Class 9 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 5 - Linear Equations in Two Variables

x + y = 5

2x - y = 10

3x + 7y = 9

-8x + 5y = 0

-13x - 22y = 104

Given: x + y = 7

3 + 4 = 7

0 + 7 = 7

7 + 0 = 7

1 + 6 = 7

2 + 5 = 7

Thus, the five solutions are (3, 4), (0, 7), (7, 0), (1, 6), (2, 5).

(i)

x + y = 4 … (I)

2x - 5y = 1 … (II)

Writing x in terms of y from equation (I)

∴ x = 4 - y

Substituting this value of x in equation (II)

∴ 2(4 - y) - 5y = 1

∴ 8 - 2y - 5y = 1

∴ 8 - 7y = 1

∴ 8 - 1 = 7y

∴ 7y = 7

∴ y = 1

Now, substituting y = 1 in equation (I)

∴ x + 1 = 4

∴ x = 4 - 1

∴ x = 3

∴ (3, 1) is the solution of the given equations.

(ii)

2x + y = 5 … (I)

3x - y = 5 … (II)

Adding equations (I) and (II)

2x+y =5

3x-y=5

∴ x= 2

Now, substituting x = 2 in equation (I)

∴ 2×2 + y = 5

∴ 4 + y = 5

∴ y = 1

∴ (2, 1) is the solution of the given equations.

(iii)

3x - 5y = 16 … (I)

x - 3y = 8 … (II)

Writing x in terms of y from equation (II)

∴ x = 8 + 3y

Substituting this value of x in equation (I)

∴ 3(8 + 3y) - 5y = 16

∴ 24 + 9y - 5y = 16

∴ 4y + 24 = 16

∴ 4y = -8

∴ y = -2

Now, substituting y = 2 in equation (II)

∴ x - 3 × (-2) = 8

∴ x + 6 = 8

∴ x = 2

∴ (2, -2) is the solution of the given equations.

(iv)

2y - x = 0 … (I)

10x + 15y = 105 … (II)

Writing x in terms of y from equation (I)

∴ x = 2y

Substituting this value of x in equation (II)

∴ 10(2y) + 15y = 105

∴ 20y + 15y = 105

∴ 35y = 105

∴ y = 3

Now, substituting y = 3 in equation (II)

∴ 2 × 3 - x = 0

∴ 6 - x = 0

∴ x = 6

∴ (6, 3) is the solution of the given equations.

(v)

2x + 3y + 4 = 0 … (I)

x - 5y = 11 … (II)

Writing x in terms of y from equation (II)

∴ x = 5y + 11

Substituting this value of x in equation (I)

∴ 2(5y + 11) + 3y + 4 = 0

∴ 10y + 22 + 3y + 4 = 0

∴ 13y + 26 = 0

∴ 13y = -26

∴ y = -2

Now, substituting y = -2 in equation (II)

∴ x - 5 × (-2) = 11

∴ x + 10 = 11

∴ x = 1

∴ (1, -2) is the solution of the given equations. 

(vi)

2x - 7y = 7 … (I)

3x + y = 22 … (II)

Writing y in terms of x from equation (II)

∴ y = 22 - 3x

Substituting this value of y in equation (I)

∴ 2x - 7(22 - 3x) = 7

∴ 2x - 154 + 21x = 7

∴ 23x = 161

∴ x = 7

Now, substituting x = 7 in equation (II)

∴ 3 × 7 + y = 22

∴ 21 + y = 22

∴ y = 1

∴ (7, 1) is the solution of the given equations.

Let the number of 5 rupee notes be x, number of 10 rupee notes be y.

By first condition, we have

5x + 10y = 350

Dividing both sides of the above equation by 5

∴ x + 2y = 70 … (I)

By the second condition, we have

x = 2y - 10 … (II)

Substituting this value of x in equation (I)

∴ 2y - 10 + 2y = 70

∴ 4y = 70 + 10

∴ 4y = 80

∴ y = 20

Substituting y 20 in equation (II)

∴ x = 2×20 - 10

∴ x = 40 - 10

∴ x = 30

Thus, the number of 5 rupee notes is 30 and number of 10 rupee notes is 20.

Let 'x' be the numerator and 'y' be the denominator of a fraction.

