Linear Equations In Two Variables

For equation, 2x + 3y = 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, x - 2y = 8

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (4, - 2), which is the intersecting point of the two lines.

For equation, 3x + 2y = 4

First, take x = 0 and find the value of y.

Now similarly solve for equation, 2x - 3y = 7

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2, - 1), which is the intersecting point of the two lines.

We can rewrite the equations as:

2x + 3y = 8

x - 2y = - 3

For equation, 2x + 3y = 8

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, x - 2y = - 3

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (1,2), which is the intersecting point of the two lines.

We can rewrite the equations as:

2x – 5y = - 4

& 2x + y = 8

For equation, 2x – 5y = - 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + y = 8

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (3,2), which is the intersecting point of the two lines.

For equation, 3x + 2y = 12

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 5x – 2y = 4

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,3), which is the intersecting point of the two lines.

We can rewrite the equations as:

3x + y = - 1

& 2x – 3y = - 8

For equation, 3x + y = - 1

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x – 3y = - 8

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is ( - 1,2), which is the intersecting point of the two lines.

We can rewrite the equations as:

2x + 3y = - 5

& 3x + 2y = 12

For equation, 2x + 3y = - 5

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x + 2y = 12

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2, - 3), which is the intersecting point of the two lines.

We can rewrite the equations as:

2x – 3y = - 13

& 3x – 2y = - 12

For equation, 2x – 3y = - 13

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x – 2y = - 12

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is ( - 2,3), which is the intersecting point of the two lines.

We can rewrite the equations as:

2x + 3y = 4

& 3x – y = - 5

For equation, 2x + 3y = 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x – y = - 5

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is ( - 1,2), which is the intersecting point of the two lines.

We can rewrite the equations as:

x + 2y = - 2

& 3x + 2y = 2

For equation, x + 2y = - 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x + 2y = 2

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2, - 2), which is the intersecting point of the two lines.

We can rewrite the equations as:

x – y = - 3

& 2x + 3y = 4

For equation, x – y = - 3

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + 3y = 4

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is ( - 1,2), which is the intersecting point of the two lines.

The vertices of the formed triangle ABC by these lines and the x - axis in the graph are A( - 1,2), B( - 3,0) and C(2,0).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = BO + OC = 3 + 2 = 5 units & height = 2 units]

Area(∆ABC) = 5 sq. units

We can rewrite the equations as:

2x – 3y = - 4

& x + 2y = 5

For equation, 2x – 3y = - 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, x + 2y = 5

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (1,2), which is the intersecting point of the two lines.

The vertices of the formed triangle ABC by these lines and the x - axis in the graph are A(1,2), B( - 2,0) and C(5,0).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = BO + OC = 2 + 5 = 7 units & height = 2 units]

Area(∆ABC) = 7 sq. units

We can rewrite the equations as:

4x – 3y = - 4

& 4x + 3y = 20

For equation, 4x – 3y = - 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 4x + 3y = 20

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,4), which is the intersecting point of the two lines.

The vertices of the formed triangle ABC by these lines and the x - axis in the graph are A(2,4), B( - 1,0) and C(5,0).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = BO + OC = 1 + 5 = 6 units & height = 4 units]

Area(∆ABC) = 12 sq. units

We can rewrite the equations as:

x – y = - 1

& 3x + 2y = 12

For equation, x – y = - 1

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x + 2y = 12

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,3), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the x - axis in the graph are A(2,3), B( - 1,0) and C(4,0).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = BO + OC = 1 + 4 = 5 units & height = 3 units]

Area(∆ABC) = 7.5 sq. units

We can rewrite the equations as:

x – 2y = - 2

& 2x + y = 6

For equation, x – 2y = - 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + y = 6

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,2), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the x - axis in the graph are A(2,2), B( - 2,0) and C(3,0).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) = 1/2 ×5×2

[∵ Base = BO + OC = 2 + 3 = 5 units & height = 2 units]

Area(∆ABC) = 2 sq. units

We can rewrite the equations as:

2x – 3y = - 6

& 2x + 3y = 18

For equation, 2x – 3y = - 6

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + 3y = 18

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (3,4), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(3,4), B(0,6) and C(0,2).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = OB – OC = 6 – 2 = 4 units & height = 3 units]

Area(∆ABC) = 6 sq. units

We can rewrite the equations as:

4x – y = 4

& 3x + 2y = 14

For equation, 4x – y = - 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x + 2y = 14

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,4), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(2,4), B(7,0) and C(0, - 4).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) =

[∵ Base = OB + OC = 7 + 4 = 11 units & height = 4 units]

Area(∆ABC) = 22 sq. units

We can rewrite the equations as:

x – y = 5

& 3x + 5y = 15

For equation, x – y = 5

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x + 5y = 15

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (5,0), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(5,0), B(0,3) and C(0, - 5).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) = 1/2 × 8 × 5

[∵ Base = OB + OC = 3 + 5 = 8 units & height = 5 units]

Area(∆ABC) = 20 sq. units

We can rewrite the equations as:

2x – 5y = - 4

& 2x + y = 8

For equation, 2x – 5y = - 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + y = 8

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (3,2), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(3,2), B(0,8) and C(0,0.8).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) = 1/2 × 7.2 × 3

[∵ Base = OB – OC = 8 - 0.8 = 7.2 units & height = 3 units]

Area(∆ABC) = 10.8 sq. units

We can rewrite the equations as:

5x – y = 7

& x – y = - 1

For equation, 5x – y = 7

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, x – y = - 1

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (2,3), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(2,3), B(0,1) and C(0, - 7).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) = 1/2 × 8 × 2

[∵ Base = OB + OC = 1 + 7 = 8 units & height = 2 units from the y - axis to the point A]

Area(∆ABC) = 8 sq. units

We can rewrite the equations as:

2x – 3y = 12

& x + 3y = 6

For equation, 2x – 3y = 12

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, x + 3y = 6

Plot the values in a graph and find the intersecting point for the solution.

Hence, the solution so obtained from the graph is (6,0), which is the intersecting point of the two lines.

The vertices of the formed triangle by these lines and the y - axis in the graph are A(6,0), B(0,2) and C(0, - 4).

Clearly, from the graph we can identify base and height of the triangle.

Now, we know

Area of Triangle = 1/2 × base × height

Thus, Area(∆ABC) = 1/2 × 6 × 6

[∵ Base = OB + OC = 2 + 4 = 6 units & height = 6 units]

Area(∆ABC) = 18 sq. units

For equation, 2x + 3y = 6

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 4x + 6y = 12

Plot the values in a graph and find the intersecting point for the solution.

The lines coincide on each other, this indicates that there are number of intersection points on the line since a line consists of infinite points.

Hence, the graph shows that the system of equations have infinite number of solutions.

For equation, 3x – y = 5

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 6x - 2y = 10

Plot the values in a graph and find the intersecting point for the solution.

The lines coincide on each other, this indicates that there are number of intersection points on the line since a line consists of infinite points.

Hence, the graph shows that the system of equations have infinite number of solutions.

For equation, 2x + y = 6

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 6x + 3y = 18

Plot the values in a graph and find the intersecting point for the solution.

The lines coincide on each other, this indicates that there are number of intersection points on the line since a line consists of infinite points.

Hence, the graph shows that the system of equations have infinite number of solutions.

For equation, x – 2y = 5

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x – 6y = 15

Plot the values in a graph and find the intersecting point for the solution.

The lines coincide on each other, this indicates that there are number of intersection points on the line since a line consists of infinite points.

Hence, the graph shows that the system of equations have infinite number of solutions.

For equation, x – 2y = 6

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 3x – 6y = 0

Plot the values in a graph and find the intersecting point for the solution.

The equation line x - 2y = 6 will pass through points (0, - 3) and (6,0).

But the equation line 3x - 6y = 0 will pass through x - axis and y - axis, which does not actually intersect the line, x - 2y = 6. Hence, the graph shows that the system of equations have no solutions.

For equation, 2x + 3y = 4

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 4x + 6y = 12

Plot the values in a graph and find the intersecting point for the solution.

The set of equations are parallel to each other in the graph.

Parallel lines never meet each other even if they are extended.

Hence, the graph shows that the system of equations have no solutions.

For equation, 2x + y = 6

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 6x + 3y = 20

Plot the values in a graph and find the intersecting point for the solution.

The set of equations are parallel to each other in the graph.

Parallel lines never meet each other even if they are extended.

Hence, the graph shows that the system of equations have no solutions.

For equation, 2x + y = 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + y = 6

Plot the values in a graph and find the intersecting point for the solution.

Since, the line 2x + y = 6 cuts the line y - axis at A(0,6) and x - axis at B(3,0)

& the line 2x + y = 2 cuts the x - axis at C(1,0) and y - axis at D(0,2).

Thus, it is clear from the graph that ABCD forms a trapezium.

And the coordinates joining this trapezium are (0,6),(3,0),(1,0) and (0,2).

