NCERT Solutions Class 10th Maths Chapter – 3 Pair of Linear Equations in Two Variables Exercise – 3.5

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

Solution: 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) 2x + y = 5 and 3x + 2y = 8

Solution: 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) 3x – 5y = 20 and 6x – 10y = 40

Solution: 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) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solution: 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.

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

Solution: 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) 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: 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.

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.

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.

Solution: 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) 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.

Solution: 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) 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?

Solution: 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) 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?

Solution: 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) 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: 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……………………………(i)
(x + 3) (y + 2) = xy + 67
2x + 3y – 61 = 0…………………………..(ii)
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