NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables (Exercise 3.5)

Q.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: 

Cross Multiplication Method

(i) Given, x – 3y – 3 =0 and 3x – 9y - 2 =0
a₁ / a₂ = 1 / 3, b₁ / b₂ = -3 / -9 = 1 / 3, c₁ / c₂ = -3 / -2 = 3 / 2
(a₁ / a₂) = (b₁ / b₂) ≠ (c₁ / c₂)
Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.
(ii) Given, 2x + y = 5 and 3x + 2y = 8
a₁ / a₂ = 2 / 3, b₁ / b₂ = 1 / 2, c₁ / c₂ = -5 / -8
(a₁ / a₂) ≠ (b₁ / b₂)
Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:
x / (b₁c₂ - c₁b₂) = y / (c₁a₂ – c₂a₁) = 1 / (a₁b₂ - a₂b₁)
x / (-8 - (-10)) = y / (15 + 16) = 1 / (4 - 3)
x / 2 = y / 1 = 1
∴ x = 2 and y = 1
(iii) Given, 3x – 5y = 20 and 6x – 10y = 40
(a₁ / a₂) = 3 / 6 = 1 / 2
(b₁ / b₂) = -5 / -10 = 1 / 2
(c₁ / c₂) = 20 / 40 = 1 / 2
a₁ / a₂ = b₁ / b₂ = c₁ / c₂
Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.
(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0
(a₁ / a₂) = 1 / 3
(b₁ / b₂) = -3 / -3 = 1
(c₁ / c₂) = -7 / -15
a₁ / a₂ ≠ b₁ / b₂
Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.
By cross multiplication,
x / (45 - 21) = y / (-21 + 15) = 1 / (-3 + 9)
x / 24 = y / -6 = 1 / 6
x / 24 = 1 / 6 and y / -6 = 1 / 6
∴ x = 4 and y = 1.

Q.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
(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
(2k – 1) x + (k – 1) y = 2k + 1
Solution: (i) 3y + 2x - 7 = 0
(a + b)y + (a-b)y – (3a + b -2) = 0
a₁ / a₂ = 2 / (a - b), b₁ / b₂ = 3 / (a + b), c₁ / c₂ = -7 / -(3a + b - 2)
For infinitely many solutions,
a₁ / a₂ = b₁ / b₂ = c₁ / c₂
Thus 2 / (a - b) = 7 / (3a + b – 2)
6a + 2b – 4 = 7a – 7b
a – 9b = -4  ………………….(i)
2 / (a - b) = 3 / (a + b)
2a + 2b = 3a – 3b
a – 5b = 0 …………………(ii)
Subtracting (i) from (ii), we get
4b = 4
b =1
Substituting this eq. in (ii), we get
a - 5 x 1= 0
a = 5
Thus at a = 5 and b = 1 the given equations will have infinite solutions.
(ii) 3x + y - 1 = 0
(2k - 1)x  +  (k - 1)y – 2k -1 = 0
a₁ / a₂ = 3 / (2k - 1), b₁ / b₂ = 1 / (k - 1), c₁ / c₂ = -1 / (-2k - 1) = 1 / ( 2k + 1)
For no solutions
a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂
3 / (2k - 1) = 1 / (k - 1)   ≠ 1 / (2k + 1)
3 / (2k – 1) = 1 / (k - 1)
3k - 3 = 2k - 1
k = 2
Therefore, for k = 2 the given pair of linear equations will have no solution.

Q.3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution: 8x + 5y = 9 …………………..(1)
3x + 2y = 4 ……………….….(2)
From equation (2) we get
x = (4 – 2y ) / 3  ……………………. (3)
Using this value in equation 1, we get
8(4 - 2y) / 3 + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ………………………(4)
Using this value in equation (2), we get
3x + 10 = 4
x = -2
Thus, x = -2 and y = 5.
Now, Using Cross Multiplication method:
8x + 5y – 9 = 0
3x + 2y – 4 = 0
x / (-20 + 18) = y / (-27 + 32 ) = 1 / (16 - 15)
-x / 2 = y / 5 =1 / 1
∴ x = -2 and y = 5.

Q.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 Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.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 charge of food per day.
According to the question,
x + 20y = 1000……………….. (i)
x + 26y = 1180………………..(ii)
Subtracting (i) from  (ii) we get
6y = 180
y = Rs.30
Using this value in equation (ii) we get
x = 1180 - 26 x 30
x = Rs.400.
Therefore, fixed charges is Rs.400 and charge per day is Rs.30.
(ii) Let the fraction be x / y.
So, as per the question given,
(x - 1) / y = 1 / 3 => 3x – y = 3…………………(1)
x / (y + 8) = 1 / 4  => 4x – y = 8 ………………..(2)
Subtracting equation (1) from (2) , we get
x = 5 ……………………………(3)
Using this value in equation (2), we get,
(4 × 5) – y = 8
y = 12
Therefore, the fraction is 5 / 12.
(iii) Let the number of right answers is x and number of wrong answers be y
According to the given question;
3x − y = 40……..(1)
4x − 2y = 50
⇒ 2x − y = 25…….(2)
Subtracting equation (2) from equation (1), we get;
x = 15 ….….(3)
Putting this in equation (2), we obtain;
30 – y = 25
Or y = 5
Therefore, number of right answers = 15 and number of wrong answers = 5
Hence, total number of questions = 20
(iv) Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.
If the car travels in the same direction,
5x – 5y = 100
x – y = 20 …………………(i)
If the car travels in the opposite direction,
x + y = 100……………(ii)
Solving equation (i) and (ii), we get
x = 60 km/h………………………(iii)
Using this in equation (i), we get,
60 – y = 20
y = 40 km/h
Therefore, the speed of car from point A = 60 km/h
Speed of car from point B = 40 km/h.
(v) Let,
The length of rectangle = x unit
And breadth of the rectangle = y unit
Now, as per the question given,
(x – 5) (y + 3) = xy - 9
3x – 5y – 6 = 0……………………………(1)
(x + 3) (y + 2) = xy + 67
2x + 3y – 61 = 0…………………………..(2)
Using cross multiplication method, we get,
x / (305 + 18) = y / (-12 + 183) = 1 / (9 + 10)
x / 323 = y / 171 = 1 / 19
Therefore, x = 17 and y = 9.
Hence, the length of rectangle = 17 units
And breadth of the rectangle = 9 units