We can find the area of trapezium ABCD.

The formula to calculate area of a trapezium ABCD is:

Area(trap. ABCD) = Area(∆OAB) – Area(∆OCD)

= (1/2 × 3 × 6) – (1/2 × 1 × 2)

[∵ base(∆OAB) = 3 units & height(∆OAB) = 6 units

= 9 - 1 base(∆OCD) = 1 units & height(∆OCD) = 2units]

= 8 sq. units

We have,

x + y = 3 …eq.1

4x – 3y = 26 …eq.2

To solve these equations, we need to make one of the variables in each equation have same coefficient.

Lets multiply eq.1 by 4, so that variable x in both the equations have same coefficient.

Recalling equations 1 & 2,

x + y = 3 [×4]

4x – 3y = 26

⇒ 4x + 4y = 12

4x – 3y = 26

On solving the two equations we get,

7y = - 14

⇒ 7y = - 14

⇒ y = - 2

Substitute y = - 2 in eq.1/eq.2, as per convenience of solving.

Thus, substituting in eq.1, we get

x + ( - 2) = 3

⇒ x = 3 + 2

⇒ x = 5

Hence, we have x = 5 and y = - 2.

We have,

x – y = 3 …eq.1

…eq.2

Let us first simplify eq.2, by taking LCM of denominator,

⇒

⇒ 2x + 3y = 36 …eq.3

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets, multiply eq.1 by 2, so that variable x in both the equations have same coefficient.

Recalling equations 1 & 2,

x – y = 3 [×2

2x + 3y = 36

⇒ 2x – 2y = 6

2x + 3y = 36

On solving we get,

⇒ - 5y = - 30

⇒ y = 6

Substitute y = 6 in eq.1/eq.3, as per convenience of solving.

Thus, substituting in eq.1, we get

x – (6) = 3

⇒ x = 3 + 6

⇒ x = 9

Hence, we have x = 9 and y = 6.

We have,

2x + 3y = 0 …eq.1

3x + 4y = 5 …eq.2

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets, multiply eq.1 by 3 and eq.2 by 2, so that variable x in both the equations have same coefficient.

Recalling equations 1 & 2,

2x + 3y = 0 [×3

3x + 4y = 5 [×2

⇒ 6x + 9y = 0

6x + 8y = 10

On solving the two equations we get,

y = - 10

Substitute y = - 10 in eq.1/eq.2, as per convenience of solving.

Thus, substituting in eq.1, we get

2x + 3( - 10) = 0

⇒ 2x = 30

⇒ x = 15

Hence, we have x = 15 and y = - 10.

We have,

2x – 3y = 13 …eq.1

7x – 2y = 20 …eq.2

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.1 by 2 and eq.2 by 3, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

2x – 3y = 13 [×2]

7x – 2y = 20 [×3]

⇒ 4x – 6y = 26

21x – 6y = 60

On solving the two equations we get,

- 17x = - 34

⇒ x = 2

Substitute x = 2 in eq.1/eq.2, as per convenience of solving.

Thus, substituting in eq.1, we get

2(2) – 3y = 13

⇒ - 3y = 13 – 4

⇒ - 3y = 9

⇒ y = - 3

Hence, we have x = 2 and y = - 3.

Rearranging the equations, we have

3x – 5y = 19 (1)

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

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply (1) by 7 and (2) by 3, so that variable x in both the equations have same coefficient.

(3x – 5y = 19)×7

(-7x + 3y = - 1)×3

21x - 35y = 133 (3)
-21x + 9y = -3 (4)

⇒ -26y = 130

⇒ y = - 5

Substitute y = - 5 in (1)
3x - 5(-5) = 19
⇒ 3x + 25 = 19
⇒ 3x = -6
⇒ x = -2
Hence, x = -2 and y = -5 is the solution of given pair of equations.

Rearranging the equations, we have

2x – y = - 3 …eq.1

3x – 7y = - 10 …eq.2

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.1 by 7, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

2x – y = - 3 [×7

3x – 7y = - 10

⇒ 14x – 7y = - 21

3x – 7y = - 10

On solving the above two equations we get,

⇒ 11x = - 11

⇒ x = - 1

Substitute x = - 1 in eq.1/eq.2, as per convenience of solving.

Thus, substituting in eq.1, we get

2( - 1) – y = - 3

⇒ - 2 – y = - 3

⇒ y = - 2 + 3

⇒ y = 1

Hence, we have x = - 1 and y = 1.

We have,

…eq.1

…eq.2

Let us first simplify eq.1 & eq.2, by taking LCM of denominators,

Eq.1 ⇒

⇒

⇒ 9x – 2y = 108 …eq.3

Eq.2 ⇒

⇒

⇒ 3x + 7y = 105 …eq.4

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.3 by 7 and eq.4 by 2, so that variable y in both the equations have same coefficient.

Recalling equations 3 & 4,

9x – 2y = 108 [×7

3x + 7y = 105 [×2

⇒ 63x – 14y = 756

6x + 14y = 210

On adding the above the two equations we get,

69x + 0 = 966

⇒ 69x = 966

⇒ x = 14

Substitute x = 14 in eq.3/eq.4, as per convenience of solving.

Thus, substituting in eq.4, we get

3(14) + 7y = 105

⇒ 7y = 105 - 42

⇒ 7y = 63

⇒ y = 9

Hence, we have x = 14 and y = 9.

We have,

…eq.1

…eq.2

Let us first simplify eq.1 & eq.2, by taking LCM of denominators,

Eq.1 ⇒

⇒

⇒ 4x + 3y = 132 …eq.3

Eq.2 ⇒

⇒

⇒ 5x – 2y = - 42 …eq.4

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.3 by 2 and eq.4 by 3, so that variable y in both the equations have same coefficient.

Recalling equations 3 & 4,

4x + 3y = 132 [×2

5x – 2y = - 42 [×3

⇒ 8x + 6y = 264

15x – 6y = - 126

23x + 0 = 138

⇒ 23x = 138

⇒ x = 6

Substitute x = 6 in eq.3/eq.4, as per convenience of solving.

Thus, substituting in eq.4, we get

5(6) – 2y = - 42

⇒ 30 – 2y = - 42

⇒ 2y = 30 + 42

⇒ 2y = 72

⇒ y = 36

Hence, we have x = 6 and y = 36.

We have,

4x – 3y = 8 …eq.1

…eq.2

Let us first simplify eq.2 by taking LCM of denominator,

Eq.2 ⇒

⇒ 18x – 3y = 29 …eq.3

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

And it is so that the equations 1 & 3 have variable y having same coefficient already, so we need not multiply or divide it with any number.

Recalling equations 1 & 3,

4x – 3y = 8

18x – 3y = 29

⇒ 4x – 3y = 8

18x – 3y = 29

On solving the above equations we get,

⇒ - 14x = - 21

⇒

⇒

Substitute in eq.1/eq.3, as per convenience of solving.

Thus, substituting in eq.1, we get

4 – 3y = 8

⇒ 6 – 3y = 8

⇒ 3y = 6 – 8

⇒ 3y = - 2

⇒

Hence, we have and

We have,

…eq.1

5x = 2y + 7 or 5x – 2y = 7 …eq.2

Let us first simplify eq.1 by taking LCM of denominator,

Eq.2 ⇒

⇒ 8x – 3y = 12 …eq.3

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.2 by 3 and eq.3 by 2, so that variable y in both the equations have same coefficient.

Recalling equations 2 & 3,

5x – 2y = 7 [×3]

8x – 3y = 12 [×2]

⇒ 15x – 6y = 21

16x – 6y = 24

On solving the above equations we get,

- x – 0 = - 3

⇒ - x = - 3

⇒ x = 3

Substitute x = 3 in eq.2/eq.3, as per convenience of solving.

Thus, substituting in eq.2, we get

5(3) – 2y = 7

⇒ 15 – 2y = 7

⇒ 2y = 15 – 7

⇒ 2y = 8

⇒ y = 4

Hence, we have x = 3 and y = 4

We have,

…eq.1

…eq.2

Let us first simplify eq.1 & eq.2 by taking LCM of denominators,

Eq.1

⇒ 6x + 15y = 8 …eq.3

Eq.2

⇒ 18x – 12y = 5 …eq 4

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.3 by 18 and eq.4 by 6, so that variable x in both the equations have same coefficient.

Recalling equations 3 & 4,

6x + 15y = 8 [×18]

18x – 12y = 5 [×6]

⇒ 108x + 270y = 144

108x – 72y = 30

On solving these two equations we get,

⇒ 342y = 114

⇒

⇒

Substitute in eq.3/eq.4, as per convenience of solving.