So, the required fraction is  

By the first condition, we have

y = 2x - 1 … (I)

By the second condition, we have

Substituting value of y from equation (I) in equation (II)

∴ 5x - 3(2x - 1) = -2

∴ 5x - 6x + 3 = -2

∴ -x = -2 - 3

∴ -x = -5

∴ x = 5

Substituting x = 1 in equation (I)

∴ y = 2 × 5 - 1

∴ y = 10 - 1

∴ y = 9

∴ The required fraction is  

Let Priyanka's age be 'x' years and Deepika's age be 'y' years.

By the first condition, we have

x + y = 34 … (I)

By the second condition, we have

x = y + 6

∴ x - y = 6 … (II)

Adding equations (I) and (II)

x+ y=34

x - y=6

∴ x = 20

Substituting x = 20 in equation (I), we get

20 + y = 34

∴ y = 34 - 20

∴ y = 14

Hence, Priyanka's present age is 20 years and Deepika's present age is 14 years.

Let the number of lions be 'x' and number of peacocks be 'y'.

By the first condition, we have

x + y = 50 … (I)

A lion has four legs and a peacock has 2 legs.

∴ By the second condition, we have

4x + 2y = 140 … (II)

Multiplying equation (I) by 2, we get

2x + 2y = 100 … (III)

Subtracting equation (III) from equation (II)

4x+2y=140

2x + 2y =100

∴ x = 20

Substituting x = 20 in equation (I), we get

20 + y = 50

∴ y = 50 - 20

∴ y = 30

Hence, the number of lions and peacocks in the zoo are 20 and 30 respectively.

Let Sanjay's original salary be Rs 'x' and his yearly increment be Rs 'y'.

By the first condition, we have

x + 4y = 4500 … (I)

After 10 years, his monthly salary became 5400 rupees. 

∴ By the second condition, we have

x + 10y = 5400 … (II)

Subtracting equation (I) from equation (II)

X+10y = 5400

X+4y = 4500

∴ y = 150

Substituting y = 150 in equation (I), we get

x + 4 × 150 = 4500

∴ x + 600 = 4500

∴ x = 3900

Hence, Sanjay's original salary is Rs 3900 and his yearly increment is Rs 150.

Let the price of one chair be Rs 'x' and price of one table be Rs 'y'.

By the first condition, we have

3x + 2y = 4500 … (I)

By the second condition, we have

5x + 3y = 7000 … (II)

Expressing y in terms of x from equation (I)

∴ 2y = 4500 - 3x

Substituting this value of y in equation (II), we get

Multiplying both sides of the above equation by 2

∴ 10x + 3(4500 - 3x) = 14000

∴ 10x + 13500 - 9x = 14000

∴ x = 14000 - 13500

∴ x = 500

Substituting x = 500 in equation (I), we get

3 × 500 + 2y = 4500

∴ 1500 + 2y = 4500

∴ 2y = 3000

∴ y = 1500

Now, 2x + 2y = 2×500 + 2×1500

 = 1000 + 3000

 = 4000

Hence, the total price of 2 chairs and 2 tables is Rs. 4000.

Let the digit at ones place be 'x' and digit at tens place be 'y'.

So, the number will be 10y + x.

The number obtained after interchanging the digits is 10x + y.

By the first condition, we have

x + y = 9 … (I)

By the second condition, we have

10x + y = 10y + x + 27

∴ 10x - x + y - 10y = 27

∴ 9x - 9y = 27

Dividing both sides of the above equation by 3, we get

x - y = 3 … (II)

Adding equations (I) and (II), we have

Substituting x = 6 in equation (I), we get

6 + y = 9

∴ y = 9 - 6

∴ y = 3

∴ 10y + x = 10×3 + 6 = 36

∴ The required number is 36.

Let the measure of ∠B be 'x°' and measure of ∠C be 'y°'.

By the first condition, we have

m∠A = m∠B + m∠C

∴ m∠A = x° + y°

Since, the sum of the measures of the angles of a triangle is 180°.

In ∆ABC,

m∠A + m∠B + m∠C = 180° 

∴ x + y + x + y = 180

∴ 2x + 2y = 180

Dividing both sides by 2, we get

x + y = 90 … (I)

By the second condition, the ratio of the measures of ∠B and ∠C is 4 : 5.

∴ 5x =4y

∴ 5x - 4y = 0 … (II)

Multiplying equation (I) by 4, we get

4x + 4y = 360 … (III)

Adding equations (II) and (III),

5x-4y= 0

4x+4y=360

∴ x = 40

Substituting x = 40 in equation (I)

∴ 40 + y = 90

∴ y = 90 - 40

∴ y = 50

∴ m∠A = x° + y° = 40° + 50° = 90° 

∴ The measures of ∠A, ∠B and ∠C are 90°, 40°, and 50° respectively.

