# Linear equations in two variables – Exercise 3.5 – Class 10

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

3x  –  9y – 2 = 0

(ii) 2x + y = 5

3x + 2y = 8

(iii) 3x – 5y  = 20

6x – 10y = 40

(iv) x –  3y – 7 = 0

3x – 3y – 15 = 0

Solution:

(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

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5

3x + 2y = 8

a1/a2 = 2/3 ; b1/b2 = 1/2 ; c1/c2 = -5/-8

a1/a2 ≠ b1/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication method,

x/(b1c2-b2c1) = y/(c1a2-c2a1) = 1/(a1b2-a2b1)

x/[-8-(-10)] = y/[-15+16] = 1/(4-3)

x/2 = y/1 = 1

x/2 = 1 ;  y/1 = 1

x = 2 ; y = 1

Therefore, x = 2, y = 1

(iii) 3x – 5y  = 20

6x – 10y = 40

a1/a2 = 3/6 = ½ ; b1/b2 = -5/-10 = 1/2 ; c1/c2 = -20/-40 = 1/2

a1/a2 = b1/b2 = c1/c2

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x –  3y – 7 = 0

3x – 3y – 15 = 0

a1/a2 = 1/3  ; b1/b2 = -3/-3 = 1 ; c1/c2 = -7/-15 = 7/15

a1/a2 ≠ b1/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

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

Thus, x = 4 and y = -1

2.

(i) For which values of a and b will 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) 2x + 3y  -7 = 0

(a-b)x + (a+b)y – (3x +b -2) = 0

a1/a2 = 2/(a-b) ; b1/b2 = 3/(a+b) ; c1/c2 = -7/-(3a+b-2) = 7/(3a+b-2) ;

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

2/(a-b) = 7/(3a+b-2)

6a+2b-4=7a-7b

a-9b = -4 ————(1)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ————(2)

Subtracting (1) from (2), we obtain

4b = 4

b = 1

Substituting this in equation(2), we obtain

a – 5×1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many 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 solution;

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

Hence, for k = 2, the given equation has no solution

1. 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 (ii), we get

x = (4-2y)/3 ————(3)

Substituting this value in equation (1), we get

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

32 – 16y + 15y = 27

-y = -5

y = 5

Substituting the value of y in equation (2)

3x + 10 = 4

x = -2

Hence x = -2 , y = 5

Again by cross multiplication method, we get

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 = 1 and y/5 = 1

x = -2 and y = 5

1. 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 ¼ 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 of the food and y be the charge for food per day.

Given

x + 20y = 1000 ————-(1)

x + 26 y = 1180 ———–(2)

Subtracting equation (1) from equation (2), we get

6y = 180

y = 30

Substituting this value in equation (1),

x + 20 x 30 = 1000

x = 1000 – 600

x = 400

Hence, fixed charge = Rs 400

And charge per day = Rs 30

(ii)Let the fraction be x/y.

Given

x-1/y = 1/3 i.e., 3x – y = 3 ———–(1)

x/(y+8) = ¼ i.e., 4x – y = 8 ————(2)

Subtracting equation (1) from equation (2),

x = 5 ———(3)

Putting this value in equation (1), we obtain

15 – y = 3

y = 12

Hence the fraction is 5/12.

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

Given,

3x – y = 40 ————-(1)

4x – 2y = 50

i.e., 2x – y = 25 ————-(2)

Subtracting equation (2) from equation (1),

x = 15 ———–(3)

Substituting the value of x in equation (2),

30 – y = 25

y = 5

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions = 20

(iv)Let the speed of 1st car and 2nd car be u km/h and v km/h.

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

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

Given

5 ( u – v) = 100

i.e., u – v = 20 ————-(1)

1(u+v) = 100 ————-(2)

Adding both the equations,

2u = 120

u = 60 km/h ———(3)

Substituting the value of u in equation (2),

v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

Given,

(x – 5)(y+3) = xy – 9

i.e., 3x – 5y – 6 = 0 ———-(1)

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

i.e., 2x + 3y – 61 = 0 ———-(2)

By cross-multiplication method, we obtain

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

x/323 = y/171 = 1/19

x = 17, y = 9

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

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