Thus, substituting in eq.3, we get

6x + = 8

⇒ 6x + 5 = 8

⇒ 6x = 8 – 5

⇒ 6x = 3

⇒

Hence, we have and

After rearrangement, we have

2x + 3y = - 1 …eq.1

…eq.2

Let us first simplify eq.2 by taking LCM of denominator,

Eq.1 ⇒

⇒ 7 – 4x = 3y

⇒ 4x + 3y = 7 …eq.3

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

And it is so that the equations 1 & 3 have variable y having same coefficient already, so we need not multiply or divide it with any number.

Recalling equations 1 & 3,

2x + 3y = - 1

4x + 3y = 7

On solving these two equations we get,

⇒ x = 4

Substitute x = 4 in eq.1/eq.3, as per convenience of solving.

Thus, substituting in eq.3, we get

4(4) + 3y = 7

⇒ 16 + 3y = 7

⇒ 3y = 7 – 16

⇒ 3y = - 9

⇒ y = - 3

Hence, we have x = 4 and y = - 3

We have

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Lets simplify these equations. We can rewrite them as,

⇒ 4x + 3y = 17 …eq.1

⇒ 7x – 2y = 8 …eq.2

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.1 by 2 & eq.2 by 3, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

4x + 3y = 17 , on multiplying equation with 2

7x – 2y = 8 , on multiplying equation with 3

We get,

8x + 6y = 34

21x – 6y = 24

On solving the equation, we get,

x = 2

Substitute x = 2 in eq.1/eq.2, as per convenience of solving.

Thus, substituting in eq.2, we get

7(2) – 2y = 8

⇒ 14 – 2y = 8

⇒ 2y = 14 – 8

⇒ 2y = 6

⇒ y = 3

Hence, we have x = 2 and y = 3.

We have

0.3x + 0.5y = 0.5

0.5x + 0.7y = 0.74

Lets simplify these equations. We can rewrite them as,

⇒ 3x + 5y = 5 …(i)

⇒ 50x + 70y = 74 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by 14, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

3x + 5y = 5 [×14

50x + 70y = 74

⇒ - 8x = - 4

⇒

⇒

⇒ x = 0.5

Substitute in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

⇒

⇒ 3 + 10y = 10

⇒ 10y = 10 – 3

⇒ 10y = 7

⇒

⇒ y = 0.7

Hence, we have x = 0.5 and y = 0.7

We have

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2

Lets simplify these equations. We can rewrite them,

7(y + 3) – 2(x + 2) = 14

⇒ 7y + 21 – 2x – 4 = 14

⇒ 7y – 2x + 17 = 14

⇒ 2x – 7y = 3 …(i)

4(y – 2) + 3(x – 3) = 2

⇒ 4y – 8 + 3x – 9 = 2

⇒ 3x + 4y – 17 = 2

⇒ 3x + 4y = 19 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3 and eq.(ii) by 2, so that variable x in both the equations have same coefficient.

Recalling equations (i) & (ii),

2x – 7y = 3 [×3

3x + 4y = 19 [×2

⇒ - 29y = - 29

⇒ y = 1

Substitute y = 1 in eq.(i) or eq.(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

2x – 7(1) = 3

⇒ 2x – 7 = 3

⇒ 2x = 7 + 3

⇒ 2x = 10

⇒ x = 5

Hence, we have x = 5 and y = 1

Since, if a = b = c ⇒ a = b & b = c

Thus, we have

6x + 5y = 7x + 3y + 1

2(x + 6y – 1) = 7x + 3y + 1

Lets simplify these equations. We can rewrite them,

6x + 5y = 7x + 3y + 1

⇒ 7x – 6x + 3y – 5y = - 1

⇒ x – 2y = - 1 …(i)

2(x + 6y – 1) = 7x + 3y + 1

⇒ 2x + 12y – 2 = 7x + 3y + 1

⇒ 7x – 2x + 3y – 12y = - 2 – 1

⇒ 5x – 9y = - 3 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 5, so that variable x in both the equations have same coefficient.

Recalling equations (i) & (ii),

x – 2y = - 1 [×5

5x – 9y = - 3

⇒ - y = - 2

⇒ y = 2

Substitute y = 2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

x – 2(2) = - 1

⇒ x – 4 = - 1

⇒ x = - 1 + 4

⇒ x = 3

Hence, we have x = 3 and y = 2

Since, if a = b = c ⇒ a = b & b = c

Thus, we have

and

Lets simplify these equations. We can rewrite them,

⇒ 3(x + y – 8) = 2(x + 2y – 14)

⇒ 3x + 3y – 24 = 2x + 4y – 28

⇒ 3x – 2x + 3y – 4y = - 28 + 24

⇒ x – y = - 4 …(i)

⇒ 3(3x + y – 12) = 11(x + 2y – 14)

⇒ 9x + 3y – 36 = 11x + 22y – 154

⇒ 11x – 9x + 22y – 3y = 154 – 36

⇒ 2x + 19y = 118 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 19, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

x – y = - 4 [×19

2x + 19y = 118

⇒ 21x = 42

⇒ x = 2

Substitute x = 2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

2 – y = - 4

⇒ y = 2 + 4

⇒ y = 6

Hence, we have x = 2 and y = 6

We have

and

Lets simplify these equations. Assuming 1/x = z, we can rewrite them,

⇒ 5z + 6y = 13 …(i)

⇒ 3z + 4y = 7 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3 and eq.(ii) by 5, so that variable z in both the equations have same coefficient.

Recalling equations (i) & (ii),

5z + 6y = 13 [×3

3z + 4y = 7 [×5

⇒ - 2y = 4

⇒ y = - 2

Substitute y = - 2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(ii), we get

3z + 4( - 2) = 7

⇒ 3z – 8 = 7

⇒ 3z = 7 + 8

⇒ 3z = 15

⇒ z = 5

Thus, z = 5 and y = - 2

As z = 1/x,

⇒ 5 = 1/x

⇒ x = 1/5

Hence, we have x = 1/5 and y = - 2

We have

and

Lets simplify these equations. Assuming 1/y = z, we can rewrite them,

⇒ x + 6z = 6 …(i)

⇒ 3x – 8z = 5 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3, so that variable “x” in both the equations have same coefficient.

Recalling equations (i) & (ii),

x + 6z = 6 [×3

3x – 8z = 5

⇒ 26z = 13

⇒ z = 13/26

⇒ z = 1/2

Substitute z = 1/2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

x + 6(1/2) = 6

⇒ x + 3 = 6

⇒ x = 3

Thus, z = 1/2 and x = 3

As z = 1/y,

⇒

⇒ y = 2

Hence, we have x = 3 and y = 2

We have

and

where y≠0

Lets simplify these equations. Assuming 1/y = z, we can rewrite them,

⇒ 2x – 3z = 9 …(i)

⇒ 3x + 7z = 2 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3 and eq.(ii) by 2, so that variable x in both the equations have same coefficient.

Recalling equations (i) & (ii),

2x – 3z = 9 [×3

3x + 7z = 2 [×2

⇒ - 23z = 23

⇒ z = - 1

Substitute z = - 1 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

2x – 3( - 1) = 9

⇒ 2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

Thus, z = - 1 and x = 3

As z = 1/y,

⇒ - 1 = 1/y

⇒ y = - 1

Hence, we have x = 3 and y = - 1

We have

and

where x≠0 and y≠0

Lets simplify these equations. Assuming 1/x = p and 1/y = q, we can rewrite them,

⇒ 3p – q = - 9 …(i)

⇒ 2p + 3q = 5 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

3p – q = - 9 [×3

2p + 3q = 5

⇒ 11p = - 22

⇒ p = - 2

Substitute p = - 2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3( - 2) – q = - 9

⇒ - 6 – q = - 9

⇒ q = 9 – 6

⇒ q = 3

Thus, p = - 2 and q = 3

As p = 1/x,

⇒ - 2 = 1/x

⇒ x = - 1/2

And q = 1/y

⇒ 3 = 1/y

⇒ y = 1/3

Hence, we have x = - 1/2 and y = 1/3

We have

and

where x≠0 and y≠0

Lets simplify these equations. Assuming 1/x = p and 1/y = q, we can rewrite them,

⇒ 9p – 4q = 8 …(i)

⇒ 13p + 7q = 101 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 7 and eq.(ii) by 4, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

9p – 4q = 8 [×7

13p + 7q = 101 [×4

⇒ 115p = 460

⇒ p = 4

Substitute p = 4 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

9(4) – 4q = 8

⇒ 36 – 4q = 8

⇒ 4q = 36 – 8 = 28

⇒ q = 7

Thus, p = 4 and q = 7

As p = 1/x,

⇒ 4 = 1/x

⇒ x = 1/4

And q = 1/y

⇒ 7 = 1/y

⇒ y = 1/7

Hence, we have x = 1/4 and y = 1/7

We have

and

where x≠0 and y≠0

Lets simplify these equations. Assuming 1/x = p and 1/y = q, we can rewrite them,

⇒ 5p – 3q = 1 …(i)

⇒

⇒ 9p + 4q = 30 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 4 and eq.(ii) by 3, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

5p – 3q = 1 [×4

9p + 4q = 30 [×3

⇒ 47p = 94

⇒ p = 2

Substitute p = 2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

5(2) – 3q = 1

⇒ 10 – 3q = 1

⇒ 3q = 10 – 1 = 9

⇒ q = 3

Thus, p = 2 and q = 3

As p = 1/x,

⇒ 2 = 1/x

⇒ x = 1/2

And q = 1/y

⇒ 3 = 1/y

⇒ y = 1/3

Hence, we have x = 1/2 and y = 1/3

We have

and

where x≠0 and y≠0

Lets simplify these equations. Assuming 1/x = p and 1/y = q, we can rewrite them,

⇒ 3p + 2q = 12 …(i)

⇒ 2p + 3q = 13 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 2 and eq.(ii) by 3, so that variable p in both the equations have same coefficient.