Let the length of the smaller part of the rope be 'x' cm and that of the larger part be 'y' cm.

By the first condition, we have

x + y = 560 … (I)

Now, twice the length of the smaller part = 2x

Also, of the length of the larger part =  

By the second condition, we have

∴ 6x = y

∴ 6x - y = 0 … (II)

Adding equations (I) and (II)

X+y=560

6x-y=0

∴ x= 80 

Substituting x = 80 in equation (II)

∴ 6×80 - y = 0

∴ 480 - y = 0

∴ y = 480

∴ The length of the larger part of the rope is 480 cm.

Let Yashwant got 'x' questions right and 'y' questions wrong.

By the first condition, total number of questions in the examination are 60.

∴ x + y = 60 … (I)

Since, Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer.

∴ Total marks he got = 2x - y

By the second condition, we have

2x - y = 90 … (II)

Adding equations (i) and (ii)

X+y=60

2x-y=90

∴ =50 

Substituting x = 50 in equation (I)

∴ 50 + y = 60

∴ y = 60 - 50

∴ y = 10

Thus, Yashwant got 10 questions wrong.

(A)

3x + 5y = 9 … (I)

5x + 3y = 7 … (II)

Adding equations (I) and (II)

3x+5y = 9

Dividing both sides of the above equation by 8, we get

x + y = 2 

(C)

Let x and y be the length and breadth of the rectangle respectively.

Now, perimeter of the rectangle = 2(length + breadth)

 = 2(x + y)

By the condition given in the question, we have

2[(x - 5) + (y - 5)] = 26

∴ 2(x - 5) + 2(y - 5) = 26

∴ 2x - 10 + 2y - 10 = 26

∴ 2x + 2y = 26 + 10 + 10

∴ 2x + 2y = 46

Dividing both the sides of the above equation by 2

∴ x + y = 23

(C)

Let Ajay's age be 'x' years and Vijay's age be 'y' years.

By the first condition, we have

y = x + 5 … (I)

By the second condition, we have

x + y = 25 … (II)

Substituting the value of y from (I) in (II)

∴ x + x + 5 = 25

∴ 2x = 25 - 5

∴ 2x = 20

∴ x = 10

Substituting this value of x in (I), we get

y = 15

Hence, Ajay's age is 10 years and Vijay's age is 15 years.

(i)

2x + y = 5 … (I)

3x - y = 5 … (II)

Adding equation (I) and (II)

2x+y=5

∴ x = 2

Substituting x = 2 in equation (I)

∴ 2×2 + y = 5

∴ 4 + y = 5

∴ y = 1

∴ (2, 1) is the solution of the given linear equations.

(ii)

x - 2y = -1

∴ x = -1 + 2y … (I)

2x - y = 7 … (II)

Substituting the value of x from (I) in equation (II)

∴ 2(-1 + 2y) - y = 7

∴ -2 + 4y - y = 7

∴ 3y = 7 + 2

∴ 3y = 9

∴ y = 3

Substituting y = 3 in (I)

∴ x = -1 + 2×3

∴ x = -1 + 6

∴ x = 5

∴ (5, 3) is the solution of the given linear equations. 

(iii)

x + y = 11

∴ x = 11 - y … (I)

2x - 3y = 7 … (II)

Substituting the value of x from (I) in equation (II)

∴ 2(11 - y) - 3y = 7

∴ 22 - 2y - 3y = 7

∴ -5y = 7 - 22

∴ -5y = -15

∴ y = 3

Substituting y = 3 in (I)

∴ x = 11 - 3

∴ x = 8

∴ (8, 3) is the solution of the given linear equations. 

(iv)

2x + y = -2

∴ y = -2 - 2x … (I)

3x - y = 7 … (II)

Substituting the value of y from (I) in equation (II)

∴ 3x - (-2 - 2x) = 7

∴ 3x + 2 + 2x = 7

∴ 5x = 7 - 2

∴ 5x = 5

∴ x = 1

Substituting x = 1 in (I)

∴ y = -2 - 2×1

∴ y = -4

∴ (1, -4) is the solution of the given linear equations. 

(v)

2x - y = 5

∴ y = 2x - 5 … (I)

3x + 2y = 11 … (II)

Substituting the value of y from (I) in equation (II)

∴ 3x + 2(2x - 5) = 11

∴ 3x + 4x - 10 = 11

∴ 7x = 21

∴ x = 3

Substituting x = 3 in (I)

∴ y = 2×3 - 5

∴ y = 1

∴ (3, 1) is the solution of the given linear equations. 

