food is ₹ 400 and per day charge of food ₹ 30.
(ii) Let the fraction is x/y.
  According to question : 
     
  (x - 1)/y = 1/3  
  3x - 3 = y  
  3x - y - 3 = 0  .......(i)

  x/(y + 8) = 1/4 
  4x = y + 8
  4x - y - 8 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 3, b₁ = -1, c₁ = -3
  a₂ = 4, b₂ = -1, c₂ = -8

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(8 - 3) = y/(-12 + 24) = 1/(-3 + 4)
  x/5 = y/12 = 1/1
  Now,
  x/5 = 1/1; y/12 = 1/1
  x = 5; y = 12

  Hence, the fraction is 5/12.
(iii) Let the right answer is x and wrong answer is y.
  According to question : 
    
  3x - y = 40  
  3x - y - 40 = 0  .......(i)
 
  4x - 2y = 50
  2(2x - y) = 50
  2x - y = 25
  2x - y - 25 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 3, b₁ = -1, c₁ = -40
  a₂ = 2, b₂ = -1, c₂ = -25

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(25 - 40) = y/(-80 + 75) = 1/(-3 + 2)
  x/(-15) = y/(-5) = -1/1
  Now,
  x/(-15) = -1/1; y/(-5) = -1/1
  x = 15; y = 5

  Hence, the no. of right answers and wrong answers are 15 and 5 respectively.
  Total no. of questions are 15 + 5 = 20.
(iv) Let the speed of 1st car and 2nd car are x and y repectively.
  According to question : 
     
  5(x - y) = 100  
  x - y = 20  
  x - y - 20 = 0  .......(i)

  x + y = 100 
  x + y - 100 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 1, b₁ = -1, c₁ = -20
  a₂ = 1, b₂ = 1, c₂ = -100

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(100 + 20) = y/(-20 + 100) = 1/(1 + 1)
  x/120 = y/(80) = 1/2
  Now,
  x/120 = 1/2; y/80 = 1/2
  x = 60; y = 40

  Hence, the speed of 1st car is 60 km/h and speed of 2nd car is 40 km/h.
(v) Let the length and breadth of rectangle is x and y respectively.
  Area of rectangle = xy
  According to question : 
     
  (x - 5)(y + 3) = xy - 9  
  xy + 3x - 5y - 15 = xy - 9
  3x - 5y - 15 + 9 = 0   
  3x - 5y - 6 = 0  .......(i)

  (x + 3)(y + 2) = xy + 67 
  xy + 2x + 3y + 6 = xy + 67
  2x + 3y + 6 - 67 = 0 
  2x + 3y - 61 = 0  ......(ii)
  
  By using of Cross Multipication method:
  a₁ = 3, b₁ = -5, c₁ = -6
  a₂ = 2, b₂ = 3, c₂ = -61

  x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
  x/(305 + 18) = y/(-12 + 183) = 1/(9 + 10)
  x/323 = y/171 = 1/19
  Now,
  x/323 = 1/19; y/171 = 1/19
  x = 17; y = 9

  Hence, the length and the breadth of rectangle is 17 units and 9 units.
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