Recalling equations (i) & (ii),

3p + 2q = 12 [×2

2p + 3q = 13 [×3

⇒ - 5q = - 15

⇒ q = 3

Substitute q = 3 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3p + 2(3) = 12

⇒ 3p + 6 = 12

⇒ 3p = 12 – 6 = 6

⇒ p = 2

Thus, p = 2 and q = 3

As p = 1/x,

⇒ 2 = 1/x

⇒ x = 1/2

And q = 1/y

⇒ 3 = 1/y

⇒ y = 1/3

Hence, we have x = 1/2 and y = 1/3

We have

4x + 6y = 3xy

and 8x + 9y = 5xy

where x≠0 and y≠0

Lets simplify these equations.

4x + 6y = 3xy

Dividing the equation by xy throughout,

⇒

Assuming p = 1/y and q = 1/x, we get

4p + 6q = 3 …(i)

Also, 8x + 9y = 5xy

Dividing the equation by xy throughout,

⇒

Assuming p = 1/y and q = 1/x, we get

⇒ 8p + 9q = 5 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 2, so that variable p in both the equations have same coefficient.

Recalling equations (i) & (ii),

4p + 6q = 3 [×2

8p + 9q = 5

⇒ 3q = 1

⇒ q = 1/3

Substitute q = 1/3 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

4p + 6(1/3) = 3

⇒ 4p + 2 = 3

⇒ 4p = 3 – 2 = 1

⇒ p = 1/4

Thus, p = 1/4 and q = 1/3

As q = 1/x,

⇒ 1/3 = 1/x

⇒ x = 3

And p = 1/y

⇒ 1/4 = 1/y

⇒ y = 4

Hence, we have x = 3 and y = 4

We have

x + y = 5xy

and 3x + 2y = 13xy

where x≠0 and y≠0

Lets simplify these equations.

x + y = 5xy

Dividing the equation by xy throughout,

⇒

Assuming p = 1/y and q = 1/x, we get

p + q = 5 …(i)

Also, 3x + 2y = 13xy

Dividing the equation by xy throughout,

⇒

Assuming p = 1/y and q = 1/x, we get

⇒ 3p + 2q = 13 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 2, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

p + q = 5 [×2]

3p + 2q = 13

⇒ - p = - 3

⇒ p = 3

Substitute p = 3 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3 + q = 5

⇒ q = 5 – 3

⇒ q = 2

Thus, p = 3 and q = 2

As q = 1/x,

⇒ 2 = 1/x

⇒ x = 1/2

And p = 1/y

⇒ 3 = 1/y

⇒ y = 1/3

Hence, we have x = 1/2 and y = 1/3

We have

and

Lets simplify these equations. Assuming p = 1/(x + y) and q = 1/(x – y),

5p – 2q = - 1 …(i)

Also,

⇒ 15p + 7q = 10 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3, so that variable p in both the equations have same coefficient.

Recalling equations (i) & (ii),

5p – 2q = - 1 [×3

15p + 7q = 10

⇒ - 13q = - 13

⇒ q = 1

Substitute q = 1 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

5p – 2(1) = - 1

⇒ 5p – 2 = - 1

⇒ 5p = 2 – 1 = 1

⇒ p = 1/5

Thus, p = 1/5 and q = 1

As p = 1/(x + y),

⇒

⇒ x + y = 5 …(iii)

And q = 1/(x – y)

⇒

⇒ x – y = 1 …(iv)

Adding equations (iii) and (iv) to obtain x and y,

(x + y) + (x – y) = 5 + 1

⇒ 2x = 6

⇒ x = 3

Putting the value of x in equation (iii), we get

3 + y = 5

⇒ y = 2

Hence, we have x = 3 and y = 2

We have

and

Lets simplify these equations. Assuming p = 1/(x + y) and q = 1/(x – y),

3p + 2q = 2 …(i)

Also,

⇒ 9p – 4q = 1 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3, so that variable p in both the equations have same coefficient.

Recalling equations (i) & (ii),

3p + 2q = 2 [×3

9p – 4q = 1

⇒ 10q = 5

⇒ q = 1/2

Substitute q = 1/2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3p + 2(1/2) = 2

⇒ 3p + 1 = 2

⇒ 3p = 2 – 1 = 1

⇒ p = 1/3

Thus, p = 1/3 and q = 1/2

As p = 1/(x + y),

⇒

⇒ x + y = 3 …(iii)

And q = 1/(x – y)

⇒

⇒ x – y = 2 …(iv)

Adding equations (iii) and (iv) to obtain x and y,

(x + y) + (x – y) = 3 + 2

⇒ 2x = 5

⇒ x = 5/2

Putting the value of x in equation (iii), we get

5/2 + y = 3

⇒ y = 3 – 5/2

⇒ y = 1/2

Hence, we have x = 5/2 and y = 1/2

We have

and

where x≠ - 1 and y≠1

Lets simplify these equations. Assuming p = 1/(x + 1) and q = 1/(y – 1),

5p – 2q = 1/2

10p – 4q = 1 …(i)

Also,

⇒ 10p + 2q = 5/2

⇒ 20p + 4q = 5 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

The variable q in both the equations have same coefficient.

⇒ 30p = 6

⇒ p = 1/5

Substitute p = 1/5 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

10(1/5) – 4q = 1

⇒ 2 – 4q = 1

⇒ 4q = 2 – 1 = 1

⇒ q = 1/4

Thus, p = 1/5 and q = 1/4

As p = 1/(x + 1),

⇒

⇒ x + 1 = 5

⇒ x = 4

And q = 1/(y – 1)

⇒

⇒ y – 1 = 4

⇒ y = 5

Hence, we have x = 4 and y = 5

We have

and

Lets simplify these equations. Assuming p = 1/(x + y) and q = 1/(x – y),

44p + 30q = 10 …(i)

Also,

⇒ 55p + 40q = 13 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 4 and eq.(ii) by 3, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

44p + 30q = 10 [×4

55p + 40q = 13 [×3

⇒ 11p = 1

⇒ p = 1/11

Substitute p = 1/11 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

44(1/11) + 30q = 10

⇒ 4 + 30q = 10

⇒ 30q = 10 – 4 = 6

⇒ q = 1/5

Thus, p = 1/11 and q = 1/5

As p = 1/(x + y),

⇒

⇒ x + y = 11 …(iii)

And q = 1/(x – y)

⇒

⇒ x – y = 5 …(iv)

Adding equations (iii) and (iv) to obtain x and y,

(x + y) + (x – y) = 11 + 5

⇒ 2x = 16

⇒ x = 8

Putting the value of x in equation (iii), we get

8 + y = 11

⇒ y = 11 – 8

⇒ y = 3

Hence, we have x = 8 and y = 3

We have

and

Lets simplify these equations. Assuming p = 1/(x + y) and q = 1/(x – y),

10p + 2q = 4 …(i)

Also,

⇒ 15p – 9q = - 2 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 9 and eq.(ii) by 2, so that variable q in both the equations have same coefficient.

Recalling equations (i) & (ii),

10p + 2q = 4 [×9]

15p – 9q = - 2 [×2]

⇒ 120p = 32

⇒ p = 4/15

Substitute p = 4/15 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(ii), we get

15(4/15) – 9q = - 2

⇒ 4 – 9q = - 2

⇒ 9q = 4 + 2 = 6

⇒ q = 2/3

Thus, p = 4/15 and q = 2/3

As p = 1/(x + y),

⇒

⇒ 4x + 4y = 15 …(iii)

And q = 1/(x – y)

⇒

⇒ 2x – 2y = 3 …(iv)

Multiplying eq.(iv) by 2, we get

4x – 4y = 6 …(v)

and then adding equations (iii) and (v) to obtain x and y,

(4x + 4y) + (4x – 4y) = 6 + 15

⇒ 8x = 21

⇒ x = 21/8

Putting the value of x in equation (iv), we get

2(21/8) – 2y = 3

⇒ 21/4 – 2y = 3

⇒ 2y = 21/4 – 3 = 9/4

⇒ y = 9/8

Hence, we have x = 21/8 and y = 9/8

We have,

71x + 37y = 253 …(i)

37x + 71y = 287 …(ii)

To solve these equations, we need to simplify them.

