ROUTERA


Chapter 3 Linear Equations in Two Variables

Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution


Linear Equations in Two Variables Exercise Ex. 3A

Solution 1

Solution 2

Given equations are   and  .

On a graph paper, draw X-axis and Y-axis respectively.

Graph of  :

Thus, we have the following table:

On the graph paper, plot the points A(3, 1.5), B(0, 6), C(-2, 9) and join them.

Graph of  :

Thus, we have the following table:

On the graph paper, plot the points P(0, 1), Q(-2, -1), R(3, 4) and join them.

  

The two graphs intersect at point M (2,3).

Therefore,  is the solution of the given equations.

Solution 3

ch3/3.4.jpg

Solution 4




Solution 5

  

Since the two graphs intersect at (2, 3),

x = 2 and y = 3.

Solution 6




Solution 7

  

  

Since the two graphs intersect at (2, -3),

x = 2 and y = -3.

Solution 8

  

Since the two graphs intersect at (-2, 3),

x = -2 and y = 3.

 

Solution 9

  

Since the two graphs intersect at (-1, 2),

x = -1 and y = 2.

Solution 10


Solution 11

  

Solution 12

  


Since the two graphs intersect at (1, 2), x = 1, y = 2 is the solution of the given system of equations.


The vertices of the triangle formed by these lines and the X-axis are (1, 2), (-2, 0) and (5, 0).


So, height of the triangle = distance of (1, 2) from the X-axis = 2 units

And, base = 7 units


Therefore,


Area space of space triangle equals 1 half cross times base cross times height equals 1 half cross times 7 cross times 2 equals 7 space sq. space units

Solution 13

Solution 14


 

Solution 15

  

  

  

 

Solution 16

  

Solution 17

  

 

Solution 18

  

 

Solution 19

SInce the two lines intersect at A(3,2), x = 3, y = 2 is the solution of system of given equations.

Height of triangle ARS = 3 units

Base of triangle ARS = 7.2 units

Therefore,

Area space of space triangle space ARS equals 1 half cross times 7.2 cross times 3 equals 10.8 space sq. space units

 

 

Solution 20

   


Solution 21

  

Solution 22

  

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions.

Solution 23



Solution 24


Solution 25

  

  

Since the graph of the system of equations is coincident lines, the system has infinitely many solutions. 

Solution 26

  

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.

Solution 27

  

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.

Solution 28

  

Since the graph of the system of equations is parallel lines, the system has no solution and hence is inconsistent.

Solution 29

  




The coordinates of the vertices of the trapezium formed by these lines are (0,6), (3,0), (1,0) and (0,2). 

Area space of space trapezium space ABCD equals straight A left parenthesis increment OAD right parenthesis minus straight A left parenthesis increment OBC right parenthesis
equals 1 half cross times 3 cross times 6 minus 1 half cross times 1 cross times 2
equals 9 minus 1
equals 8 space sq. space units






Linear Equations in Two Variables Exercise Ex. 3B

Solution 1

Solution 2

Solution 3

ch3/3.23.jpg

Solution 4

ch3/3.24.jpg

Solution 5

Solution 6

The given equations are

  

Multiplying equation (i) by 3, we get

Subtracting equation (ii) from equation (iii),

Substituting   in equation (i),

Therefore, solution is  .

Solution 7

  

Solution 8


Solution 9


Solution 10


Solution 11


Solution 12


Solution 13

Solution 14

Solution 15


Solution 16

The given equations are:

6x + 5y = 7x + 3y + 1 = 2(x + 6y - 1)

Solution 17

 


Solution 18

Putting the given equations become

5u + 6y = 13---(1)

3u + 4y = 7 ----(2)

Multiplying (1) by 4 and (2) by 6, we get

20u + 24y = 52---(3)

18u + 24y = 42---(4)

Subtracting (4) from (3), we get

2u = 10 u = 5

Putting u = 5 in (1), we get

5 × 5 + 6y = 13

6y = 13 - 25

6y = -12

y = -2

Solution 19

The given equations are and

Putting

x + 6v = 6 ----(1)