(vi)

x - 2y = -2

∴ x = -2 + 2y … (I)

x + 2y = 10 … (II) 

Substituting the value of x from (I) in equation (II)

∴ -2 + 2y + 2y = 10

∴ 4y = 12

∴ y = 3

Substituting y = 3 in (I)

∴ x = -2 + 2×3

∴ x = 4

∴ (4, 3) is the solution of the given linear equations 

(i)

3x - 4y = 7 … (I)

5x + 2y = 3 … (II)

Multiplying equation (II) by 2, we get

10x + 4y = 6 …(III)

Adding equations (I) and (III),

∴ x = 1 

Substituting x = 1 in equation (I)

∴ 3×1 - 4y = 7

∴ -4y = 7 - 3

∴ -4y = 4

∴ y = -1

∴ (1, -1) is the solution of the given equations.

(ii)

5x + 7y = 17 … (I)

3x - 2y = 4 … (II)

Multiplying equation (I) by 2, we get

10x + 14y = 34 … (III)

Multiplying equation (II) by 7, we get

21x - 14y = 28 … (IV)

Adding equations (III) and (IV)

∴ x = 2 

Substituting x = 2 in equation (II)

∴ 3×2 - 2y = 4

∴ 6 - 2y = 4

∴ -2y = 4 - 6

∴ -2y = -2

∴ y = 1

∴ (2, 1) is the solution of the given equations.

(iii)

x - 2y = -10 … (I)

3x - 5y = -12 … (II)

Multiplying equation (I) by 3, we get

3x - 6y = -30 … (III)

Subtracting equation (II) from (III)

∴ y = 18 

Substituting y = 18 in equation (I)

∴ x - 2×18 = -10

∴ x - 36 = -10

∴ x = -10 + 36

∴ x = 26

∴ (26, 18) is the solution of the given equations.

(iv)

4x + y = 34 … (I)

x + 4y = 16 … (II)

Multiplying equation (I) by 4,

16x + 4y = 136 … (III)

Subtracting equation (II) from (III)

∴ x= 8 

Substituting x = 8 in equation (I)

∴ 4×8 + y = 34

∴ 32 + y = 34

∴ y = 34 - 32

∴ y = 2

∴ (8, 2) is the solution of the given equations.

(i)

Multiplying both sides by 12, we get

4x + 3y = 48 … (I)

Multiplying both sides by 8, we get

4x - 2y = 8 … (II)

Subtracting equation (II) from (I)

∴ y=8 

Substituting y = 8 in equation (II)

∴ 4x - 2×8 = 8

∴ 4x - 16 = 8

∴ 4x = 8 + 16

∴ 4x = 24

∴ x = 6

∴ (6, 8) is the solution of the given equations.

(ii)

Multiplying both sides by 3, we get

x + 15y = 39 … (I)

Multiplying both sides by 2, we get

4x + y = 38 … (II)

Multiplying equation (I) by 4

∴ 4x + 60y = 156 … (III)

Subtracting equation (II) from (III)

∴ y=2 

Substituting y = 2 in equation (I)

∴ x+ 15×2 = 39

∴ x + 30 = 39

∴ x = 9

∴ (9, 2) is the solution of the given equations.

(iii)

Multiplying both sides by 5, we get

Multiplying both sides by 2, we get

 … (II)

Substituting the value of from equation (I) in equation (II)

Substituting the value of y in equation (I)

 is the solution of the given equations.

Let the digit at ones place be 'x' and digit at tens place be 'y'.

So, the number will be 10y + x.

By the first condition, we have

10y + x = 4(x + y) + 3

∴ 10y + x = 4x + 4y + 3

∴ 10y + x - 4x - 4y = 3

∴ -3x + 6y = 3

Dividing both sides by 3, we get

-x + 2y = 1 … (I)

The number obtained after interchanging the digits is 10x + y.

By the second condition, we have

10y + x + 18 = 10x + y

∴ 10y + x - 10x - y = -18

∴ -9x + 9y = -18

Dividing both sides of the above equation by -9, we get

x - y = 2 … (II)

Adding equations (I) and (II), we have

Substituting y = 3 in equation (II), we get

x - 3 = 2

∴ x = 5

∴ 10y + x = 10×3 + 5 = 35

∴ The required number is 35.