So, by adding equations (i) and (ii), we get

(71x + 37y) + (37x + 71y) = 253 + 287

⇒ (71x + 37x) + (37y + 71y) = 540

⇒ 108x + 108y = 540

Now dividing it by 108, we get

x + y = 5 …(iii)

Similarly, subtracting equations (i) and(ii),

(71x + 37y) – (37x + 71y) = 253 – 287

⇒ (71x – 37x) + (37y – 71y) = - 34

⇒ 34x – 34y = - 34

Dividing the equation by 34, we get

x – y = - 1 …(iv)

To solve equations (iii) and (iv), we need to make one of the variables (in both the equations) have same coefficient.

Here the variables x & y in both the equations have same coefficients.

⇒ 2x = 4

⇒ x = 2

Substitute x = 2 in eq.(iii)/eq.(iv), as per convenience of solving.

Thus, substituting in eq.(iii), we get

2 + y = 5

⇒ y = 3

Hence, we have x = 2 and y = 3.

We have,

217x + 131y = 913 …(i)

131x + 217y = 827 …(ii)

To solve these equations, we need to simplify them.

So, by adding equations (i) and (ii), we get

(217x + 131y) + (131x + 217y) = 913 + 827

⇒ (217x + 131x) + (131y + 217y) = 1740

⇒ 348x + 348y = 1740

Now dividing it by 348, we get

x + y = 5 …(iii)

Similarly, subtracting equations (i) and (ii),

(217x + 131y) – (131x + 217y) = 913 – 827

⇒ (217x – 131x) + (131y – 217y) = 86

⇒ 86x – 86y = 86

Dividing the equation by 86, we get

x – y = 1 …(iv)

To solve equations (iii) and (iv), we need to make one of the variables (in both the equations) have same coefficient.

Here the variables x & y in both the equations have same coefficients.

⇒ 2x = 6

⇒ x = 3

Substitute x = 3 in eq.(iii)/eq.(iv), as per convenience of solving.

Thus, substituting in eq.(iii), we get

3 + y = 5

⇒ y = 2

Hence, we have x = 3 and y = 2.

We have,

23x – 29y = 98 …(i)

29x – 23y = 110 …(ii)

To solve these equations, we need to simplify them.

So, by adding equations (i) and (ii), we get

(23x – 29y) + (29x – 23y) = 98 + 110

⇒ (23x + 29x) – (29y + 23y) = 208

⇒ 52x – 52y = 208

Now dividing it by 52, we get

x – y = 4 …(iii)

Similarly, subtracting equations (i) and(ii),

(23x – 29y) – (29x – 23y) = 98 – 110

⇒ (23x – 29x) – (29y – 23y) = - 12

⇒ - 6x – 6y = - 12

Dividing the equation by - 6, we get

x + y = 2 …(iv)

To solve equations (iii) and (iv), we need to make one of the variables (in both the equations) have same coefficient.

Here the variables x & y in both the equations have same coefficients.

⇒ 2x + 0 = 6

⇒ 2x = 6

⇒ x = 3

Substitute x = 3 in eq.(iii)/eq.(iv), as per convenience of solving.

Thus, substituting in eq.(iv), we get

3 + y = 2

⇒ y = - 1

Hence, we have x = 3 and y = - 1.

We have

and

Lets simplify these equations. Assuming p = 1/x and q = 1/y,

⇒

2q + 5p = 6 …(i)

Also,

⇒

⇒ 4q – 5p = - 3 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Coefficients of p in both equations are already same.

⇒ 6q = 3

⇒ q = 1/2

Substitute q = 1/2 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(ii), we get

4(1/2) – 5p = - 3

⇒ 2 – 5p = - 3

⇒ 5p = 5

⇒ p = 1

Thus, p = 1 and q = 1/2

As p = 1/x,

⇒ 1 = 1/x

⇒ x = 1

And q = 1/y,

⇒

⇒ y = 2

Hence, we have x = 1 and y = 2

We have

and

Lets simplify these equations. Assuming p = 1/(3x + y) and q = 1/(3x – y),

p + q = 3/4

⇒ 4p + 4q = 3 …(i)

Also,

⇒ p/2 – q/2 = - 1/8

⇒ 4p – 4q = - 1 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

The variable p and q in both the equations have same coefficient.

⇒ 8p = 2

⇒ p = 1/4

Substitute p = 1/4 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(ii), we get

4(1/4) – 4q = - 1

⇒ 1 – 4q = - 1

⇒ 4q = 2

⇒ q = 1/2

Thus, p = 1/4 and q = 1/2

As p = 1/(3x + y),

⇒

⇒ 3x + y = 4 …(iii)

And q = 1/(3x – y)

⇒

⇒ 3x – y = 2 …(iv)

Adding equations (iii) and (iv) to obtain x and y,

(3x + y) + (3x – y) = 4 + 2

⇒ 6x = 6

⇒ x = 1

Putting the value of x in equation (iv), we get

3(1) – y = 2

⇒ 3 – y = 2

⇒ y = 1

Hence, we have x = 1 and y = 1

We have

and

Where x + 2y ≠ 0 and 3x - 2y ≠ 0

Lets simplify these equations. Assuming and

⇒

Multiply it with 6, we get

3p + 10q = -9 …(i)

Also,

⇒

Multiply it with 20, we get

25p – 12q = 61/3

⇒ 75 – 36q = 61 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Multiply equation (i) by 36 and equation (ii) by 10, so that the variables p and q in both the equations have same coefficients.

Recalling equations (i) and (ii),

3p + 10q = -9 [×36

75p – 36q = 61 [×10

⇒ 858p = 286

⇒ p = 286/858 = 1/3

Substitute p = 1/3 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3(1/3) + 10q = -9

⇒ 1 + 10q = -9

⇒ 10q = -9-1 = -10

⇒ q = -1

Thus, p=1/3 and q=-1

As p = 1/(x + 2y),

⇒

⇒ x + 2y = 3 …(iii)

And q = 1/(3x – 2y)

⇒

⇒ 2y – 3x = 1 …(iv)

Subtracting equations (iii) and (iv) to obtain x and y,

(x + 2y) – (2y – 3x) = 3 – 1

⇒ x + 2y – 2y + 3x = 2

⇒ 4x = 2

⇒ x = 1/2

Putting the value of x in equation (iv), we get

2y – 3(1/2) = 1

⇒ 4y – 3 = 2

⇒ 4y = 2 + 3 = 5

⇒ y = 5/4

Hence, we have x=1/2 and y=5/4

We have

and

Lets simplify these equations. Assuming p = 1/(3x + 2y) and q = 1/(3x – 2y),

2p + 3q = 17/5

⇒ 10p + 15q = 17 …(i)

Also,

⇒ 5p + q = 2 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Multiply equation (ii) by 2, so that the variable p in both the equations have same coefficient.

Recalling equations (i) and (ii),

10p + 15q = 17

5p + q = 2 [×2

⇒ 13q = 13

⇒ q = 1

Substitute q = 1 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(ii), we get

5p + 1 = 2

⇒ 5p = 1

⇒ p = 1/5

Thus, p = 1/5 and q = 1

As p = 1/(3x + 2y),

⇒

⇒ 3x + 2y = 5 …(iii)

And q = 1/(3x – 2y)

⇒

⇒ 3x – 2y = 1 …(iv)

Adding equations (iii) and (iv) to obtain x and y,

(3x + 2y) + (3x – 2y) = 5 + 1

⇒ 6x = 6

⇒ x = 1

Putting the value of x in equation (iv), we get

3(1) – 2y = 1

⇒ 3 – 2y = 1

⇒ 2y = 2

⇒ y = 1

Hence, we have x = 1 and y = 1

We have

3(2x + y) = 7xy

And 3(x + 3y) = 11xy

Lets simplify these equations.

3(2x + y) = 7xy

Dividing throughout by xy, and assuming p = 1/x and q = 1/y,

⇒

⇒

6q + 3p = 7 …(i)

Also, 3(x + 3y) = 11xy

Dividing throughout by xy, and assuming p = 1/x and q = 1/y,

⇒

⇒

⇒ 3q + 9p = 11 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Multiply equation (ii) by 2, so that the variable q in both the equations have same coefficient.

Recalling equations (i) and (ii),

6q + 3p = 7

3q + 9p = 11 [×2

⇒ - 15p = - 15

⇒ p = 1

Substitute p = 1 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

6q + 3(1) = 7

⇒ 6q = 7 – 3

⇒ 6q = 4

⇒ q = 2/3

Thus, p = 1 and q = 2/3

As p = 1

⇒ 1 = 1/x

⇒ x = 1

And q = 1/y,

⇒

⇒ y = 3/2

Hence, we have x = 1 and y = 3/2

We have,

x + y = a + b …(i)

ax – by = a² – b²…(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by b, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

x + y = a + b [×b

ax – by = a² – b²

⇒ bx + ax = ab + a²

⇒ (b + a)x = a(a + b)

⇒ x = a

Substitute x = a in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

a + y = a + b

⇒ y = b

Hence, we have x = a and y = b.