3x - 8v = 5---(2)

Multiplying (1) by 4 and (2) by 3

4x + 24v = 24---(3)

9x - 24v = 15 ---(4)

Adding (3) and (4)

13x = 24 + 15 = 39

Puttingx = 3 in (1)

3 + 6v = 6

6v = 6 - 3 = 3

solution is x = 3, y = 2

Solution 20

Putting in the given equation

2x - 3v = 9 ---(1)

3x + 7v = 2 ---(2)

Multiplying (1) by7 and (2) by 3

14x - 21v = 63 ---(3)

9x + 21v = 6 ---(4)

Adding (3) and (4), we get

Putting x= 3 in (1), we get

2 × 3 - 3v = 9

         -3v = 9 - 6

     -3v= 3

       v = -1

 

the solution is x = 3, y = -1

Solution 21

The given equations are

Putting   in the given equations,

Multiplying equation (iv) by 4, we get

Adding equations (iii) and (v),

Putting   in equation (ii),

Now,

Therefore, solution is  .

Solution 22

Putting in the equation

9u - 4v = 8 ---(1)

13u + 7v = 101---(2)

Multiplying (1) by 7 and (2) by 4, we get

63u - 28v = 56---(3)

52u + 28v = 404---(4)

Adding (3) and (4), we get

Putting u = 4 in (1), we get

9 × 4 - 4v = 8

36 - 4v = 8

-4v = 8 - 36

-4v = -28

Solution 23

Putting in the given equation, we get

5u - 3v = 1 ---(1)

Multiplying (1) by 4 and (2) by 3, we get

20u - 12v = 4----(3)

27u + 12v = 90---(4)

Adding (3) and (4), we get

Putting u = 2 in (1), we get

(5 × 2) - 3v = 1

10 - 3v = 1

-3v = 1 - 10 -3v = -9 

v = 3

Solution 24

Solution 25

4x + 6y = 3xy

Putting in (1) and (2), we get

4v + 6u = 3---(3)

8v + 9u = 5---(4)

Multiplying (3) by 9 and (4) by 6, we get

36v + 54u = 27 ---(5)

48v + 54u = 30 ---(6)

Subtracting (3) from (4), we get

12v = 3

Putting in (3), we get

the solution is x = 3, y = 4

Solution 26

Solution 27

Solution 28

Putting

3u + 2v = 2----(1)

9u - 4v = 1----(2)

Multiplying (1) by 2 and (2) by 1. We get

6u + 4v = 4----(3)

9u - 4v = 1----(4)

Adding (3) and (4), we get

Adding (5) and (6), we get

Putting in (5). We get

the solution is

Solution 29

The given equations are

Putting

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5

Solution 30

Putting in the equation, we get

44u + 30v = 10----(1)

55u + 40v = 13----(2)

Multiplying (1) by 4 and (2) by 3, we get

176u + 120v = 40---(3)

165u + 120v = 39---(4)

Subtracting (4) from (3), we get

Putting in (1) we get

Adding (5) and (6), we get

Putting x = 8 in (5), we get

8 + y = 11 y = 11 - 8 = 3

the solution is x = 8, y = 3

Solution 31


Substituting straight x equals 21 over 8 in (iii), we get y equals 9 over 8

Therefore,  straight x equals 21 over 8 and y equals 9 over 8

Solution 32

The given equations are

71x + 37y = 253---(1)

37x + 71y = 287---(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

----(3)

Subtracting (2) from (1)

34x - 34y = 253 - 287 = -34

34(x - y) = -34

---(4)

Adding (3) and (4)

2x = 5 - 1= 4

 

Subtracting (4) from (3)

2y = 5 + 1 = 6


solution is x = 2, y = 3

Solution 33

217x + 131y = 913---(1)

131x + 217y = 827---(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

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

Subtracting (2) from (1), we get

86x - 86y = 86

86(x - y) = 86

   x - y = 1---(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 - 3 = 2

solution is x = 3, y = 2

Solution 34

The given equations are

Adding equations (i) and (ii),

Subtracting equation (i) from (ii),

Adding equation (iii) and (iv),

Substituting   in equation (iv),

Therefore, solution is  .

