Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 3 - Linear Equations in Two Variables

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.

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

x = 2 and y = 3.

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

x = 2 and y = -3.

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

x = -2 and y = 3.

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

x = -1 and y = 2.

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,

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,

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

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

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

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

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

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

The given equations are

Multiplying equation (i) by 3, we get

Subtracting equation (ii) from equation (iii),

Substituting  in equation (i),

Therefore, solution is .

The given equations are:

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

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

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

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

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 .

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

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

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

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

The given equations are

Putting

Adding (1) and (2)

Putting value of u in (1)

Hence the required solution isx = 4, y = 5

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

Substituting  in (iii), we get 

Therefore,   and 

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

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

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 .

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

Given equations are as follows:

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  

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

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 .

The given equations are

Adding equations (i) and (ii),

Substituting x=7 in equation (i),

Therefore,

And,

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

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

By cross multiplication, we have

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

3x - 2y + 3 = 0

4x + 3y - 47 = 0

By cross multiplication we have

the solution is x = 5, y = 9

6x - 5y - 16 = 0

7x - 13y + 10 = 0

By cross multiplication we have

the solution is x = 6, y = 4

3x + 2y + 25 = 0

2x + y + 10 = 0

By cross multiplication, we have

the solution is x = 5,y = -20

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

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

By cross multiplication we have

the solution is x = 3, y = -1

2x + y - 35 = 0

3x + 4y - 65 = 0

By cross multiplication, we have

Taking

u + v - 7 = 0

2u + 3v - 17 = 0

By cross multiplication, we have

the solution is

Let in the equation

5u - 2v + 1 = 0

15u + 7v - 10 = 0

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

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

By cross multiplication, we have

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

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

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

These equations are of the form

Hence the given system of equations has infinitely many solutions

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

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.

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

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

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

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

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.

(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

(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

The given equations are

These equations are of the form

Where

If system of equations have infinite number of solutions, then

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

The given equations are

These equations are of the form

Where

If system of equations have no solution, then

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

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

Comparing the equation with

These equations have a non - zero solution if

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.

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

2 × 25 + 3y = 92

50 + 3y = 92

 3y = 92 - 50

y = 14

Hence, the first number is 25 and second is 14

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

3 × 40 + y = 142

y = 142 - 120

y = 22

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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

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.

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

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

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

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

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

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.

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.

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

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.

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.

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

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

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.

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.

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.

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.

Let the required fraction be .

Then,

Also,

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

Hence, the required fraction is .

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.

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.

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.

Correct option: (d)

Given equations are

Adding equations (i) and (ii),

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),

Correct option: (c)

Correct option: (b)

Now,

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 . 

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