We have,

⇒ bx + ay = 2ab …(i)

ax – by = a² – b²…(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by b and (ii) by a, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

bx + ay = 2ab [×b

ax – by = a² – b² [×a

⇒ b²x + a²x = 2ab² + a³ – ab²

⇒ (b² + a²)x = a (2b² + a² – b²)

⇒ (b² + a²)x = a(b² + a²)

⇒ x = a

Substitute x = a in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

ab + ay = 2ab

⇒ ay = 2ab – ab = ab

⇒ y = b

Hence, we have x = a and y = b.

We have,

px + qy = p – q …(i)

qx – py = p + q …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by p and (ii) by q, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

px + qy = p – q [×p]

qx – py = p + q [×q]

⇒ p²x + q²x = p² + q²

⇒ (p² + q²)x = p² + q²

⇒ x = 1

Substitute x = 1 in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

p + qy = p – q

⇒ qy = - q

⇒ y = - 1

Hence, we have x = 1 and y = - 1.

We have,

⇒ bx – ay = 0 …(i)

ax + by = a² + b²…(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by b and (ii) by a, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

bx – ay = 0 [×b

ax + by = a² + b² [×a

b²x + a²x = a³ + ab²

⇒ (b² + a²)x = a (a² + b²)

⇒ (b² + a²)x = a(b² + a²)

⇒ x = a

Substitute x = a in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

ab – ay = 0

⇒ ay = ab

⇒ y = b

Hence, we have x = a and y = b.

We have,

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

⇒ 6ax + 6by = 3a + 2b …(i)

6(bx – ay) = 3b – 2a

⇒ 6bx – 6ay = 3b – 2a …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by a and (ii) by b, so that variable y in both the equations have same coefficient.

Recalling equations 1 & 2,

6ax + 6by = 3a + 2b [×a]

6bx – 6ay = 3b – 2a [×b]

⇒ 6a²x + 6b²x + 0 = 3a² + 3b²

⇒ 6(a² + b²)x = 3(a² + b²)

⇒ 2x = 1

⇒ x = 1/2

Substitute x = 1/2 in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

6a(1/2) + 6by = 3a + 2b

⇒ 3a + 6by = 3a + 2b

⇒ 6by = 2b

⇒ y = 1/3

Hence, we have x = 1/2 and y = 1/3.

We have,

ax – by = a² + b² …(i)

x + y = 2a …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (ii) by b, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

ax – by = a² + b²

x + y = 2a [×b

⇒ ax + bx = a² + b² + 2ab

⇒ (a + b)x = (a + b)²

⇒ x = a + b

Substitute x = (a + b) in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (ii), we get

a + b + y = 2a

⇒ y = 2a – a – b

⇒ y = a – b

Hence, we have x = (a + b) and y = (a – b).

We have,

⇒ b²x – a²y + a²b + ab² = 0

⇒ a²y – b²x = a²b + ab² …(i)

bx – ay = - 2ab

⇒ ay – bx = 2ab …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (ii) by b, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

a²y – b²x = a²b + ab²

ay – bx = 2ab [×b]

⇒ a²y – aby = a²b – ab²

⇒ (a² – ab)y = a²b – ab²

⇒ a(a – b)y = ab(a – b)

⇒ y = b

Substitute y = b in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (ii), we get

a(b) – bx = 2ab

⇒ ab – bx = 2ab

⇒ bx = ab – 2ab = - ab

⇒ x = - a

Hence, we have x = - a and y = b.

We have,

⇒ b²x + a²y = a³b + ab³ …(i)

x + y = 2ab …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (ii) by a², so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

b²x + a²y = a³b + ab³

x + y = 2ab [×a²]

⇒ b²x – a²x = - a³b + ab³

⇒ (b² – a²)x = ab(b² – a²)

⇒ x = ab

Substitute x = ab in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (ii), we get

(ab) + y = 2ab

⇒ y = ab

Hence, we have x = ab and y = ab.

We have,

x + y = a + b …(i)

ax – by = a² – b²…(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by b, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

x + y = a + b [×b

ax – by = a² – b²

⇒ bx + ax = ab + a²

⇒ (b + a)x = a(b + a)

⇒ x = a

Substitute x = a in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

a + y = a + b

⇒ y = b

Hence, we have x = a and y = b.

We have,

a²x + b²y = c² …(i)

b²x + a²y = d²…(ii)

To solve these equations, we need to simplify them.

So, by adding equations (i) and (ii), we get

(a²x + b²y) + (b²x + a²y) = c² + d²

⇒ (a²x + b²x) + (b²y + a²y) = c² + d²

⇒ (a² + b²)x + (a² + b²)y = c² + d²

Now dividing it by (a² + b²), we get

x + y = (c² + d²)/( a² + b²) …(iii)

Similarly, subtracting equations (i) and (ii),

(a²x + b²y) – (b²x + a²y) = c² – d²

⇒ (a²x – b²x) – (b²y – a²y) = c² – d²

⇒ (a² – b²)x – (a² – b²)y = c² – d²

Dividing the equation by (a² – b²), we get

x – y = (c² – d²)/ (a² – b²) …(iv)

To solve equations (iii) and (iv), we need to make one of the variables (in both the equations) have same coefficient.

Here the variables x in both the equations have same coefficients.

⇒

⇒

⇒

⇒

⇒

Substitute in eq.(iii)/eq.(iv), as per convenience of solving.

Thus, substituting in eq.(iii), we get

⇒

⇒

⇒

⇒

Hence, we have and .

We have,

⇒ bx + ay = a²b + ab² …(i)

b²x + a²y = 2a²b²…(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply equation (i) by a, so that variable y in both the equations have same coefficient.

Recalling equations (i) & (ii),

bx + ay = a²b + ab² [×a

b²x + a²y = 2a²b²

⇒ abx – b²x = a³b – a²b²

⇒ b(a – b)x = a²b(a – b)

⇒ x = a²

Substitute x = a² in equations (i)/(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

b(a²) + ay = a²b + ab²

⇒ a²b + ay = a²b + ab²

⇒ ay = ab²

⇒ y = b²

Hence, we have x = a² and y = b².

We have,

x + 2y + 1 = 0 …(i)

2x – 3y – 12 = 0 …(ii)

From equation (i), we get a₁ = 1, b₁ = 2 and c₁ = 1

And from equation (ii), we get a₂ = 2, b₂ = - 3 and c₂ = - 12

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 3 and y = - 2

Thus, x = 3, y = - 2

We have,

3x – 2y + 3 = 0 …(i)

4x + 3y – 47 = 0 …(ii)

From equation (i), we get a₁ = 3, b₁ = - 2 and c₁ = 3

And from equation (ii), we get a₂ = 4, b₂ = 3 and c₂ = - 47

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 5 and y = 9

Thus, x = 5, y = 9

We have,

6x – 5y – 16 = 0 …(i)

7x – 13y + 10 = 0 …(ii)

From equation (i), we get a₁ = 6, b₁ = - 5 and c₁ = - 16

And from equation (ii), we get a₂ = 7, b₂ = - 13 and c₂ = 10

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 6 and y = 4

Thus, x = 6, y = 4

We have,

3x + 2y + 25 = 0 …(i)

2x + y + 10 = 0 …(ii)

From equation (i), we get a₁ = 3, b₁ = 2 and c₁ = 25

And from equation (ii), we get a₂ = 2, b₂ = 1 and c₂ = 10

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ x = 5 and y = - 20

Thus, x = 5, y = - 20

We have,

2x + 5y – 1 = 0 …(i)

2x + 3y – 3 = 0 …(ii)

From equation (i), we get a₁ = 2, b₁ = 5 and c₁ = - 1

And from equation (ii), we get a₂ = 2, b₂ = 3 and c₂ = - 3

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 3 and y = - 1

Thus, x = 3, y = - 1

We have,

2x + y – 35 = 0 …(i)

3x + 4y – 65 = 0 …(ii)

From equation (i), we get a₁ = 2, b₁ = 1 and c₁ = - 35

And from equation (ii), we get a₂ = 3, b₂ = 4 and c₂ = - 65

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 15 and y = 5

Thus, x = 15, y = 5

We have,

7x – 2y – 3 = 0 …(i)

22x – 3y – 16 = 0 …(ii)

From equation (i), we get a₁ = 7, b₁ = - 2 and c₁ = - 3

And from equation (ii), we get a₂ = 22, b₂ = - 3 and c₂ = - 16

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 1 and y = 2

Thus, x = 1, y = 2

We have,

…(i)

…(ii)

By simplifying, we get

From equation (i),

⇒ 5x + 2y – 120 = 0 …(iii)

