Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.5 pdf || UP Board

July 08, 2021 July 31, 2021

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………………(1)

3x – 9y – 2 = 0………………(2)

Here, a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3 and c1/c2 = -3/-2 = 3/2

a1/a2 = b1/b2 ≠ c1/c2 Hence the pair of equations has no solution.

(ii) 2x + y – 5 = 0…………..(1)

3x + 2y – 8 = 0………………(2)

Here a1/a2 = 2/3, b1/b2 = ½ and c1/c2 = -5/-8

a1/a2 ≠ b1/b2 Therefore the pair of equations has a unique solution.

By cross-multiplication method,

x/((1×-8)-(2×-5))=y/((-5×3)- (-8×2))=1/((2×2)- (3×1))

x/2 = y/1 = 1/1

x/2 = 1/1 and y/1 = 1/1

x = 2 and y = 1

Hence, x = 2, y = 1

(iii) 3x – 5y – 20 = 0…………….(1)

6x – 10y – 40 = 0……………..(2)

Here, a1/a2 = 3/6 = 1/2, b1/b2 = -5/-10 = 1/2 and c1/c2 = -20/-40 =1/2

a1/a2 = b1/b2 = c1/c2 Therefore, the pair of equations has infinitely many solutions.

(iv) x – 3y – 7 = 0…………(1)

3x – 3y – 15 = 0…………….(2)

यहाँ a1/a2 = 1/3, b1/b2 = -3/-3 = 1/1

a1/a2 ≠ b1/b2 Therefore the pair of equations has a unique solution.

By cross-multiplication method,

x/((-3×-15)-(-3×-7))=y/((-7×3)-(-15×1))=1/((1×-3)- (3×-3))

x/24 = y/-6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

x = 24/6 = 4 and y = -6/6 = -1

Hence, x = 4, y = -1

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solution?

2x + 3y = 7

(a – b)x + (a + b)y = 3a + b – 2

Solution:

2x + 3y – 7 = 0…………………..(1)

(a – b)x + (a + b)y – (3a + b – 2) = 0………………(2)

Here, a1/a2 = 2/(a - b), b1/b2 = 3/(a + b) and c1/c2 = -7/-(3a + b – 2) = 7/(3a + b – 2)

A pair of linear equations will have infinitely many solutions, if

a1/a2 = b1/b2 = c1/c2

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

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

2(a + b) = 3(a - b) और 3(3a + b – 2) = 7(a + b)

2a + 2b = 3a – 3b और 9a + 3b – 6 = 7a + 7b

a = 5b……..(3) और 2a = 4b + 6………(4)

Substituting the value of 'a' in equation (4),

2×5b = 4b + 6

10b = 4b + 6

6b = 6

b = 1

Substituting the value of 'b' in equation (3),

a = 5×1 = 5

Hence, a = 5, b = 1

(ii) For which value of k will the following pair of linear equation have no solution?

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

Solution:

3x + y – 1 = 0……………..(1)

(2k – 1)x + (k – 1)y – (2k + 1) = 0…………..(2)

Here, a1/a2 = 3/(2k - 1), b1/b2 = 1/(k - 1) and c1/c2 = -1/-(2k + 1) = 1/(2k + 1)

Pairs of linear equations will have no solution, if

a1/a2 = b1/b2 ≠ c1/c2

3/(2k - 1) = 1/(k - 1) ≠ 1/(2k + 1)

3/(2k - 1) = 1/(k - 1)

3(k - 1) = 1(2k - 1)

3k – 3 = 2k – 1

3k – 2k = 3 – 1

k = 2

Hence, k = 2

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)

By Substitution method,

From equation (1),

y = (9 – 8x)/5 …………..(3)

Substituting the value of 'y' in equation (2),

3x + 2× (9 – 8x)/5 = 4

15x + 18 – 16x = 20

-x = 2

x = -2

Substituting the value of 'x' in equation (1),

8×-2 + 5y = 9

-16 + 5y = 9

5y = 25

y = 25

Hence, x = -2, y = 5

By cross-multiplication method,

x/((5×-4)-(2×-9))=y/((-9×3)-(-4×8))=1/((8×2)- (3×5))

x/-2 = y/5 = 1/1

x/-2 = 1/1 and y/5 = 1/1

x = -2 and y = 5

Hence, x = -2, 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 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.