Solution 35

Solution 36

Solution 37

The given equations are

Multiplying (1) by 6 and (2) by 20, we get

Multiplying (3) by 6 and (4) by 5, we get

18u + 60v = -54---(5)

125u - 60v = ---(6)

Adding (5) and (6), we get

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Given equations are as follows:

 

px plus qy equals straight p minus straight q space space space space space space space..... left parenthesis straight i right parenthesis
qx minus py equals straight p plus straight q space space space space space space space..... left parenthesis ii right parenthesis

 

Solution 43

Solution 44

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b ---(1)

6(bx - ay) = 3b - 2a

6bx - 6ay = 3b- 2a ---(2)


6ax + 6by = 3a + 2b ---(1)

6bx - 6ay = 3b - 2a ---(2)


Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting  in(1), we get

Hence, the solution is  

Solution 45

Solution 46

Solution 47

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

solution is x = ab, y = ab

Solution 48

Solution 49

The given equations are

Multiplying equation (i) by   and equation (ii) by  ,

Subtracting equations (iv) from equation (iii),

Multiplying equation (i) by b2 and equation (ii) by  ,

Subtracting equations (v) from equation (vi),

Therefore, solution is  .

Solution 50

Solution 51

The given equations are

Adding equations (i) and (ii),

Substituting x=7 in equation (i),

Therefore,

And,

Linear Equations in Two Variables Exercise Ex. 3C

Solution 1

x + 2y + 1 = 0 ---(1)

2x - 3y - 12 = 0 ---(2)

By cross multiplication, we have

Hence, x = 3 and y = -2 is the solution

Solution 2

3x - 2y + 3 = 0

4x + 3y - 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9

Solution 3

6x - 5y - 16 = 0

7x - 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4

Solution 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20

Solution 5

2x + 5y - 1 = 0 ---(1)

2x + 3y - 3 = 0---(2)

By cross multiplication we have

the solution is x = 3, y = -1

Solution 6

2x + y - 35 = 0

3x + 4y - 65 = 0

By cross multiplication, we have

Solution 7

  

Solution 8

Solution 9

Taking

u + v - 7 = 0

2u + 3v - 17 = 0

By cross multiplication, we have

the solution is

Solution 10

Let in the equation

5u - 2v + 1 = 0

15u + 7v - 10 = 0

Solution 11

Solution 12

2ax + 3by - (a + 2b) = 0

3ax+ 2by - (2a + b) = 0

By cross multiplication, we have

Solution 13

  

Linear Equations in Two Variables Exercise Ex. 3D

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

4x - 5y - k = 0, 2x - 3y - 12 = 0

These equations are of the form

Thus, for all real value of k the given system of equations will have a unique solution

Solution 9

kx + 3y - (k - 3) = 0

12x + ky - k = 0

These equations are of the form

Thus, for all real value of k other than , the given system of equations will have a unique solution

Solution 10

2x - 3y - 5 = 0, 6x - 9y - 15 = 0

These equations are of the form

Hence the given system of equations has infinitely many solutions

Solution 11

Solution 12

kx + 2y - 5 = 0

3x - 4y - 10 = 0

These equations are of the form

This happens when

Thus, for all real value of k other that , the given system equations will have a unique solution

(ii) For no solution we must have

Hence, the given system of equations has no solution if

Solution 13

x + 2y - 5= 0

3x + ky + 15 = 0

These equations are of the form of

Thus for all real value of k other than 6, the given system ofequation will have unique solution

(ii) For no solution we must have

k = 6

Hence the given system will have no solution when k = 6.

Solution 14

x + 2y - 3 = 0, 5x + ky + 7 = 0

These equations are of the form

(i)For a unique solution we must have

Thus, for all real value of k other than 10

The given system of equation will have a unique solution.

