TOPICS

Exercise - 3.5

Pair of Linear Equations in two Variables

**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

(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

(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

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.

(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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