Pair Of Linear Equations In Two Variables

Let no. of plates of gol - gappa = x

and no. of plates of dhai - bhalla = y

Cost of 1 plate gol - gappa = Rs. 10

Cost of 1 plate dhai - bhalla = Rs. 5

Total money spent = Rs. 20

According to the question,

…(1)

10x + 5y = 60 …(2)

From eqⁿ (1), we get

2x – y = 0 …(3)

Now, table for 2x – y = 0

Now, table for 10x + 5y = 60

On plotting points on a graph paper and join them to get a straight line representing .

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 10x + 5y = 60.

Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e. (3,6).

Let the cost of one pencil = Rs x

and cost of one eraser = Rs y

Romila spent = Rs. 9

Sonali spent = Rs. 18

According to the question

2x + 3y = 9 …(1)

4x + 6y = 18 …(2)

Now, table for 2x + 3y = 9

Now, table for 4x + 6y = 18

On plotting points on a graph paper and join them to get a straight line representing 2x + 3y = 9.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 4x + 6y = 18.

Here, we can see that both the lines coincide. This is so, because, both the equations are equivalent, i.e.2(2x + 3y) = 2×9 equation (2) is derived from the other.

Let the present age of son = x year

and the age of his father = y year

According to the question

y = 2x + 30

or, 2x – y = – 30 …(1)

After 10 years,

Age of son = (x + 10)year

Age of father = (y + 10)year

So, According to the question

y + 10 = 3(x + 10)

y + 10 = 3x + 30

y = 3x + 20

or, 3x – y = – 20 …(2)

Now, table for y = 2x + 30

Now, table for y = 3x + 20

On plotting points on a graph paper and join them to get a straight line representing y = 2x + 30.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing y = 3x + 20.

Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e. (10,50).

So, the age of son is 10years and age of his father is 50years.

The given equation is

x + 2y – 4 = 0

and

2x + 4y – 12 = 0

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x + 2y – 4 = 0

Now, table for 2x + 4y – 12 = 0

From the graph, it is clear that lines represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0 are parallel.

The given equation is

x – y = 1

and

2x + 3y = 12

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x – y = 1 or y = x – 1

Now, table for 2x + 3y = 12

From the graph, it is clear that lines represented by the equations x – y = 1 and 2x + 3y – 12 = 0 are intersecting at a point A i.e. (3,2)

The given points are at which Person A walks (0,3) and (1,3)

and the points at which person B walks (0,4) and (1,5)

Now, we plot these points on a same graph as shown in the following figure.

Given equation is

i) Justification

On substituting x = 0, y = 8 in LHS of given equation, we get

LHS = 4(0) – 3(8) + 24 = 0 – 24 + 24 = 0 = RHS

Hence, x = 0, y = 8 is a solution of the equation 4x – 3y + 24 = 0

ii) Justification

On substituting x = – 6, y = 0 in LHS of given equation, we get

LHS = 4( – 6) – 3(0) + 24 = – 24 + 24 = 0 = RHS

Hence, x = – 6, y = 0 is a solution of the equation 4x – 3y + 24 = 0

iii) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

iv) Justification

On substituting x = – 3, y = 4 in LHS of given equation, we get

LHS = 4( – 3) – 3(4) + 24 = – 12 – 12 + 24 = 0 = RHS

Hence, x = – 3, y = 4 is a solution of the equation 4x – 3y + 24 = 0

v) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

vi) Justification

On substituting x = – 4, y = 2 in LHS of given equation, we get

LHS = 4( – 4) – 3(2) + 24 = – 16 – 6 + 24 = – 22 + 24 = 2 ≠ RHS

Hence, x = – 4, y = 2 is not a solution of the equation 4x – 3y + 24 = 0

The given equation is 5x – 3y + 30 = 0

(i) Given A ( – 6,0). Here x = – 6 and y = 0

On substituting x = – 6, y = 0 in LHS of given equation, we get

LHS = 5( – 6) – 3(0) + 30 = – 30 + 30 = 0 = RHS

So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.

Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.

(ii) Given B (0,10). Here x = 0 and y = 10

On substituting x = 0, y = 10 in LHS of given equation, we get

LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS

So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0

Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.

(iii) Given C (3, – 5). Here x = 3 and y = – 5

On substituting x = 3, y = – 5 in LHS of given equation, we get

LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS

So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(iv) Given D (4,2). Here x = 4 and y = 2

On substituting x = 4, y = 2 in LHS of given equation, we get

LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS

So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(v) Given E ( – 9,5). Here x = – 9 and y = 5

On substituting x = – 9, y = 5 in LHS of given equation, we get

LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS

So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(vi) Given F ( – 3,5). Here x = – 3 and y = 5

On substituting x = – 3, y = 5 in LHS of given equation, we get

LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS

So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.

(vii) Given G ( – 9, – 5). Here x = 3 and y = – 5

On substituting x = – 9, y = – 5 in LHS of given equation, we get

LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS

So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.

Or Graphically

Here, we can see through the graph also that Point A, B, F and G lie on the graph of the linear equation 5x – 3y + 30 = 0

The given pair of linear equations is

3x + y = 2 or 3x + y – 2 = 0

and 6x + 2y = 1 or 6x + 2y – 1 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = 1 and c₁ = – 2

and a_{2 =} 6, b₂ = 2 and c₂ = – 1

The lines representing the given pair of linear equations are parallel.

The given pair of linear equations is

2x – 3y + 13 = 0

and 3x – 2y + 12 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = – 3 and c₁ = 13

and a_{2 =} 3, b₂ = – 2 and c₂ = 12

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. ( – 2,3)

Hence, the unique solution is x = – 2 and y = 3.

The given pair of linear equations is

3x + 2y = 14 or 3x + 2y – 14 = 0

and x – 4y = – 14 or x – 4y + 14 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = 2 and c₁ = – 14

and a_{2 =} 1, b₂ = – 4 and c₂ = 14

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. (2,4)

Hence, the unique solution is x = 2 and y = 4.

The given pair of linear equations is

2x – 3y = 1 or 2x – 3y – 1 = 0

and 3x – 4y = 1 or 3x – 4y – 1 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = – 3 and c₁ = – 1

and a_{2 =} 3, b₂ = – 4 and c₂ = – 1

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. ( – 1, – 1)

Hence, the unique solution is x = – 1 and y = – 1.

The given pair of linear equations is

2x – y = 9 or 2x – y – 9 = 0

and 5x + 2y = 27 or 5x + 2y – 27 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = – 1 and c₁ = – 9

and a_{2 =} 5, b₂ = 2 and c₂ = – 27

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. (5,1)

Hence, the unique solution is x = 5 and y = 1.

The given pair of linear equations is

x + 3y = 5 or x + 3y – 5 = 0

and 2x – y = 3 or 2x – y – 3 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 1, b₁ = 3 and c₁ = – 5

and a_{2 =} 2, b₂ = – 1 and c₂ = – 3

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. (2,1)

Hence, the unique solution is x = 2 and y = 1.