(ii)For no solution we must have

Hence the given system of equations has no solution if

For infinite number of solutions we must have

This is never possible since

There is no value of k for which system of equations has infinitely many solutions

Solution 15

2x + 3y - 7 = 0

(k - 1)x + (k + 2)y - 3k = 0

These are of the form

 

This hold only when

Now the following cases arises

Case : I

 

 

Case: II


 

Case III

 

For k = 7, there are infinitely many solutions of the given system of equations

Solution 16

2x + (k - 2)y - k = 0

6x + (2k - 1)y - (2k + 5) = 0

These are of the form

For infinite number of solutions, we have

This hold only when

Case (1)

 

Case (2)

 

Case (3)

Thus, for k = 5 there are infinitely many solutions

Solution 17

kx + 3y - (2k +1) = 0

2(k + 1)x + 9y - (7k + 1) = 0

These are of the form

For infinitely many solutions, we must have

This hold only when

Now, the following cases arise

Case - (1)

 

Case (2)

 

Case (3)

Thus, k = 2, is the common value for which there are infinitely many solutions

Solution 18

5x + 2y - 2k = 0

2(k +1)x + ky - (3k + 4) = 0

These are of the form

For infinitely many solutions, we must have

These hold only when

Case I

Thus, k = 4 is a common value for which there are infinitely by many solutions.

Solution 19

(k - 1)x - y - 5 = 0

(k + 1)x + (1 - k)y - (3k + 1) = 0

These are of the form

For infinitely many solution, we must now

 

k = 3 is common value for which the number of solutions is infinitely many

Solution 20

Solution 21

(a - 1)x + 3y - 2 = 0

6x + (1 - 2b)y - 6 = 0

These equations are of the form

For infinite many solutions, we must have

Hence a = 3 and b = -4

Solution 22

The given equations are

These equations are of the form

  

Where

If system of equations have infinite number of solutions, then

Solution 23

2x - 3y - 7 = 0

(a + b)x + (a + b - 3)y - (4a + b) = 0

These equation are of the form

For infinite number of solution

Putting a = 5b in (2), we get

Putting b = -1 in (1), we get

Thus, a = -5, b = -1

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

The given equations are

These equations are of the form

  

Where

If system of equations have no solution, then

Solution 31

We have 5x - 3y = 0 ---(1)

2x + ky = 0---(2)

Comparing the equation with

These equations have a non - zero solution if

Linear Equations in Two Variables Exercise Ex. 3E

Solution 1

Solution 2

Solution 3

Solution 4

Let the two numbers be x and y respectively.

 

Given:

x + y = 137---(1)

x - y = 43---(2)

 

Adding (1) and (2), we get

2x = 180

 

Putting x = 90 in (1), we get

90 + y = 137

y = 137 - 90

  = 47

 

Hence, the two numbers are 90 and 47.

Solution 5

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92---(1)

4x - 7y = 2 ---(2)


Multiplying (1) by 7 and (2) by 3, we get

14 x+ 21y = 644---(3)

12x - 21y = 6---(4)


Adding (3) and (4), we get

 

Putting x = 25 in (1), we get

× 25 + 3y = 92

50 + 3y = 92

 3y = 92 - 50

y = 14


Hence, the first number is 25 and second is 14

Solution 6

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142---(1)

4x - y = 138---(2)

 

Adding (1) and (2), we get

Putting x = 40 in (1), we get

× 40 + y = 142

y = 142 - 120

y = 22


Hence, the first and second numbers are 40 and 22.

Solution 7

Let the greater number be x and smaller be y respectively.

According to the question:

2x - 45 = y

2x - y = 45---(1)

and

2y - x = 21

 -x + 2y = 21---(2)


Multiplying (1) by 2 and (2) by 1

4x - 2y = 90---(3)

-x + 2y = 21 ---(4)


Adding (3) and (4), we get

3x = 111

 

Putting x = 37 in (1), we get

2 × 37 - y = 45

 74 - y = 45

y = 29

 

Hence, the greater and the smaller numbers are 37 and 29.

Solution 8

Let the larger number be x and smaller be y respectively.