Let the cost of one book be Rs. x and the cost of one pen be Rs. y.

By the first condition, we have

8x + 5y = 420 … (I)

By the second condition, we get

5x + 8y = 321 … (II)

Multiplying equation (I) by 5

∴ 40x + 25y = 2100 … (III)

Multiplying equation (II) by 8, we get

40x + 64y = 2568 … (IV)

Subtracting equation (III) from (IV)

∴ y=12 

Substituting y = 12 in equation (I)

∴ 8x + 5×12 = 420

∴ 8x + 60 = 420

∴ 8x = 420 - 60

∴ 8x = 360

∴ x = 45

Cost of 1 book and 2 pens = x + 2y

= 45 + 2×12

= 45 + 24

= Rs. 69

∴ The cost of 1 book and 2 pens is Rs. 69.

Let the income of first person be Rs. x and that of second person be Rs y.

By the first condition, we have

∴ 7x = 9y

∴ 7x - 9y = 0 … (I)

Since, each person saves Rs 200.

Expenses of first person = Income - Savings = x - 200

Expenses of second person = y - 200

By the second condition,

∴ 3(x - 200) = 4(y - 200)

∴ 3x - 600 = 4y - 800

∴ 3x - 4y = - 800 + 600

∴ 3x - 4y = -200 … (II)

Multiplying equation (I) by 4, we get

28x - 36y = 0 … (III)

Multiplying equation (II) by 9, we get

27x - 36y = -1800 … (IV)

Subtracting equation (IV) from (III),

Substituting x = 1800 in equation (I)

∴ 7 × 1800 - 9y = 0

∴ 9y = 7 × 1800

∴ y= 1400

∴ The income of first person is Rs 1800 and that of second person is Rs 1400.

Let length of the rectangle be 'x' units and breadth of the rectangle be 'y' units.

∴ Area of the rectangle = xy sq. units

Length of the rectangle is reduced by 5 units

∴ Reduced length = x - 5

Breadth of the rectangle is increased by 3 units

∴ Increased breadth = y + 3

Now, area of the rectangle is reduced by 9 square units

∴ Reduced area of the rectangle = xy - 9

By the first condition, we have

(x - 5)(y + 3) = xy - 9

∴ xy + 3x - 5y - 15 = xy - 9

∴ 3x - 5y = -9 + 15

∴ 3x - 5y = 6 … (I)

Length of the rectangle is reduced by 3 units

∴ Reduced length = x - 3

Breadth of the rectangle is increased by 2 units

∴ Increased breadth = y + 2

Now, area of the rectangle is increased by 67 square units

∴ Increased area of the rectangle = xy + 61

By the second condition, we have

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

∴ xy + 2x - 3y - 6 = xy + 67

∴ 2x - 3y = 67 + 6

∴ 2x - 3y = 73 … (II)

Multiplying equation (I) by 3, we get

9x - 15y = 18 … (III)

Multiplying equation (II) by 5, we get

10x - 15y = 365 … (IV)

Subtracting equation (III) from equation (IV)

Substituting x = 347 in equation (II)

∴ 2×(347) - 3y = 73

∴ 694 - 73 = 3y

∴ 621 = 3y

∴ y = 207

∴ The length and breadth of rectangle are 347 units and 207 units respectively.

Let speed of the first car which is starting from A be 'x' km/hr and speed of second car which is starting from B be 'y' km/hr. (x > y)

By the first condition,

Distance covered by the first car in 7 hours = 7x km

Distance covered by the second car in 7 hours = 7y km

If the cars are travelling in the same direction, we have

7x - 7y = 70

Dividing both sides by 7, we get

x - y = 10 … (I)

By the second condition,

Distance covered by the first car in 1 hour = x km

Distance covered by the second car in 1 hour = y km

If the cars are travelling in the opposite direction, we have

x + y = 70 … (II)

Adding equations (I) and (II)

∴ x=40 

Substituting x = 40 in equation (II)

∴ 40 + y = 70

∴ y = 70 - 40

∴ y = 30

∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.

Let the digit at ones place be 'x' and digit at tens place be 'y'.

So, the number will be 10y + x.

The number obtained after interchanging the digits is 10x + y.

By the condition given in the question, we have

10y + x + 10x + y = 99

∴ 11x + 11y = 99

Dividing both sides by 99, we get

x + y = 9 … (I)

If y = 1, then x = 8

If y = 2, then x = 7

If y = 3, then x = 6 and so on.

∴ The number can be 18, 27, 36, … etc.