The given pair of linear equations is

3x – 5y = – 1 or 3x – 5y + 1 = 0

and 2x – y = – 3 or 2x – y + 3 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = – 5 and c₁ = 1

and a_{2 =} 2, b₂ = – 1 and c₂ = 3

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. ( – 2, – 1)

Hence, the unique solution is x = – 2 and y = – 1.

The given pair of linear equations is

2x – 6y + 10 = 0

and 3x – 9y + 15 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = – 6 and c₁ = 10

and a_{2 =} 3, b₂ = – 9 and c₂ = 15

The lines representing the given pair of linear equations will coincide.

The given pair of linear equations is

3x + y – 11 = 0

and x – y – 1 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = 1 and c₁ = – 11

and a_{2 =} 1, b₂ = – 1 and c₂ = – 1

The lines representing the given pair of linear equations will intersect at a point.

Now, table for 3x + y – 11 = 0 or y = 11 – 3x

Now, table for x – y – 1 = 0 or y = x – 1

Here, the lines intersecting at point B i.e. (3,2)

Hence, the unique solution is x = 3 and y = 2.

The given equation is 3x – 5y = 19

and 3y – 7x + 1 = 0 or 7x – 3y = 1

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

From the graph, it is clear that lines represented by the equations 3x – 5y = 19 and 7x – 3y – 1 = 0 are intersecting at a point C i.e. ( – 2, – 5).

Yes, point (4,9) lie on 3y – 7x + 1 = 0.

The given equation is

2x – 3y = 1

and 3x – 4y = 1

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for or

Now, table for or

Here, the lines intersecting at point F i.e. ( – 1, – 1)

Yes, point (3,2) lie on the line 3x – 4y = 1

Given, pair of equations

2x + 3y = 7

and (a + b)x + (2a – b)y = 3(a + b + 1)

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = 3 and c₁ = – 7

and a_{2 =} (a + b), b₂ = (2a – b) and c₂ = – (a + b + 1)

For infinitely many solutions,

Here,

On taking I and II terms, we get

⇒ 2(2a – b) = 3(a + b)

⇒ 4a – 2b = 3a + 3b

⇒ 4a – 3a – 3b – 2b = 0

⇒ a – 5b = 0 …(1)

On taking I and III terms, we get

⇒ 6(a + b + 1) = 7(a + b)

⇒ 6a + 6b + 6 = 7a + 7b

⇒ 6a – 7a + 6b – 7b = – 6

⇒ – a – b = – 6

⇒ a + b = 6 …(2)

Solving eqⁿ (1) and (2), we get

⇒ b = 1

Now, substituting the value of b in eqⁿ (2), we get

⇒ a + b = 6

⇒ a + 1 = 6

⇒ a = 5

The given equation is

x – 2y = – 3

and 2x + y = 4

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect the x axis are B ( – 3,0) and E (2,0)

The given equation is

2x + 3y = 8

and x – 2y = – 3

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect at x axis are B (4,0) and E ( – 3,0).

The given equation is

x + 2y = 5

and 2x – 3y = – 4

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect the x axis are B (5,0) and E ( – 2,0)

The given equation is

x – y + 1 = 0

and

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point C i.e. (3,4)

The points which intersect the x axis are B ( – 1,0) and E (6,0)

The given equation is

x + 2y = 1

and x – 2y = 7

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point B i.e. (4, – 1.5)

The points which intersect the x axis are B (1,0) and E (7,0)

The given equation is

x + 2y = 1

and x – 2y = – 7

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point B i.e. ( – 3,2)

The points which intersect the x axis are B (1,0) and E ( – 7,0)

The given equation is

2x – y = 4

and 3y – x = 3

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 2x – y = 4 or y = 2x – 4

Now, table for

Here, the lines intersecting at point C, i.e. (3,2)

The point which intersects at y axis are A (0, – 4) and D (0,1)

The given equation is

2x + 3y = 12

and 2x – y – 4 = 0

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at point C, i.e. (3,2)

The points which intersects at y axis is A (0,4) and D (0, – 4)

The given equation is

2x – y = 5

and x – y = 3

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 2x – y = 5 or y = 2x – 5

Now, table for x – y = 3 or y = x – 3

Here, the lines intersecting at point C, i.e. (2, – 1)

The point which intersects at y axis are A (0, – 5) and D (0, – 3)

The given equation is

2x – y = 4

and x + y + 1 = 0

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 2x – y = 4 or y = 2x – 4

Now, table for x + y + 1 = 0 or y = – (x + 1)

Here, the lines intersecting at point C, i.e. (1, – 2)

The point which intersects at y axis are A (0, – 4) and D (0, – 1)

The given equation is

3x + y = 5

and 2x – y = 5

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 3x + y = 5 or y = 5 – 3x

Now, table for 2x – y = 5 or y = 2x + 5

Here, the lines intersecting at point C i.e. (2, – 1)

The point which is intersect at y axis are A (0,5) and D (0, – 5)

The given equation is

3x + 2y = 4

and 2x – 3y = 7

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at a point C i.e. (2, – 1).

The given equation is

3x – 2y = 1

and 2x – 3y = – 6

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines intersecting at a point C, i.e. (3,4).

The given equation is

2x + y = 6

and 2x – y = 0

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 2x + y = 6 or y = 2x – 6

Now, table for 2x – y = 0 or y = 2x

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(),O(0,0) and B(3,0).

Area =

The given equation is

2x + 3y = – 5

and 3x – 2y = 12

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(),B(0) and D(4,0)

()

The given equation is

4x – 3y = – 4

and 4x + 3y = 20

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(), A(0) and D(5,0)

(∵, base = (5–(–1)) = 6)

= 12 sq.units

The given equation is

2x + y = 6

and 2x – y = – 2

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 2x + y = 6 or y = 2x – 6

Now, table for 2x – y = – 2 or y = 2x + 2

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(), A(3,0) and D( – 1,0)

()

= 8 sq.units

The given equation is

x – y = 1

and 2x + y = 8

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x – y = 1 or y = x – 1

Now, table for 2x + y = 8 or y = 2x – 8

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(), B(0, – 1) and E(0,8)

The given equation is

3x + y = 11

and x – y = 1

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 3x + y = 11 or y = 3x – 11

Now, table for x – y = 1or y = x – 1

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(), B(0, 11) and E(0, – 1)

()

= 18 sq.units

The given equation is

x + 2y = 7

and 2x – y = 4

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(),B(0, ) and E(0, – 4)

()

The given equation is

4x – y = 4

and 3x + 2y = 14

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 4x – y = 4 or y = 4x – 4

Now, table for

Here, the lines are intersecting at point C(2,4).

The given equation is

x – y = 1

and 2x + y = 8

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x – y = 1 or y = x – 1

Now, table for 2x + y = 8 or y = 8 – 2x

Here, the lines are intersecting at point C(3, 2).

The given equation is

5x – 6y = – 30

and 5x + 4y = 20

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at point C (0,5).

The coordinates of the vertices of ∆ACD are A( – 6,0), C(0,5)and D(4,0)

The given equation is

3x – y = – 9

and 3x + 4y = 6

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for 3x – y = – 9 or y = 3x + 9

Now, table for

Here, the lines are intersecting at point C ( – 2, 3).