We know,

Dividend = Divisor × Quotient + Remainder


3x = y × 4 + 8

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

And

5y = x × 3 + 5

-3x + 5y = 5---(2)

 

Adding (1) and (2), we get

y = 13

 

putting y = 13 in (1)

Hence, the larger and smaller numbers are 20 and 13 respectively.

Solution 9

Let the required numbers be x and y respectively.

Then,

 

Therefore,

2x - y =-2---(1)

11x - 5y = 24 ---(2)


Multiplying (1) by 5 and (2) by 1

10x - 5y = -10---(3)

11x - 5y = 24---(4)

 

Subtracting (3) and (4) we get

x = 34


putting x = 34 in (1), we get

 2 × 34 - y = -2

 68 - y = -2

-y = -2 - 68

y = 70

Hence, the required numbers are 34 and 70.

Solution 10

Let the numbers be x and y respectively.

 

According to the question:

x - y = 14 ---(1)


From (1), we get

x = 14 + y---(3)


putting x = 14 + y in (2), we get


Putting y = 9 in (1), we get

x - 9 = 14

  x = 14 + 9 = 23

 

Hence the required numbers are 23 and 9

Solution 11

Let the ten's digit be x and units digit be y respectively.

Then,

x + y = 12---(1)


Required number = 10x + y

Number obtained on reversing digits = 10y + x

 

According to the question:

10y + x - (10x + y) = 18

10y + x - 10x - y = 18

9y - 9x = 18

y - x = 2 ----(2)


Adding (1) and (2), we get


Putting y = 7 in (1), we get

x + 7 = 12

x = 5

 

Number= 10x + y

                  = 10 × 5 + 7

                  = 50 + 7

                  = 57

Hence, the number is 57.

Solution 12

Let the ten's digit of required number be x and its unit's digit be y respectively

Required number = 10x + y

 10x + y = 7(x + y)

     10x + y = 7x + 7y

       3x - 6y = 0---(1)


Number found on reversing the digits = 10y + x

(10x + y) - 27 = 10y + x

      10x - x + y - 10y = 27

        9x - 9y = 27

         (x - y) = 27

           x - y = 3---(2)


Multiplying (1) by 1 and (2) by 6

3x - 6y = 0---(3)

6x - 6y = 18 ---(4)

 

Subtracting (3) from (4), we get

 

Putting x = 6 in(1), we get

3 × 6 - 6y = 0

18 - 6y = 0

 

Number = 10x + y

            = 10 × 6 + 3

            = 60 + 3

            = 63

Hence the number is 63.

Solution 13

Let the ten's digit and unit's digits of required number be x and y respectively.

Then,

x + y = 15---(1)


Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

 

10y + x - (10x + y) = 9

10y + x - 10x - y = 9

       9y - 9x = 9

        

 

Add (1) and (2), we get

 

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 - 8 = 7

 

Required number = 10x + y

                         = 10 × 7 + 8

                         = 70 + 8

                         = 78

Hence the required number is 78.

Solution 14

Let the ten's and unit's of required number be x and y respectively.

Then,required number =10x + y

 

According to the given question:

 10x + y = 4(x + y) + 3

 10x + y = 4x + 4y + 3

6x - 3y = 3

  2x - y = 1 ---(1)

And

10x + y + 18 = 10y + x

9x - 9y = -18


 

 x - y = -2---(2)

 

Subtracting (2) from (1), we get

x = 3

 

Putting x = 3 in (1), we get

2 × 3 - y = 1

y = 6 - 1 = 5

x = 3, y = 5

 

Required number = 10x + y

                         = 10 × 3 + 5

                         = 30 + 5

                         = 35

 

Hence, required number is 35.

Solution 15

Let the ten's digit and unit's digit of required number be x and y respectively.

We know,

Dividend = (divisor × quotient) + remainder


According to the given questiion:

10x + y = 6 × (x + y) + 0

10x - 6x + y - 6y = 0

4x - 5y = 0 ---(1)

 

Number obtained by reversing the digits is 10y + x

 10x + y - 9 = 10y + x

9x - 9y = 9

9(x - y) =9

(x - y) = 1---(2)

 

Multiplying (1) by 1 and (2) by 5, we get

4x - 5y = 0 ---(3)

5x - 5y = 5 ---(4)

 

Subtracting (3) from (4), we get

x = 5

 

Putting x = 5 in (1), we get

x =5 and y= 4

 

Hence, required number is 54.

