# NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.5)

## Ex 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

Solutions:

(i) x – 3y – 3 =0

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)

∴The given pair of equations has no solution.

(ii) Given, 2x + y = 5

3x +2y = 8

a1/a2 = 2/3

b1/b2 = 1/2

c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

∴ The given pair of equations has unique solution.

By cross multiplication method:

x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 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

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

The given pair of equations has infinite solutions.

(iv) x – 3y – 7 = 0

3x – 3y – 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

a1/a2 ≠ b1/b2

∴ The given pair of equations has 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.

## Ex 3.5 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

(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 equation (i) from equation (ii),

4b = 4

b =1

Substituting the value of b equation in (ii),

a -5 x 1= 0

a = 5

Thus  a = 5 and b = 1 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

∴ The value of k=2 has no solutions

## Ex 3.5 Question 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 ……..(i)

3x + 2y = 4 …….….(ii)

From equation (ii)

x = (4 – 2y )/ 3  ………. (iii)

Using this value of x in equation (i),

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5

Substituting this value of y in equation (ii),

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

By 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.

## Ex 3.5 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 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.

Solutions:

(i) Let x be the fixed charge and y be the charge of food per day.

x + 20y = 1000…….. (i)

x + 26y = 1180……..(ii)

Subtracting equation (i) from equation (ii)

6y = 180

y = Rs.30

Substituting the value of y in equation (i),

x = 1180 -26 x 30

x= Rs.400.

∴ The fixed charges are Rs.400, and the charge per day is Rs.30.

(ii) Let the fraction be x/y.

(x -1)/y = 1/3 => 3x – y = 3………(i)

x/(y + 8) = 1/4  => 4x –y =8 ……..(ii)

Subtracting equation (i) from (ii) , we get

x = 5

Substituting the value of x in equation (i),

(3×5)– y = 3

y= 12

∴The fraction is 5/12.

(iii) Let the number of right answers and wrong answers be x and y respectively.

3x−y=40……..(i)

4x−2y=50

⇒2x−y=25…….(ii)

Subtracting equation (ii) from equation (i)

x = 15

Substituting the value of x in equation (ii)

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 the speed of first car and second car be u km/h and v km/h respectively.

Speed of both cars while they are travelling in the same direction = (u – v) km/h

Speed of both cars while they are travelling in opposite directions i.e. when they are travelling towards each other = (u + v) km/h

Distance travelled = Speed x Time

5(u – v) = 100

⇒ u – v =20 …….(i)

1(u + v) = 100

⇒ (u + v) = 100………(ii)

2u = 120

u = 60

Substituting the value of u in equation (ii),

v = 40

Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h

(v) Let,the length of rectangle = x unit and breadth of the rectangle = y unit

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0…………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0………..(2)

By cross multiplication method,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

∴ x = 17 and y = 9.

Thus, the length of the rectangle = 17 units and the breadth of the rectangle = 9 units.

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