The coordinates of the vertices of ∆ACE are A( – 3,0), C( – 2,3)and E(2,0)

The given equation is

3x – 4y = – 9

and 3x + y = 9

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at point C (2, 3).

The coordinates of the vertices of ∆ACD are A( – 2,0), C(2,3)and D(3,0)

The given equation is

– x + 2y = 8

– x + 5y = 14

and – 2x + y = 1

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Now, table for

The coordinates of the vertices of ∆PQR are P( – 4, 2), Q(2, 5)and R(1, 3)

The given equation is

y = x

y = 2x

and x + y = 6

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for y = x

Now, table for y = 2x

Now, table for x + y = 6 or y = 6 – x

The coordinates of the vertices of ∆PQR are P(0, 0), Q(2, 4)and R(3, 3)

The given equation is

y = x

3y = x

and x + y = 8

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for y = x

Now, table for 3y = x

Now, table for x + y = 8 or y = 8 – x

The coordinates of the vertices of ∆PQR are P(0, 0), Q(4, 4)and R(6, 2)

Given, pair of equations

(2a – 1)x – 3y = 5

and 3x + (b – 2)y = 3

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = (2a – 1), b₁ = – 3 and c₁ = – 5

and a_{2 =} 3, b₂ = b – 2 and c₂ = 3

For infinitely many solutions,

Here,

On taking I and III terms, we get

⇒ 3(2a – 1) = 15

⇒ 6a – 3 = 15

⇒ 6a = 15 + 3

On taking II and III terms, we get

⇒ – 9 = 5(b – 2)

⇒ 5b – 10 = – 9

⇒ 5b = – 9 + 10

Given, pair of equations

kx + 3y – (k – 3) = 0

and 12x + ky – k = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = k, b₁ = 3 and c₁ = – (k – 3)

and a_{2 =} 12, b₂ = k and c₂ = – k

For infinitely many solutions,

Here,

∴ …(1)

On taking I and II terms, we get

⇒ k² = 36

⇒ k = √36

⇒ k = ±6

But k = – 6 not satisfies the last two terms of eqⁿ (1)

On taking II and III terms, we get

⇒ 3k = k(k – 3)

⇒ 3k = k^{2 –} 3k

⇒ k² – 3k – 3k = 0

⇒ k(k – 6) = 0

⇒ k = 0 and 6

Which satisfies the last two terms of eqⁿ (1)

Hence, the required value of k = 0, 6

Given, pair of equations

3x + 4y = 12

and (a + b)x + 2(a – b)y = 5a – 1

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = 4 and c₁ = – 12

and a_{2 =} (a + b), b₂ = 2(a – b) and c₂ = – (5a – 1) = 1 + 5a

and

For infinitely many solutions,

Here,

∴

On taking I and II terms, we get

⇒ 6(a – b) = 4(a + b)

⇒ 6a – 6b = 4a + 4b

⇒ 6a – 4a – 6b – 4b = 0

⇒ 2a – 10b = 0

⇒ a – 5b = 0 …(1)

On taking I and III terms, we get

⇒ 3(5a – 1) = 12(a + b)

⇒ 15a – 3 = 12a + 12b

⇒ 15a – 12a – 12b = 3

⇒ 3a – 12b = 3

⇒ a – 4b = 1 …(2)

Solving eqⁿ (1) and (2), we get

⇒ b = 1

Now, substituting the value of b in eqⁿ (2), we get

⇒ a – 4b = 1

⇒ a – 4 = 1

⇒ a = 1 + 4

⇒ a = 5

Given, pair of equations

(a – 1)x + 3y = 2

and 6x + (1 – 2b)y = 6

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = (a – 1), b₁ = 3 and c₁ = – 2

and a_{2 =} 6, b₂ = 1 – 2b and c₂ = – 6

For infinitely many solutions,

=

Here,

∴

On taking I and III terms, we get

⇒ 3(a – 1) = 6

⇒ 3a – 3 = 6

⇒ 3a = 6 + 3

On taking II and III terms, we get

⇒ 9 = 1 – 2b

⇒ – 2b = 9 – 1

⇒ – 2b = 8

⇒ b = – 4

Given, pair of equations

2x + 3y = 7

and (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = 3 and c₁ = – 7

and a_{2 =} (a + b + 1), b₂ = (a + 2b + 2) and c₂ = – {4(a + b) + 1}

For infinitely many solutions,

=

Here,

∴

On taking I and II terms, we get

⇒ 2(a + 2b + 2) = 3(a + b + 1)

⇒ 2a + 4b + 4 = 3a + 3b + 3

⇒ 2a – 3a – 3b + 4b = 3 – 4

⇒ – a + b = – 1

⇒ a – b = 1 …(1)

On taking I and III terms, we get

⇒ 2{4(a + b) + 1)} = 7(a + b + 1)

⇒ 2(4a + 4b + 1) = 7a + 7b + 7

⇒ 8a – 7a + 8b – 7b = – 2 + 7

⇒ a + b = 5 …(2)

Solving eqⁿ (1) and (2), we get

⇒ a = 3

Now, substituting the value of a in eqⁿ (1), we get

⇒ a – b = 1

⇒ 3 – b = 1

⇒ b = 2

Given, pair of equations

ax + 3y = a – 2

and 12x + ay = a

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = a, b₁ = 3 and c₁ = – (a – 2)

and a_{2 =} 12, b₂ = a and c₂ = – a

For no solutions,

∴

On taking I and II terms, we get

⇒ a² = 36

⇒ a = √36

⇒a =±6

Given, pair of equations

x + 2y = 5

and 3x + ay + 15 = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 1, b₁ = 2 and c₁ = – 5

and a_{2 =} 3, b₂ = a and c₂ = 15

For no solutions,

∴

On taking I and II terms, we get

Given, pair of equations

3x + y = 1

and (2a – 1)x + (a – 1)y = 2a + 1

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = 1 and c₁ = – 1

and a_{2 =} (2a – 1), b₂ = (a – 1) and c₂ = – (2a + 1)

For no solutions,

∴

On taking I and II terms, we get

⇒ 3(a – 1) = 2a – 1

⇒ 3a – 3 = 2a – 1

⇒ 3a – 2a = – 1 + 3

⇒ a = 2

Given, pair of equations

(3a + 1)x + 3y – 2 = 0

and (a² + 1)x + (a – 2)y – 5 = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3a + 1, b₁ = 3 and c₁ = – 2

and a_{2 =} a² + 1, b₂ = a – 2 and c₂ = – 5

For no solutions,

∴

On taking I and II terms, we get

⇒ (3a + 1)(a – 2) = 3(a² + 1)

⇒ 3a² – 6a + a – 2 = 3a² + 3

⇒ – 5a = 2 + 3

⇒ a = – 1

Given, pair of equations

cx + 3y – (c – 3) = 0

and 12x + cy – c = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = c, b₁ = 3 and c₁ = – (c – 3)

and a_{2 =} 12, b₂ = c and c₂ = – c

For infinitely many solutions,

Here,

∴ …(1)

On taking I and III terms, we get

⇒ c² = 36

⇒ c = √36

⇒ c = ±6

But c = – 6 not satisfies the eqⁿ (1)

Hence, the required value of c = 6.