Solution 16

Let the ten's and unit's digits of the required number be x and y respectively.

Then, xy = 35

Required number = 10x + y


Also,

(10x + y) + 18 = 10y + x

9x - 9y = -18

9(y - x) = 18---(1)

y - x = 2

Now, 


 

Adding (1) and (2),

2y = 12 + 2 = 14

y = 7

 

Putting y = 7 in (1),

7 - x = 2

x = 5

 

Hence, the required number = 5 × 10 + 7

                                        = 57

Solution 17

Let the ten's and unit's digits of the required number be x and y respectively.

Then, xy = 18

Required number = 10x + y

Number obtained on reversing its digits = 10y + x

 

(10x + y) - 63 = (10y + x)

9x - 9y = 63

x - y = 7---(1)

Now,

 

 

Adding (1) and (2), we get

 

Putting x = 9 in (1), we get

9 - y = 7

y = 9 - 7

y =2


x = 9, y = 2

Hence, the required number = 9 × 10 + 2

                                        = 92.

Solution 18

Solution 19

 

Let the numerator and denominator of fraction be x and y respectively.

 

According to the question:

x + y = 8---(1)

And

 

Multiplying (1) be 3 and (2) by 1

3x + 3y = 24---(3)

4x - 3y = -3 ---(4)

 

Add (3) and (4), we get

 

Putting x = 3 in (1), we get

3 + y= 8

y = 8 - 3

y = 5

 

x = 3, y = 5

 

Hence, the fraction is

Solution 20

Let the numerator and denominator be x and y respectively.

Then the fraction is .


 

Subtracting (1) from (2), we get

x = 3

 

Putting x = 3 in (1), we get

2 × 3 - 4

-y = -4 -6

y = 10

 

x = 3 and y = 10

Hence the fraction is

Solution 21

Let the numerator and denominator be x and y respectively.

Then the fraction is .

 

According to the given question:

y = x + 11

y- x = 11---(1)


and

-3y + 4x = -8 ---(2)


Multiplying (1) by 4 and (2) by 1

4y - 4x = 44---(3)

-3y + 4x = -8---(4)

 

Adding (3) and (4), we get

y = 36

 

Putting y = 36 in (1), we get

y - x = 11

36 - x = 11 

  x = 25


x = 25, y = 36

Hence the fraction is

Solution 22

Let the numerator and denominator be x and y respectively.

Then the fraction is


Subtracting (1) from (2), we get

x = 15

 

Putting x = 15 in (1), we get

2 × 15 - y = 4

 30 - y = 4

y = 26

 

x = 15 and y = 26

Hence the given fraction is

Solution 23

Solution 24

Let the two numbers be x and y respectively.

 

According to the given question:

x + y = 16---(1)

And

---(2)

 

From (2),

xy = 48

 

We know,

 

Adding (1) and (3), we get

2x = 24

x = 12

 

Putting x = 12 in (1),

y = 16 - x

   = 16 - 12

   = 4

The required numbers are 12 and 4

Solution 25

Let the number of student in class room A and B be x and y respectively.

 

When 10 students are transferred from A to B:

x - 10 = y + 10

 x - y = 20---(1)


When 20 students are transferred from B to A:

 2(y - 20) = x + 20

 2y - 40 = x + 20

 

-x + 2y = 60---(2)

 

Adding (1) and (2), we get

y = 80

 

Putting y = 80 in (1), we get

x - 80 = 20

 x = 100


Hence, number of students of A and B are 100 and 80 respectively.

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

  

Solution 31

Let P and Q be the cars starting from A and B respectively and let their speeds be x km/hr and y km/hr respectively.

Case- I

When the cars P and Q move in the same direction.

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

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

Let the cars meet at point M.

        

AM = 7x km and BM = 7y km

AM - BM = AB

 7x - 7y = 70

7(x - y) = 70

x - y = 10 ----(1)


Case II

When the cars P and Q move in opposite directions.