From equation (ii),

⇒ 4x – y – 57 = 0 …(iv)

From equation (iii), we get a₁ = 5, b₁ = 2 and c₁ = - 120

And from equation (ii) we get a₂ = 4, b₂ = - 1 and c₂ = - 57

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 18 and y = 15

Thus, x = 18, y = 15

We have,

…(i)

…(ii)

Let 1/x = p and 1/y = q. Now,

From equation (i), p + q = 7

⇒ p + q – 7 = 0 …(iii)

From equation (ii), 2p – 3q = 17

⇒ 2p + 3q – 17 = 0 …(iv)

From equation (iii), we get a₁ = 1, b₁ = 1 and c₁ = - 7

And from equation (iv), we get a₂ = 2, b₂ = 3 and c₂ = - 17

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ p = 4 and q = 3

⇒ x = 1/4 and y = 1/3 [∵ p = 1/x and q = 1/y]

Thus, x = 1/4, y = 1/3

We have,

…(i)

…(ii)

Let 1/(x + y) = p and 1/(x - y) = q. Now,

From equation (i), 5p – 2q + 1 = 0 …(iii)

From equation (ii), 15p + 7q – 10 = 0 …(iv)

From equation (iii), we get a₁ = 5, b₁ = - 2 and c₁ = 1

And from equation (iv), we get a₂ = 15, b₂ = 7 and c₂ = - 10

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ p = 1/5 and q = 1

⇒ and [∵ p = 1/(x + y) and q = 1/(x - y)]

To solve these, we need to take reciprocal of these equations. By taking reciprocal, we get

x + y = 5 and x – y = 1

Rearranging them again,

x + y – 5 = 0 …(v)

x – y – 1 = 0 …(vi)

From equation (v), we get a₁ = 1, b₁ = 1 and c₁ = - 5

And from equation (vi), we get a₂ = 1, b₂ = - 1 and c₂ = - 1

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = 3 and y = 2

Thus, x = 3, y = 2

We have,

…(i)

ax – by = 2ab …(ii)

By simplifying, we get

From equation (i),

…(iii)

From equation (ii),

ax – by – 2ab = 0 …(iv)

From equation (iii), we get a₁ = a/b, b₁ = - b/a and c₁ = - (a + b)

And from equation (iv), we get a₂ = a, b₂ = - b and c₂ = - 2ab

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ x = b and y = - a

Thus, x = b, y = - a

We have,

2ax + 3by – (a + 2b) = 0 …(i)

3ax + 2by – (2a + b) = 0 …(ii)

From equation (i), we get a₁ = 2a, b₁ = 3b and c₁ = - (a + 2b)

And from equation (ii), we get a₂ = 3a, b₂ = 2b and c₂ = - (2a + b)

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ and

Thus, ,

We have,

…(i)

…(ii)

Let 1/x = p and 1/y = q. Now,

From equation (i), ap – bq = 0

⇒ ap – bq + 0 = 0 …(iii)

From equation (ii), ab²p – a²bq = (a² + b²)

⇒ ab²p – a²bq – (a² + b²) = 0 …(iv)

From equation (iii), we get a₁ = a, b₁ = - b and c₁ = 0

And from equation (iv), we get a₂ = ab², b₂ = - a²b and c₂ = - (a² + b²)

Using cross multiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒ p = 1/a and q = 1/b

Thus, x = a, y = b [∵ p = 1/x and q = 1/y]

Given: 3x + 5y = 12 – eq 1

5x + 3y = 4 – eq 2

Here,

a₁ = 3, b₁ = 5, c₁ = - 12

a₂ = 5, b₂ = 3, c₂ = - 4

= , =

Here, ≠

∴ given system of equations have unique solutions.

Now,

In – eq 1

3x = 12 – 5y

X =

Substitute x in – eq 2

we get,

5× + 3y = 4

= 4

60 – 25y + 9y = 12

60 – 16y = 12

16y = 60 – 12

16y = 48

y = = 3

∴ y = 3

Now, substitute y in – eq 1

We get,

3x + 5×3 = 12

3x + 15 = 12

3x = 12 – 15

3x = - 3

x = = - 1

∴ x = - 1

∴ x = - 1 and y = 3

Given: 2x – 3y = 17 – eq 1

4x + y = 13 – eq 2

Here,

a₁ = 2, b₁ = - 3, c₁ = 17

a₂ = 4, b₂ = 1, c₂ = 13

= , =

Here, ≠

∴ Given system of equations have unique solutions.

Now,

In – eq 1

2x = 17 + 3y

X =

Substitute x in – eq 2

we get,

4× + y = 13

= 13

68 + 12y + 2y = 26

68 + 14y = 26

14y = 26 – 68

14y = - 42

y = = - 3

∴ y = - 3

Now, substitute y in – eq 1

We get,

2x – 3×( - 3) = 17

2x + 9 = 17

2x = 17 – 9

2x = 8

x = 8/2 = 4

∴ x = 4

∴ x = 4 and y = - 3

Given: ⇒ 2x + 3y = 18 – eq 1

x - 2y = 2 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = 18

a₂ = 1, b₂ = - 2, c₂ = 2

= , =

Here, ≠

∴ Given system of equations have unique solutions.

Now,

In – eq 1

2x = 18 – 3y

X =

Substitute x in – eq 2

we get,

– 2y = 2

= 2

18 – 3y – 4y = 4

18 – 7y = 4

7y = 18 – 4

7y = 14

y = = 2

∴ y = 2

Now, substitute y in – eq 1

We get,

x – 2×(2) = 2

x – 4 = 2

x = 2 + 4

x = 6

∴ x = 6

∴ x = 6 and y = 2

Given: 4x + ky + 8 = 0 – eq 1

x + y + 1 = 0 – eq 2

Here,

a₁ = 4, b₁ = k, c₁ = 8

a₂ = 1, b₂ = 1, c₂ = 1

Given systems of equations has a unique solution

∴

≠

4 ≠ k

∴ k ≠ 4

Given: 2x + 3y - 5 = 0 – eq 1

kx - 6y - 8 = 0 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 5

a₂ = k, b₂ = - 6, c₂ = - 8

Given systems of equations has a unique solution

∴ ≠

2≠

k ≠ 2× 2 = - 4

∴ k ≠ - 4

Given: x - ky = 2 – eq 1

3x + 2y + 5 = 0 – eq 2

Here,

a₁ = 1, b₁ = - k, c₁ = - 2

a₂ = 3, b₂ = 2, c₂ = 5

Given systems of equations has a unique solution

∴

2 ≠ - 3k

- 3k ≠2

Given: 5x - 7y - 5 = 0 – eq 1

2x + ky - 1 = 0 – eq 2

Here,

a₁ = 5, b₁ = - 7, c₁ = - 5

a₂ = 2, b₂ = k, c₂ = - 1

Given systems of equations has a unique solution

∴ ≠

≠

5k ≠ _{( - 7)× 2}

5k ≠₁₄

k ≠

∴

Given: 4x – 5y = k – eq 1

2x – 3y = 12 – eq 2

Here,

a₁ = 4, b₁ = - 5, c₁ = - k

a₂ = 2, b₂ = - 3, c₂ = - 12

Given systems of equations has a unique solution

∴

Here, the system of equations have unique solutions, irrespective of the value of k.

That is solution of the given system of equations is independent of the value of k.

∴ k is any real number

Given: kx + 3y = (k – 3) – eq 1

12x + ky = k – eq 2

Here,

a₁ = k, b₁ = 3, c₁ = k – 3

a₂ = 12, b₂ = k, c₂ = k

Given systems of equations has a unique solution

∴ ≠

≠

K² ≠36

k ≠ √36

∴ k ≠ ±6

∴ k ≠ 6 and k ≠ – 6

_{That is k can be any real number other than - 6 and 6}

∴ k is any real number other than 6 and - 6

Given: 2x – 3y = 5 – eq 1

6x – 9y = 15 – eq 2

Here,

a₁ = 2, b₁ = - 3, c₁ = 5

a₂ = 6, b₂ = - 9, c₂ = 15

Here,

= =

= =

= =

Here,

= =

∴ The given system of equations has infinite number of solutions.

Given: 6x + 5y = 11 – eq 1

9x + y = 21 ⇒ 18x + 15y = 42 – eq 2

Here,

a₁ = 6, b₁ = 5, c₁ = - 11

a₂ = 18, b₂ = 15, c₂ = - 42

Here,

= =

= =

Here,

= ≠

That is give system of equations are parallel lines, that is they don’t have any solutions.

∴ The system of equations has no solution.