Given, pair of equations

2x + 3y = 2

and (c + 2)x + (2c + 1)y = 2(c – 1)

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 2, b₁ = 3 and c₁ = – 2

and a_{2 =} c + 2, b₂ = 2c + 1 and c₂ = – 2(c – 1)

For infinitely many solutions,

Here,

…(1)

On taking I and III terms, we get

⇒ 2(2c + 1) = 3(c + 2)

⇒ 4c + 2 = 3c + 6

⇒ 4c – 3c = 6 – 2

⇒ c = 4

Hence, the required value of c = 4.

The pair of equations are:

x + (c + 1) y = 5

(c + 1) x + 9y = 8c – 1

These equations can be written as:

x + (c + 1) y – 5 = 0

(c + 1) x + 9y – (8c – 1) = 0

On comparing the given equation with standard form i.e.

a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 1 , b₁ = c + 1 , c₁ = –5

a₂ = c + 1, b₂ = 9 , c₂ = –(8c – 1)

For infinitely many solutions,

So,

So,

⇒ From (I) and (II)

⇒ 9 = (c+1)²

⇒ 9 = c² + 1 + 2c

⇒ 9 – 1 = c² + 2c

⇒ 8 = c² + 2c

⇒ c² + 2c – 8 = 0

Factorize by splitting the middle term,

c² + 4c – 2c – 8 = 0

⇒ c ( c + 4 ) – 2 ( c + 4) = 0

⇒ (c+4) (c–2) = 0

⇒ c = –4, c = 2

From (II) and (III)

⇒ (c+1)(–8c=1) = –5 × 9

⇒ –8c² + c – 8c + 1 = –45

⇒ –8c² + c – 8c + 1 + 45 = 0

⇒ –8c² – 7c + 46 = 0

⇒ 8c² + 7c – 46 = 0

⇒ 8c² – 16c + 23c –46 = 0

⇒ 8c ( c–2) + 23 ( c–2) = 0

⇒ (8c+23) ( c–2) = 0

⇒ c = –23/8 and c = 2

From (I) and (III)

⇒ –8c+1 = –5(c+1)

⇒ –8c + 1 = –5c – 5

⇒ –8c + 5c = –5 –1

⇒ –3c = –6

⇒ c = 2

So the value of c = 2.

The pair of equations are:

(c – 1) x – y = 5

(c + 1) x + (1 – c) y = 3c + 1

These equations can be written as:

(c – 1) x – y – 5 = 0

(c + 1) x + (1 – c) y –( 3c + 1)

On comparing the given equation with standard form i.e.

a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = c – 1 , b₁ = –1 , c₁ = –5

a₂ = c + 1, b₂ = 1 – c , c₂ = –(3c + 1)

For infinitely many solutions,

So,

So,

From (I) and (II)

⇒ (c–1)(1–c) = – (c+1)

⇒ c – c² –1 + c = –c – 1

⇒ c – c² –1 + c + c + 1 = 0

⇒ 3c – c² = 0

⇒ c ( 3 – c ) = 0

⇒ c = 0 , c = 3

From (II) and (III)

⇒ –(–3c–1) = –5(1–c)

⇒ 3c + 1 = –5 + 5c

⇒ 3c + 1 + 5 – 5c =0

⇒ 6 – 2c = 0

⇒ 6 = 2c

⇒ c = 3

From (I) and (III)

⇒ (c–1)(–3c–1) = –5 (c+1)

⇒ –3c² – c + 3c + 1 = –5c – 5

⇒ –3c² – c + 3c + 1 + 5c + 5 = 0

⇒ –3c² + 7c + 6 = 0

⇒ 3c² – 7c – 6 = 0

⇒ 3c² – 9c + 2c – 6 = 0

⇒ 3c ( c–3) + 2 (c–3) = 0

⇒ (3c+2) ( c–3) = 0

⇒ c = –2/3 and c = 3

Hence the value of c is 3.

The given equation is

4x – 5y = 20

and 3x + 5y = 15

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

Here, the lines are intersecting at point C (5, 0).

The coordinates of the vertices of ∆ABC are A(0, – 4), B(0, 3)and C(5,0)

Given, pair of equations

ax + 2y = 5

and 3x + y = 1

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = a, b₁ = 2 and c₁ = – 5

and a_{2 =} 3, b₂ = 1 and c₂ = – 1

For unique solutions,

Here,

I II

On taking I and II terms, we get

⇒ a≠6

Thus, given lines have a unique solution for all real values of a, except 6.

Given, pair of equations

9x + py – 1 = 0

and 3x + 4y – 2 = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 9, b₁ = p and c₁ = – 1

and a_{2 =} 3, b₂ = 4 and c₂ = – 2

For unique solutions,

Here,

∴ 3 ≠ p/4
I II

On taking I and II terms, we get

p ≠ 12

Thus, given lines have a unique solution for all real values of p, except 12.

Given, pair of equations

3x + 2y = 4

and ax – y = 3

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 3, b₁ = – 2 and c₁ = – 4

and a_{2 =} a, b₂ = – 1 and c₂ = – 3

For unique solutions,

Here,

∴

On taking I and II terms, we get

Thus, given lines have a unique solution for all real values of a, except .

Given, pair of equations

4x + py + 8 = 0

and 2x + 2y + 2 = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = 4, b₁ = p and c₁ = 8

and a_{2 =} 2, b₂ = 2 and c₂ = 2

For unique solutions,

Here,

∴

On taking I and II terms, we get

⇒ p≠4

Thus, given lines have a unique solution for all real values of p, except 4.

Let the number of boys = x

and the number of girls = y

Now, table for x + y – 10 = 0

Now, table for x – y + 4 = 0

On plotting points on a graph paper and join them to get a straight line representing x + y – 10 = 0.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – y + 4 = 0.

∴ x = 3, y = 7 is the solution of the pair of linear equations.

Hence, the required number of boys is 3 and girls is 7.

Let the present age of father = x year

and the present age of his son = y year

Two years ago,

Father’s age = (x – 2) year

His son’s age = (y – 2) year

According to the question,

⇒ (x – 2) = 5(y – 2)

⇒ x – 2 = 5y – 10

⇒ x – 5y + 8 = 0 …(1)

After two years,

Father’s age = (x + 2) year

His son’s age = (y + 2) year

According to the question,

⇒ (x + 2) = 3(y + 2) + 8

⇒ x + 2 = 3y + 6 + 8

⇒ x – 3y – 12 = 0 …(2)

Now, table for x – 5y + 8 = 0

Now, table for x – 3y – 12 = 0

On plotting points on a graph paper and join them to get a straight line representing x – 5y + 8 = 0.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – 3y – 12 = 0.

∴ x = 42, y = 10 is the solution of the pair of linear equations.

Hence, the age of father is 42years and age of his son is 10 years.