Distance covered by P in 1 hour = x km

Distance covered by Q in 1 hour = y km


In this case let the cars meet at a point N.


AN = x km and BN = y km

AN + BN = AB

 x + y = 70---(2)


Adding (1) and (2), we get

2x = 80

 x = 40

 

Putting x = 40 in (1), we get

40 - y = 10

 y = (40 - 10) = 30

 

x = 40, y = 30

Hence, the speeds of these cars are 40 km/ hr and 30 km/ hr respectively.

Solution 32

Let the original speed be x km/h and time taken be y hours

Then, length of journey = xy km

 

Case I:

Speed = (x + 5)km/h and time taken = (y - 3)hour

Distance covered = (x + 5)(y - 3)km

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

xy + 5y -3x -15 = xy

5y - 3x = 15 ---(1)

 

Case II:

Speed (x - 4)km/hr and time taken = (y + 3)hours

Distance covered = (x - 4)(y + 3) km

(x - 4)(y + 3) = xy

xy -4y + 3x -12 = xy

 3x - 4y = 12 ---(2)


Multiplying (1) by 4 and (2) by 5, we get

20y - 12x = 60 ---(3)

-20y + 15x = 60 ---(4)

 

Adding (3) and (4), we get

3x = 120

or x = 40

 

Putting x = 40 in (1), we get

5y - 3 × 40 = 15 

  5y = 135  

  y = 27

 

Hence, length of the journey is (40 × 27) km = 1080 km

Solution 33

Solution 34

  

  

Solution 35

Solution 36

Let the speed of the boat in still water be   km/hr and the speed of the stream be   km/hr. Then,

speed upstream = (x-y) km/hr

speed downstream = (x+y) km/hr

Time taken to cover 30 km upstream =  hours

Time taken to cover 44 km downstream =  hours

Total time taken = 10 hours

   (i)

Again,

Time taken to cover 40 km upstream =  hours

Time taken to cover 55 km downstream =  hours

Total time taken = 13 hours

  (ii)

Putting   in (i) and (ii), we get

On multiplying (iv) by 2 and subtracting (iii) from the result, we get

Then,

Adding equations (v) and (vi),

Therefore, speed of the stream is 3 km/hr and speed of the boat in still water is 8 km/hr.

Solution 37

Let man's 1 day's work be and 1 boy's day's work be

Also let  and

 

Multiplying (1) by 6 and (2) by 5 we get

 

Subtracting (3) from (4), we get

Putting in (1), we get

x = 18, y = 36

The man will finish the work in 18 days and the boy will finish the work in 36 days when they work alone.

Solution 38

Let the length = x meters and breadth = y meters

Then,

x = y + 3

x - y = 3 ----(1)

Also,

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

 3y - 2x = 6 ----(2)

 

 

Multiplying (1) by 2 and (2) by 1

-2y + 2x = 6 ---(3)

3y - 2x = 6 ---(4)

 

Adding (3) and (4), we get

y = 12

 

Putting y = 12 in (1), we get

x - 12 = 3

x= 15


x = 15, y = 12

Hence length = 15 metres and breadth = 12 metres

Solution 39

Let the length of a rectangle be x meters and breadth be y meters.

Then, area = xy sq.m

Now,

xy - (x - 5)(y + 3) = 8

xy - [xy - 5y + 3x -15] = 8

xy - xy + 5y - 3x + 15 = 8 

3x - 5y = 7 ---(1)


And

(x + 3)(y + 2) - xy = 74

xy + 3y +2x + 6 - xy = 74

2x + 3y = 68---(2)


Multiplying (1) by 3 and (2) by 5, we get

9x - 15y = 21---(3)

10x + 15y = 340---(4)

 

Adding (3) and (4), we get

 

Putting x = 19 in (3) we get

 

x = 19 meters, y = 10 meters

Hence, length = 19m and breadth = 10m

Solution 40

Solution 41

Solution 42

Solution 43

Let the present ages of the man and his son be x years and y years respectively.