(i) Given: kx + 2y = 5 – eq 1

3x – 4y = 10 – eq 2

Here,

a₁ = k, b₁ = 2, c₁ = 5

a₂ = 3, b₂ = - 4, c₂ = 10

Given systems of equations has a unique solution

∴ ≠

≠

- 4k 6

k ≠

∴ k ≠

(ii) Given: kx + 2y = 5 – eq 1

3x – 4y = 10 – eq 2

Here,

a₁ = k, b₁ = 2, c₁ = 5

a₂ = 3, b₂ = - 4, c₂ = 10

Given that systems of equations has no solution

∴ =

Here,

=

Here,

- 4k = 6

∴

(i) Given: x + 2y = 5 – eq 1

3x + ky + 15 = 0 – eq 2

Here,

a₁ = 1, b₁ = 2, c₁ = - 5

a₂ = 3, b₂ = k, c₂ = 15

Given systems of equations has a unique solution

∴ ≠

≠

k ≠6

∴ k ≠ 6

(ii) Given: x + 2y = 5 – eq 1

3x + ky + 15 = 0 – eq 2

Here,

a₁ = 1, b₁ = 2, c₁ = - 5

a₂ = 3, b₂ = k, c₂ = 15

Given that systems of equations has no solution

∴ = ≠

Here,

=

Here,

k = 6

∴ k = 6

(i) Given: x + 2y = 3 – eq 1

5x + ky + 7 = 0 – eq 2

Here,

a₁ = 1, b₁ = 2, c₁ = - 3

a₂ = 5, b₂ = k, c₂ = 7

Given systems of equations has a unique solution

∴ ≠

≠

k ≠10

∴ k ≠ 10

(ii) Given: x + 2y = 3 – eq 1

5x + ky + 7 = 0 – eq 2

Here,

a₁ = 1, b₁ = 2, c₁ = - 3

a₂ = 5, b₂ = k, c₂ = 7

Given that system of equations has no solution

∴ = ≠

Here,

=

Here,

k = 10

∴ k = 10

For the system of equations to have infinitely many solutions

= =

= = which is wrong.

That is, for any value of k the give system of equations cannot have infinitely many solutions.

Given: 2x + 3y = 7 – eq 1

(k - 1)x + (k + 2)y = 3k – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = 7

a₂ = k - 1, b₂ = k + 2, c₂ = 3k

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

2×(k + 2) = 3×(k - 1)

2k + 4 = 3k – 3

3k – 2k = 4 + 3

K = 7

∴ k = 7

Given: 2x + (k – 2)y = k – eq 1

6x + (2k – 1)y = (2k + 5) – eq 2

Here,

a₁ = 2, b₁ = k – 2, c₁ = k

a₂ = 6 , b₂ = 2k – 1, c₂ = 2k + 5

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

2×(2k – 1) = 6×(k - 2)

4k – 2 = 6k – 12

12 – 2 = 6k – 4k

2k = 10

K = 5

∴ k = 5

Given: kx + 3y = (2k + 1) – eq 1

2(k + 1)x + 9y = (7k + 1) – eq 2

Here,

a₁ = k, b₁ = 3, c₁ = - (2k + 1)

a₂ = 2(k + 1) , b₂ = 9 , c₂ = - (7k + 1)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

9k = 6×(k + 1)

9k = 6k + 6

9K – 6k = 6

3k = 6

K =

K = 2

∴ k = 2

Given: 5x + 2y = 2k – eq 1

2(k + 1)x + ky = (3k + 4) – eq 2

Here,

a₁ = 5, b₁ = 2, c₁ = - 2k

a₂ = 2(k + 1) , b₂ = k, c₂ = - (3k + 4)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

5k = 4×(k + 1)

5k = 4k + 4

5K – 4k = 4

k = 4

∴ k = 4

Given: (k – 1)x – y = 5 – eq 1

(k + 1)x + (1 – k)y = (3k + 1) – eq 2

Here,

a₁ = (k - 1), b₁ = - 1, c₁ = - 5

a₂ = (k + 1) , b₂ = (1 - k), c₂ = - (3k + 1)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

(3k + 1) = - 5×(1 - k)

3k + 1 = - 5 + 5k

5K – 3k = 1 + 5

2k = 6

k =

k = 3

∴ k = 3

Given: (k – 3)x + 3y = k – eq 1

kx + ky = 12 – eq 2

Here,

a₁ = (k - 3), b₁ = 3, c₁ = - k

a₂ = k , b₂ = k, c₂ = - 12

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3×( - 12) = - k×(k)

- 36 = - k²

K² = 36

k = √36

k = ±6

k = 6 and k = - 6 – eq 3

Also,

=

K(k - 3) = 3k

K² - 3k = 3k

K² - 6k = 0

K(k - 6) = 0

K = 0 and k = 6 – eq 4

From – eq 3 and – eq 4

k = 6

∴ k = 6

Given: (a – 1)x + 3y = 2 – eq 1

6x + (1 – 2b)y = 6 – eq 2

Here,

a₁ = (a - 1), b₁ = 3, c₁ = - 2

a₂ = 6 , b₂ = (1 - 2b), c₂ = - 6

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3×( - 6) = (1 - 2b)×( - 2)

- 18 = - 2 + 4b

4b = - 18 + 2

4b = - 16

b =

b = - 4

Also,

=

- 6(a - 1) = - 2×6

- 6a + 6 = - 12

- 6a = - 12 - 6

- 6a = - 18

a =

a = 3

∴ a = 3

∴ a = 3, b = - 4

Given: (2a - 1)x + 3y = 5 – eq 1

3x + (b - 1)y = 2 – eq 2

Here,

a₁ = (2a - 1), b₁ = 3, c₁ = - 5

a₂ = 3 , b₂ = (b - 1), c₂ = - 2

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

- 2×(2a - 1) = 3×( - 5)

- 4a + 2 = - 15

- 4a = - 15 - 2

- 4a = - 17

b =

∴ b = -

Also,

=

3( - 2) = - 5×(b - 1)

- 6 = - 5b + 5

5b = 5 + 6

5b = 11

b =

∴b =

∴ a = , b =

Given: 2x - 3y = 7 – eq 1

(a + b)x - (a + b - 3)y = 4a + b – eq 2

Here,

a₁ = 2, b₁ = - 3, c₁ = - 7

a₂ = (a + b), b₂ = - (a + b - 3), c₂ = - (4a + b)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

- 3×( - 4a + b) = - 7× - (a + b - 3)

12a + 3b = 7a + 7b - 21

12a - 7a = - 3b + 7b - 21

5a = 4b - 21

5a – 4b + 21 = 0 eq 3

Also,

=

2× - (4a + b) = - 7×(a + b)

- 8a – 2b = - 7a – 7b

- 8a + 7a = 2b – 7b

- a = - 5b

a = 5b eq 4

substitute – eq 4 in – eq 3

5(5b) – 4b + 21 = 0

25b – 4b + 21 = 0

21b + 21 = 0

b =

b = - 1

substitute ‘b’ in – eq 4

a = 5( - 1)

a = - 5

∴ a = - 5, b = - 1

Given: 2x + 3y = 7 – eq 1

(a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 7

a₂ = (a + b + 1), b₂ = (a + 2b + 2), c₂ = - (4(a + b) + 1)

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3× - (4(a + b) + 1) = - 7×(a + 2b + 2)

- 12a - 12b - 3 = - 7a - 14b - 14

- 12a + 7a - 3 = 12b - 14b - 14

- 5a - 3 = - 2b - 14

5a - 2b - 11 = 0 eq 3

Also,

=

2× - (4(a + b) + 1) = - 7×(a + b + 1)

- 8a – 8b – 2 = - 7a – 7b – 7

- 8a + 7a = 8b – 7b – 7 + 2

- a = b – 5

a + b = 5

a = 5 – b eq 4

substitute – eq 4 in – eq 3

5(5 – b) - 2b - 11 = 0

25 – 5b - 2b - 11 = 0

- 7b + 14 = 0

b =

b = 2

substitute ‘b’ in – eq 4

a = 5 - 2

a = 3

∴a = 3, b = 2

Given: 2x + 3y = 7 – eq 1

(a + b)x + (2a - b)y = 21 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 7

a₂ = (a + b), b₂ = (2a – b), c₂ = - 21

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=

3× - 21 = - 7×(2a - b)

- 63 = - 14a + 7b

14a - 7b - 63 = 0

2a – b – 9 = 0 eq 3

Also,

=

2× - 21 = - 7×(a + b)

- 42 = - 7a – 7b

7a + 7b + 42 = 0

a + b + 6 = 0

a + b = 6

a = 6 – b eq 4

substitute – eq 4 in – eq 3

2(6 – b) – b – 9 = 0

12 – 2b – b – 9 = 0

- 3b + 3 = 0

b =

b = 1

substitute ‘b’ in – eq 4

a = 6 - 1

a = 5

∴ a = 5, b = 1

Given: 2x + 3y = 7 – eq 1

2ax + (a + b)y = 28 – eq 2

Here,

a₁ = 2, b₁ = 3, c₁ = - 7

a₂ = 2a, b₂ = (a + b), c₂ = - 28

Given that system of equations has infinitely many solution

∴ = =

= =

Here,

=