Let the number of pants = x

and the number of skirts = y

According to the question

Number of skirts = 2(Number of pants) – 2

y = 2x – 2 …(i)

Also, Number of skirts = 4(Number of pants) – 4

y = 4x – 4 …(ii)

Substituting the value of y = 4x – 4 in eqⁿ (i),we get

4x – 4 = 2x – 2

⇒ 4x – 2x – 4 + 2 = 0

⇒ 2x = 2

⇒ x = 1

Now, substitute the value of x in eqⁿ (ii), we get

y = 4(1) – 4 = 0

∴ x = 1, y = 0 is the solution of the pair of linear equations.

We can solve this problem through graphically also

Hence, the number of pants she purchased is 1 and the number of skirts, she purchased is zero i.e., she didn’t buy any skirt.

Let the cost of one pencil = Rs x

and the cost of one eraser = Rs y

According to the question

2x + 3y = 9 …(i)

x + 2y = 5 …(ii)

Now, table for 2x + 3y = 9

Now, table for x + 2y = 5

On plotting points on a graph paper and join them to get a straight line representing 2x + 3y = 9.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x + 2y = 5.

∴ x = 3, y = 1 is the solution of the pair of linear equations.

Hence, the cost of one pencil is Rs 3 and cost of one eraser is Rs 1.

Given equations are

3x – y = 3 …(i)

9x – 3y = 9 …(ii)

From eqⁿ (i), y = 3x – 3 …(iii)

On substituting y = 3x – 3 in eqⁿ (ii), we get

⇒ 9x – 3(3x – 3) = 9

⇒ 9x – 9x + 9 = 9

⇒ 9 = 9

This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.

Given equations are

7x – 15y = 2 …(i)

x + 2y = 3 …(ii)

From eqⁿ (ii), x = 3 – 2y …(iii)

On substituting x = 3 – 2y in eqⁿ (i), we get

⇒ 7(3 – 2y) – 15y = 2

⇒ 21 – 14y – 15y = 2

⇒ 21 – 29y = 2

⇒ – 29y = – 19

Now, on putting in eqⁿ (iii), we get

Thus, and is the required solution.

Given equations are

x + y = 14 …(i)

x – y = 4 …(ii)

From eqⁿ (ii), x = 4 + y …(iii)

On substituting x = 4 + y in eqⁿ (i), we get

⇒ 4 + y + y = 14

⇒ 4 + 2y = 14

⇒ 2y = 14 – 4

⇒ 2y = 10

Now, on putting in eqⁿ (iii), we get

⇒ x = 4 + 5

⇒ x = 9

Thus, x = 9 andy = 5 is the required solution.

Given equations are

0.5x + 0.8y = 3.4 …(i)

0.6x – 0.3y = 0.3 …(ii)

From eqⁿ (ii), 2x – y = 1

y = 2x – 1 …(iii)

On substituting y = 0.2x – 1 in eqⁿ (i), we get

⇒ 0.5x + 0.8(2x – 1) = 3.4

⇒ 0.5x + 1.6x – 0.8 = 3.4

⇒ 2.1x = 3.4 + 0.8

⇒ 2.1x = 4.2

Now, on putting x = 2 in eqⁿ (iii), we get

⇒ y = 2(2) – 1

⇒ y = 4 – 1

⇒ y = 3

Thus, x = 2 and y = 3 is the required solution.

Given equations are

x + y = a – b …(i)

ax – by = a² + b² …(ii)

From eqⁿ (i), x = a – b – y …(iii)

On substituting x = a – b – y in eqⁿ (ii), we get

⇒ a(a – b – y) – by = a² + b²

⇒ a² – ab – ay – by = a² + b²

⇒ – ab – y(a + b) = b²

⇒ – y(a + b) = b² + ab

Now, on putting y = – b in eqⁿ (iii), we get

⇒ x = a – b – ( – b)

⇒ x = a

Thus, x = a and y = – b is the required solution.

Given equations are

x + y = 2m …(i)

mx – ny = m² + n² …(ii)

From eqⁿ (i), x = 2m – y …(iii)

On substituting x = 2m – y in eqⁿ (ii), we get

⇒ m(2m – y) – ny = m² + n²

⇒ 2m² – my – ny = m² + n²

⇒ – y(m + n) = m² – 2m² + n²

⇒ – y(m + n) = – m² + n²

⇒ y = – (n – m) = m – n

Now, on putting y = m – n in eqⁿ (iii), we get

⇒ x = 2m – (m – n)

⇒ x = 2m – m + n

⇒ x = m + n

Thus, x = m + n and y = m – n is the required solution.

Given equations are

…(i)

…(ii)

eqⁿ (i) can be re – written as,

⇒ x + 2y = 0.8×2

⇒ x + 2y = 1.6 …(iii)

On substituting x = 1.6 – 2y in eqⁿ (ii), we get

Now, putting the y = 0.6 in eqⁿ (iii), we get

⇒ x + 2y = 1.6

⇒ x + 2(0.6) = 1.6

⇒ x + 1.2 = 1.6

⇒x = 0.4

Thus, x = 0.4 and y = 0.6 is the required solution.

Given equations are

s – t = 3 …(i)

…(ii)

From eqⁿ (i), we get

⇒ s = 3 + t …(iii)

On substituting s = 3 + t in eqⁿ (ii), we get

⇒ 6 + 2t + 3t = 6×6

⇒ 5t = 36 – 6

Now, putting the t = 6 in eqⁿ (iii), we get

⇒ s = 3 + 6

⇒s = 9

Thus, s = 9 andt = 6 is the required solution.

Given equations are

…(i)

ax – by = a² – b² …(ii)

eqⁿ (i) can be re – written as,

⇒ bx + ay = ab×2

⇒ bx + ay = 2ab

…(iii)

On substituting in eqⁿ (ii), we get

⇒ 2a² b – a² y – b² y = b(a² – b²)

⇒ 2a² b – y(a² + b² ) = a²b – b³

⇒ – y(a² + b² ) = a² b – b³ – 2a²b

⇒ – y(a² + b² ) = – a²b – b³

⇒ – y(a² + b² ) = – b(a² + b²)

Now, on putting y = b in eqⁿ (iii), we get

⇒ x = a

Thus, x = a and y = b is the required solution.

Given equations are

…(i)

x + y = 2ab …(ii)

eqⁿ (i) can be re - written as,

⇒ b² x + a² y = ab × (a² + b²)

…(iii)

Now, from eqⁿ (ii), y = 2ab – x …(iv)

On substituting y = 2ab – x in eqⁿ (iii), we get

⇒ b² x = b³ a – a³b + a²x

⇒ b²x – a²x = b³a – a³b

⇒ (b^{2 –} a²) x = ab(b² – a²)

⇒x = ab

Now, on putting x = ab in eqⁿ (iv), we get

⇒ y = 2ab – ab

⇒ y = ab

Thus, x = ab and y = ab is the required solution.

Given pair of linear equations is

3x – 5y – 4 = 0 …(i)

And 9x = 2y + 7 …(ii)

On multiplying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as

9x – 15y – 12 = 0 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

9x – 15y – 12 – 9x = 0 – 2y – 7

⇒ – 15y + 2y = – 7 + 12

⇒ – 13y = 5

On putting in Eq. (ii), we get

Hence, and , which is the required solution.