Solution:

Let fixed value of monthly expenditure = ₹x

and cost of food per day = ₹ y

When a student takes food for 20 days, ₹ 1000 has to be paid for hostel expenses.

Therefore,

x + 26y = 1000

x + 26y – 1000 = 0 …………..(1)

When a student takes food for 26 days, ₹1180 has to be paid for hostel expenses.

Therefore,

x + 26y = 1180

x + 26y – 1180 = 0 …………….(2)

By cross-multiplication method,

x/((20×-1180)-(26×-1000))=y/((-1000×1)-(-1180×1))=1/((1×26)- (1×20))

x/(-23600 + 26000) = y/(-1000 + 1180) = 1/6

x/2400 = y/180 = 1/6

x/2400 = 1/6 and y/180 = 1/6

x = 2400/6 and y = 180/6

x = 400 and y = 30

Hence, the fixed charges is ₹ 400 and the cost of food per day is ₹ 30.

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.

Solution:

Let numerator = x

and denominator = y

so fraction = x/y

According to the first condition,

(x – 1)/y = 1/3

3x – 3 = y

3x – y – 3 = 0 ……………(1)

According to the second condition,

x/(y + 8) = ¼

4x = y + 8

4x – y – 8 = 0 …………….(2)

By cross-multiplication method,

x/((-1×-8)-(-1×-3))=y/((-3×4)-(-8×3))=1/((3×-1)- (4×-1))

x/(8 – 3) = y/(-12 + 24) = 1/(-3 + 4)

x/5 = y/12 = 1/1

x/5 = 1/1 and y/12 = 1/1

x = 5 and y = 12

Hence, fraction = x/y = 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 correct answers = x

and number of wrong answers = y

According to the first condition,

3x – y = 40

3x – y – 40 = 0 ………..(1)

According to the second condition,

4x – 2y = 50

4x – 2y – 50 = 0 ………….(2)

By cross-multiplication method,

x/((-1×-50)-(-2×-40))=y/((-40×4)-(-50×3))=1/((3×-2)- (4×-1))

x/( 50 – 80) = y/(-160 + 150) = 1/(-6 + 4)

x/(-30) = y/(-10) = 1/(-2)

x/(-30) = 1/(-2) and y/(-10) = 1/(-2)

x = 15 and y = 5

Total Number questions = x + y = 15 + 5 = 20

Hence, the total number of questions in the test was 20.

iv. Places A and B are 100 km apart on a highway. One car starts from A and another form 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 tow cars?

Solution:

Let the speed of the car starting from A = x km/h

and speed of car starting from B = y km/h

five hours later,

Distance covered by car starting from A = 5x km

Distance covered by car starting from B = 5y km

therefore,

5x – 5y = 100

x - y – 20 = 0 …………….(1)

If these cars move towards each other, they meet after 1 hour.

1 hour later,

Distance covered by car starting from A = x km

Distance covered by car starting from B = y km

therefore,

x + y = 100

x + y – 100 = 0 …………..(2)

By cross-multiplication method,

x/((-1×-100)-(1×-20))=y/((-20×1)-(-100×1))=1/((1×1)- (1×-1))

x/(100 + 20) = y/(-20 + 100) = 1/2

x/120 = y/80 = 1/2

x/120 = 1/2 and y/80 = 1/2

x = 120/2 and y = 80/2

x = 60 and y = 40

Hence, speed of car starting from A = 60 km/h and speed of car starting from 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 breath 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 the rectangle = x units

and breadth of rectangle = y units

Therefore, area of the rectangle = xy square units

If its length is decreased by 5 units and breadth is increased by 3 units, then area = (x – 5)(y + 3) square units

According to Question,

(x – 5)(y + 3) = xy – 9

xy + 3x – 5y – 15 = xy – 9

3x – 5y – 6 = 0 …………..(1)

If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. so,

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

xy + 2x + 3y + 6 = xy + 67