Then,

 

Two years ago:

(x - 2) = 5(y - 2)


x - 2 = 5y - 10

  x - 5y = -8 ---(1)

 

Two years later:

(x + 2) = 3(y + 2) + 8


x + 2 = 3y + 6 + 8

x - 3y = 12 ---(2)


Subtracting (2) from (1), we get

-2y = -20

  y = 10


Putting y = 10 in (1), we get

x - 5 10 = -8

x - 50 = -8

x = 42

 

 

Hence the present ages of the man and the son are 42 years and 10 respectively.

Solution 44

Let the present ages of father and his son be x and y respectively.

According to given information

x + 2y = 70     ....(i)

And

2x + y = 95     ....(ii)

Multiplying equation (ii) by 2, we get

4x + 2y = 190   ....(iii)

Subtracting (i) from (iii), we get

Putting x = 40 in (i), we get

40 + 2y = 70

2y = 30

 y = 15

x = 40, y = 15

Hence, the ages of father and son are 40 years and 15 years respectively.

Solution 45

Let the present ages of woman and daughter be x and y respectively.

Then,

 

Their present ages:

x = 3y + 3

x - 3y = 3---(1)

 

Three years later:

 (x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

 x - 2y = 13---(2)


Subtracting (2) from (1), we get

 y = 10

 

Putting y = 10 in (1), we get

x - 3 × 10 = 3

 x = 33

 

x = 33, y = 10

Hence, present ages of woman and daughter are 33 and 10 years.

Solution 46

Solution 47

Solution 48

Let the ten's and unit's digits of the required number be   and   respectively.

Then, required number = (10x+y).

Now,

Number obtained on reversing the digits = (10y+x)

Adding equations (i) and (ii),

Hence, the required number is 36.

Solution 49

Let the required fraction be  .

Then,

Also,

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

Hence, the required fraction is  .

Solution 50

Let the present age of father be   years and the sum of the ages of his two children be   years.  

Then,

 .

After 5 years,

father's age = (x+5) years

sum of the ages of two children =   years

Then,

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

Hence, the present age of the father is 45 years.

Solution 51

Let the sides of two squares be   m and   m respectively.  

Then,

  

And,

Now,

Adding equations (i) and (ii), we get

Hence, the sides of the squares are 24 m and 8 m respectively.

Solution 52

Let the sides of two squares be   m and   m respectively.  

Then,

  

And,

Now,

Adding equations (i) and (ii), we get

Hence, the sides of the squares are 11 m and 6 m respectively.

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Linear Equations in Two Variables Exercise Ex. 3F

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Linear Equations in Two Variables Exercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Correct option: (d)

Given equations are

Adding equations (i) and (ii),

Solution 5

Solution 6

Solution 7

Solution 8

Correct option: (a)

The given equations are

Adding equations (i) and (ii),

Subtracting equation (ii) from (i),

Adding equation (iii) and (iv),

Substituting   in equation (iii),

Solution 9

Correct option: (c)

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

We space know space that comma
the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0
has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2.
So comma space the space pair space of space equations space has space no space solution.




Solution 17

We space know space that comma
the space system space of space linear space equations space ax subscript 1 plus by subscript 1 plus straight c subscript 1 equals 0 space and space ax subscript 2 plus by subscript 2 plus straight c subscript blank subscript 2 end subscript equals 0
has space no space solution space if space straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2.
So comma space the space pair space of space equations space has space no space solution.

Solution 18

Solution 19

Solution 20

Correct option: (b)

Now,

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Linear Equations in Two Variables Exercise Test Yourself

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

The given system of equations are

These equations are of the form

  

Where

 

(i) If system of equations have unique solution, then

Hence, the given system of equations will have a unique solution when  . 

 

(ii) If system of equations have no solution, then

Clearly,   also satisfies the condition  .

Hence, the given system of equations will have no solution when  . 

Solution 14

Solution 15

Solution 16

  

  

Since the intersection of the lines is the point with coordinates (-1, -1), x = -1 and y = -1.

Solution 17

Solution 18

Solution 19

Solution 20