Given pair of linear equations is

3x + 4y = 10 …(i)

And 2x – 2y = 2 …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of x equal, we get the equation as

6x + 8y = 20 …(iii)

6x – 6y = 6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

6x – 6y – 6x – 8y = 6 – 20

⇒ – 14y = – 14

⇒ y = 1

On putting y = 1 in Eq. (ii), we get

2x – 2(1) = 2

⇒ 2x – 2 = 2

⇒ x = 2

Hence, x = 2 and y = 1 , which is the required solution.

Given pair of linear equations is

x + y = 5 …(i)

And 2x – 3y = 4 …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

2x + 2y = 10 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

2x + 2y – 2x + 3y = 10 – 4

⇒ 5y = 6

On putting in Eq. (i), we get

x + y = 5

Hence, and , which is the required solution.

Given pair of linear equations is

2x + 3y = 8 …(i)

And 4x + 6y = 7 …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

4x + 6y = 16 …(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

4x + 6y – 4x – 6y = 16 – 7

⇒ 0 = 9

Which is a false equation involving no variable.

So, the given pair of linear equations has no solution i.e. this pair of linear equations is inconsistent.

Given pair of linear equations is

8x + 5y = 9 …(i)

And 3x + 2y = 4 …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of y equal, we get the equation as

16x + 10y = 18 …(iii)

15x + 10y = 20 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

15x + 10y – 16x – 10y = 20 – 18

⇒ – x = 2

⇒ x = – 2

On putting x = – 2 in Eq. (ii), we get

3x + 2y = 4

⇒ 3( – 2) + 2y = 4

⇒ – 6 + 2y = 4

⇒ 2y = 4 + 6

⇒ 2y = 10

Hence, and , which is the required solution.

Given pair of linear equations is

2x + 3y = 46 …(i)

And 3x + 5y = 74 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as

6x + 9y = 138 …(iii)

6x + 10y = 148 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

6x + 10y – 6x – 9y = 148 – 138

⇒ y = 10

On putting y = 10 in Eq. (ii), we get

3x + 5y = 74

⇒ 3x + 5(10) = 74

⇒ 3x + 50 = 74

⇒ 3x = 74 – 50

⇒ 3x = 24

⇒ x = 8

Hence, x = 8 and y = 10 , which is the required solution.

Given pair of linear equations is

0.4x – 1.5y = 6.5 …(i)

And 0.3x + 0.2y = 0.9 …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equation as

1.2x – 4.5y = 19.5 …(iii)

1.2x + 0.8y = 3.6 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

1.2x + 0.8y – 1.2x + 4.5y = 3.6 – 19.5

⇒ 5.3y = – 15.9

⇒ y = – 3

On putting y = – 3 in Eq. (ii), we get

0.3x + 0.2y = 0.9

⇒ 0.3x + 0.2( – 3) = 0.9

⇒ 0.3x – 0.6 = 0.9

⇒ 0.3x = 1.5

⇒ x = 1.5/0.3

⇒ x = 5

Hence, x = 5 and y = – 3 , which is the required solution.

Given pair of linear equations is

√2 x – √3 y = 0 …(i)

And √5 x + √2 y = 0 …(ii)

On multiplying Eq. (i) by √2 and Eq. (ii) by √3 to make the coefficients of y equal, we get the equation as

2x – √6 y = 0 …(iii)

√15 x + √6 y = 0 …(iv)

On adding Eq. (iii) and (iv), we get

2x – √6 y + √15 x + √6 y = 0

⇒ 2x + √15 x = 0

⇒ x(2 + √15) = 0

⇒ x = 0

On putting x = 0 in Eq. (i), we get

√2 x – √3 y = 0

⇒ √2(0) – √3 y = 0

⇒ – √3 y = 0

⇒ y = 0

Hence, x = 0 and y = 0 , which is the required solution.

Given pair of linear equations is

2x + 5y = 1 …(i)

And 2x + 3y = 3 …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

2x + 3y – 2x – 5y = 3 – 1

⇒ – 2y = 2

⇒ y = – 1

On putting y = – 1 in Eq. (ii), we get

2x + 3( – 1) = 3

⇒ 2x – 3 = 3

⇒ 2x = 6

⇒ x = 6/2

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (ii) by 2 to make the coefficients of equal, we get the equation as

…(iii)

On adding Eq. (i) and Eq. (ii), we get

⇒ x = 2

On putting in Eq. (ii), we get

⇒ y = – 3

Hence, x = 2 and y = – 3 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (ii) by to make the coefficients of equal, we get the equation as

…(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

⇒ y = 15

On putting y = 15 in Eq. (ii), we get

⇒ x = 18

Hence, x = 18 and y = 15 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 3 to make the coefficients of equal, we get the equation as

…(iii)

On subtracting Eq. (ii) from Eq. (iii), we get

⇒ y = 2

On putting y = 2 in Eq. (i), we get

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Given pair of linear equations is

37x + 43y = 123 …(i)

And 43x + 37y = 117 …(ii)

On multiplying Eq. (i) by 43 and Eq. (ii) by 37 to make the coefficients of x equal, we get the equation as

1591x + 1849y = 5289 …(iii)

1591x + 1369y = 4329 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 1591x + 1369y – 1591x – 1849y = 4329 – 5289

⇒ – 480y = – 960

⇒ y = 2

On putting y = 2 in Eq. (ii), we get

⇒ 43x + 37(2) = 117 ⇒ 43x + 74 = 117

⇒ 43x = 117 – 74

⇒ 43x = 43

⇒ x = 1

Hence, x = 1 and y = 2 , which is the required solution.

Given pair of linear equations is

217x + 131y = 913 …(i)

And 131x + 217y = 827 …(ii)

On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as

28427x + 17161y = 119603 …(iii)

28427x + 47089y = 179459 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603

⇒ 47089y – 17161y = 179459 – 119603

⇒ 29928y = 59856

⇒ y = 2

On putting y = 2 in Eq. (ii), we get

⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827

⇒ 131x = 393

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Given pair of linear equations is

99x + 101y = 499 …(i)

And 101x + 99y = 501 …(ii)

On multiplying Eq. (i) by 101 and Eq. (ii) by 99 to make the coefficients of x equal, we get the equation as

9999x + 10201y = 50399 …(iii)

9999x + 9801y = 49599 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 9999x + 9801y – 9999x – 10201y = 49599 – 50399

⇒ 9801y – 10201y = 49599 – 50399

⇒ – 400y = – 800

⇒ y = 2

On putting y = 2 in Eq. (i), we get

⇒ 99x + 101(2) = 499 ⇒ 99x + 202 = 499

⇒ 99x = 297

⇒ x = 297/99

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Given pair of linear equations is

29x – 23y = 110 …(i)

And 23x – 29y = 98 …(ii)

On multiplying Eq. (i) by 23 and Eq. (ii) by 29 to make the coefficients of x equal, we get the equation as

667x – 529y = 2530 …(iii)

667x – 841y = 2842 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 667x – 841y – 667x + 529y = 2842 – 2530

⇒ – 312y = 312

⇒ y = – 1

On putting y = 2 in Eq. (ii), we get

⇒ 29x – 23( – 1) = 110 ⇒ 29x + 23 = 110

⇒ 29x = 110 – 23

⇒ 29x = 87

⇒ x = 3

Hence, x = 3 and y = – 1 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

Adding Eq. (i) and Eq. (ii), we get

On putting in Eq. (ii), we get

Hence, and , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 5 and Eq. (ii) by 2 to make the coefficients of equal, we get the equation as

…(iii)

…(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

On putting in Eq. (ii), we get

Hence, and , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of equal, we get the equation as

…(iii)

…(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

On putting in Eq. (ii), we get

Hence, and , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (ii) by 3 to make the coefficients of equal, we get the equation as

…(iii)

On subtracting Eq. (i) from Eq. (iii), we get

⇒ y = b

On putting y = b in Eq. (ii), we get

⇒ x = – a

Hence, x = – a and y = b , which is the required solution.

Given pair of linear equations is

Or 2x + 5y = 6xy …(i)

And

Or 4x – 5y = – 3xy …(ii)

On adding Eq. (i) and Eq. (ii), we get

2x + 5y + 4x – 5y = 6xy – 3xy

⇒ 6x = 3xy

⇒ y = 2 and x = 0

On putting y = 2 in Eq. (ii), we get

2x + 5(2) = 6xy

⇒ 2x + 10 = 6x(2)

⇒ 2x + 10 = 12x ⇒ 2x – 12x = – 10

⇒ – 10x = – 10

⇒ x = 1

On putting x = 0 , we get y = 0

Hence, x = 0,1 and y = 0,2 , which is the required solution.

Given pair of linear equations is

x + y = 2xy …(i)

And x – y = 6xy …(ii)

On adding Eq. (i) and Eq. (ii), we get

x + y + x – y = 2xy + 6xy

⇒ 2x = 8xy

On putting in Eq. (ii), we get

⇒ 4x – 1 = 6x ⇒ – 1 = 6x – 4x

⇒ – 1 = 2x

On putting x = 0 , we get y = 0

Hence, and , which is the required solution.

Given pair of linear equations is

5x + 3y = 19xy …(i)

And 7x – 2y = 8xy …(ii)

On multiplying Eq. (i) by 2 and Eq. (ii) by 3 to make the coefficients of y equal, we get the equation as

10x + 6y = 38xy …(iii)

And 21x – 6y = 24xy …(iv)

On adding Eq. (i) and Eq. (ii), we get

10x + 6y + 21x – 6y = 38xy + 24xy

⇒ 31x = 62xy

On putting in Eq. (ii), we get

7x – 2y = 8xy

⇒ 7x – 1 = 4x ⇒ – 1 = 4x – 7x

⇒ – 1 = – 3x

On putting x = 0 , we get y = 0

Hence, and , which is the required solution.

Given pair of linear equations is

x + y = 7xy …(i)

And 2x – 3y = – xy …(ii)

On multiplying Eq. (i) by 2 to make the coefficients of x equal, we get the equation as

2x + 2y = 14xy …(iii)

On subtracting Eq. (ii) and Eq. (iii), we get

2x + 2y – 2x + 3y = 14xy + xy

⇒ 2y + 3y = 15xy

⇒ 5y = 15xy

On putting in Eq. (ii), we get

2x – 3y = – xy

On putting x = 0 , we get x = 0

Hence, and , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 7 and Eq. (ii) by to make the coefficients equal of first term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

…(a)

On multiplying Eq. (ii) by to make the coefficients equal of second term, we get the equation as

…(v)

On substracting Eq. (i) from Eq. (iv), we get

…(b)

From Eq. (a) and (b), we get

⇒ 2(4 + 2y) = 3(7 – 3y)

⇒ 8 + 4y = 21 – 9y

⇒ 4y + 9y = 21 – 8

⇒ 13y = 13

⇒ y = 1

On putting the value of y = 1 in Eq. (b), we get

Hence, x = 2 and y = 1 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients equal of first term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

⇒ y + 1 = 3

⇒ y = 3 – 1

⇒ y = 2

On putting the value of y = 2 in Eq. (ii), we get

⇒ x – 1 = 2

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 4 and Eq. (ii) by 3 to make the coefficients equal of second term, we get the equation as

…(iii)

…(iv)

On substracting Eq. (iii) from Eq. (iv), we get

⇒ x + y = 11 …(a)

On putting the value of x + y = 11 in Eq. (1), we get

⇒ 6(x – y) = 30

⇒ x – y = 5 …(b)

Adding Eq. (a) and (b), we get

⇒ 2x = 16

⇒ x = 8

On putting value of x = 8 in eq. (a), we get

8 + y = 11

⇒ y = 3

Hence, x = 8 and y = 3 , which is the required solution.

Given pair of linear equations is

…(i)

And …(ii)

On multiplying Eq. (i) by 3 to make the coefficients equal of second term, we get the equation as

…(iii)

On adding Eq. (ii) and Eq. (iii), we get

⇒ x – 1 = 3

⇒ x = 3 + 1

⇒ x = 4

On putting the value of x = 4 in Eq. (ii), we get

⇒ (y – 2) = 3

⇒ y = 3 + 2

⇒ y = 5

Hence, x = 4 and y = 5 , which is the required solution.

Let the present age of father i.e. Aftab = x yr

And the present age of his daughter = y yr

Seven years ago,

Aftab’s age = (x – 7)yr

Daughter’s age = (y – 7)yr

According to the question,

(x – 7) = 7(y – 7)

⇒ x – 7 = 7y – 49

⇒ x – 7y = – 42 …(i)

After three years,

Aftab’s age = (x + 3)yr

Daughter’s age = (y + 3)yr

According to the question,

(x + 3) = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x – 3y = 6 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 3y – x + 7y = 6 – ( – 42)

⇒ – 3y + 7y = 6 + 42

⇒ 4y = 48

⇒ y = 12

On putting y = 12 in Eq. (ii) we get

x – 3(12) = 6

⇒ x – 36 = 6

⇒ x = 42

Hence, the age of Aftab is 42years and age of his daughter is 12years.

Let the present age of Nuri = x yr

And the present age of Sonu = y yr

Five years ago,

Nuri’s age = (x – 5)yr

Sonu’s age = (y – 5)yr

According to the question,

(x – 5) = 3(y – 5)

⇒ x – 5 = 3y – 15

⇒ x – 3y = – 10 …(i)

After ten years,

Aftab’s age = (x + 10)yr

Daughter’s age = (y + 10)yr

According to the question,

(x + 10) = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x – 2y = 10 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 2y – x + 3y = 10 – ( – 10)

⇒ – 2y + 3y = 10 + 10

⇒ y = 20

On putting y = 20 in Eq. (i) we get

x – 3(20) = – 10

⇒ x – 60 = – 10

⇒ x = 50

Hence, the age of Nuri is 50 years and age of Sonu is 20 years.

Let the one number = x

And the other number = y

According to the question,

x – y = 26 …(i)

and x = 3y

or x – 3y = 0 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 3y – x + y = 0 – 26

⇒ – 3y + y = – 26

⇒ – 2y = – 26

⇒ y = 13

On putting y = 13 in Eq. (ii) we get

x – 3(13) = 0

⇒ x – 39 = 0

⇒ x = 39

Hence, the two numbers are 39 and 13.

Given, pair of equations is

8x + 5y – 9 = 0 and 3x + 2y – 4 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = – 2

On taking II and III ratio, we get

⇒ y = 5

Given, pair of equations is

2x + 3y – 46 = 0 and 3x + 5y – 74 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 8

On taking II and III ratio, we get

⇒ y = 10

Given, pair of equations is

x + 4y + 9 = 0 and 5x – 3y – 1 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = – 1

On taking II and III ratio, we get

⇒ y = – 2

Given, pair of equations is

2x + 3y – 7 = 0 and 6x + 5y – 11 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Given, pair of equations is

Let uand v

So, Eq. (1) and (2) reduces to

2u + 3v – 13 = 0

5u – 4v + 2 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ u = 2

On taking II and III ratio, we get

⇒ v = 3

So, u

and v

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 18

On taking II and III ratio, we get

⇒ y = 15

Given, pair of equations is

ax + by = a – b ⇒ ax + by –(a – b) = 0

bx – ay = a + b ⇒ bx –ay – (a + b) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 1

On taking II and III ratio, we get

⇒ y = – 1

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

x = a²

On taking II and III ratio, we get

⇒ y = b²

Given, pair of equations is

x – y = a + b ⇒ x – y –(a + b) = 0

ax + by = a² –b²⇒ ax + by – (a² –b² ) = 0

By cross – multiplication method, we have

On taking I and III ratio, we get

⇒ x = a

On taking II and III ratio, we get

⇒ y = – b

Given, pair of equations is

And

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒x = 2a

On taking II and III ratio, we get

⇒ y = – 2b

2ax + 3by = a + 2b

3ax + 2by = 2a + b

Given, pair of equations is

2ax + 3by = a + 2b ⇒ 2ax + 3by –(a + 2b) = 0

3ax + 2by = 2a + b ⇒ 3ax + 2by –(2a + b) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

ax + by =

Given, pair of equations is

ax + by = a² –b²⇒ ax + by – (a² –b² ) = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = a

On taking II and III ratio, we get

⇒ y = b

The given system of equations can be re - written as

ax + ay + bx – by – a² + ab – b² = 0

⇒(a + b)x + (a – b)y – (a² – ab + b² ) = 0 …(1)

and ax + ay – bx + by – a² – ab – b² = 0

⇒ (a – b)x + (a + b)y – (a² + ab + b² ) = 0 …(2)

Now, by cross – multiplication method, we have

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Given pair of linear equations

x – 3y – 7 = 0

3x – 3y – 15 = 0

⇒ x – y – 5 = 0 …(ii)

As we can see that a₁ = 1, b₁ = – 3 and c₁ = – 7

and a_{2 =} 1, b₂ = – 1 and c₂ = – 5

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 4

On taking II and III ratio, we get

⇒ y = – 1

Given pair of linear equations

2x + y = 5

3x + 2y = 8

As we can see that a₁ = 2, b₁ = 1 and c₁ = – 5

and a_{2 =} 3, b₂ = 2 and c₂ = – 8

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 2

On taking II and III ratio, we get

⇒ y = 1

Given pair of linear equations

3x – 5y = 20

6x – 10y = 40

⇒ 3x – 5y = 20 …(ii)

As we can see that a₁ = 3, b₁ = – 5 and c₁ = – 20

and a_{2 =} 3, b₂ = – 5 and c₂ = – 20

∴ Given pair of equations has infinitely many solutions.

Given pair of linear equations

x – 3y – 3 = 0

3x – 9y – 2 = 0

As we can see that a₁ = 1, b₁ = – 3 and c₁ = – 3

and a_{2 =} 3, b₂ = – 9 and c₂ = – 2

∴ Given pair of equations has no solution

Given pair of linear equations

x + y = 2

2x + 2y = 4

⇒ x + y – 2 = 0

As we can see that a₁ = 1, b₁ = 1 and c₁ = – 2

and a_{2 =} 1, b₂ = 1 and c₂ = – 2

∴ Given pair of equations has infinitely many solution

Given pair of linear equations

x + y = 2

2x + 2y = 6

⇒ x + y – 3 = 0

As we can see that a₁ = 1, b₁ = 1 and c₁ = – 2

and a_{2 =} 1, b₂ = 1 and c₂ = – 3

∴ Given pair of equations has no solution

Given, pair of equations is

…(1)

…(2)

Let u and v

Now, the Eq. (1) and (2) reduces to

5u + 2v + 1 = 0

15u + 7v – 10 = 0

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

⇒ v = 13

So, u …(a)

and v …(b)

On adding Eq. (a) and (b), we get

On putting the value of in Eq. (a), we get

Given, pair of equations is

ax – ay = 2

(a – 1)x + (a + 1)y = 2(a² + 1)

By cross - multiplication method, we have

On taking I and III ratio, we get

On taking II and III ratio, we get

Let the cost of one pencil = Rs x

and cost of one eraser = Rs y

According to the question

2x + 3y = 9 …(1)

4x + 6y = 18

⇒2(2x + 3y) = 18

⇒2x + 3y = 9 …(2)

As we can see From Eq. (1) and (2)

∴Given pair of linear equations has infinitely many solutions.

Given paths traced by the wheel of two trains are

x + 2y – 4 = 0 …(i)

2x + 4y – 12 = 0

⇒ x + 2y – 6 = 0 …(ii)

As we can see that a₁ = 1, b₁ = 2 and c₁ = – 4

and a_{2 =} 1, b₂ = 2 and c₂ = – 6

∴ Given pair of equations has no solution

Hence, two paths will not cross each other.

Given ratio of incomes = 9:7

And the ratio of their expenditures = 4:3

Saving of each person = Rs. 2000

Let incomes of two persons = 9x and 7x

And their expenditures = 4y and 3y

According to the question,

9x – 4y = 2000

⇒9x – 4y – 2000 = 0 …(i)

7x – 3y = 2000

⇒7x – 3y – 2000 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratios, we get

⇒ x = 2000

On taking II and III ratios, we get

⇒ y = 4000

Hence, the monthly incomes of two persons are 9(2000) = Rs18000 and 7(2000) = Rs14000

Let unit’s digit = y

and the ten’s digit = x

So, the original number = 10x + y

The sum of the number = 10x + y

The sum of the digit = x + y

reversing number = x + 10y

According to the question,

10x + y + x + 10y = 66

⇒11x + 11y = 66

⇒ x + y = 6 …(i)

x – y = 2 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get

⇒ x = 4

On taking II and III ratio, we get

⇒ y = 2

So, the original number = 10x + y

= 10(4) + 2

= 42

Reversing the number = x + 10y

= 24

Hence, the two digit number is 42 and 24. These are two such numbers.

Let the numerator = x

and the denominator = y

So, the fraction

According to the question,

Condition I:

⇒ x + 1 = y – 1

⇒ x – y = – 2

⇒ x – y + 2 = 0 …(i)

Condition II:

⇒ 2x = y + 1

⇒ 2x – y = 1

⇒ 2x – y – 1 = 0 …(ii)

By cross - multiplication method, we have

On taking I and III ratio, we get