Exercise 3.1 Pair Of Linear Equations In Two Variables Class 10th Mathematics KC Sinha Solution

Exercise 3.1

1. Sudha went to market with her friends. They wanted to eat `gol - gappa' as well…
2. Romila went to a stationary shop and purchased 2 pencils and 3 erasers for Rs.…
3. Present age of father is 30 years more than twice that of his son. After 10…
4. The path of a wheel of train A is given by the equation x + 2y — 4 = 0 and the…
5. The path of highway number 1 and 2 are given by the equations x — y = 1 and 2x +…
6. Person A walks along the path joining points (0, 3) and (1, 3) and person B…
7. Examine which of the following pair of values of x and y is a solution of…
8. Examine which of the following points lie on the graph of the linear equation 5x…
9. 3x + y = 2 6x + 2y = 1 Solve graphically the following system of linear…
10. 2x — 3y + 13 = 0 3x — 2y + 12 = 0 Solve graphically the following system of…
11. 3x + 2y = 14 x — 4y = — 14 Solve graphically the following system of linear…
12. 2x — 3y = 1 3x — 4y = 1 Solve graphically the following system of linear…
13. 2x — y = 9 5x + 2y = 27 Solve graphically the following system of linear…
14. 3y = 5 — x 2x = y + 3 Solve graphically the following system of linear…
15. 3x — 5y = —1 2x — y = —3 Solve graphically the following system of linear…
16. 2x — 6y + 10 = 0 3x — 9y + 15 = 0 Solve graphically the following system of…
17. 3x + y — 11 = 0 x — y — 1 = 0 Solve graphically the following system of linear…
18. 3x — 5y = 19, 3y — 7x + 1 = 0 Does the point (4, 9) lie on any of the lines?…
19. Solve the following system of linear equations graphically: 2x — 3y = 1, 3x —…
20. 2x + 3y = 7, (a + b) x + (2a - b) y = 3(a + b + 1) Find the values of a and b for which…
21. x — 2y = - 3 2x + y = 4 Solve the following system of equations graphically.…
22. 2x + 3y = 8 x — 2y = — 3 Solve the following system of equations graphically.…
23. x + 2y = 5 2x — 3y = — 4 Solve the following system of equations graphically.…
24. x — y + 1 = 0 4x + 3y = 24 Solve the following system of equations…
25. x + 2y = 1 x— 2y = 7 Solve the following system of equations graphically. Also…
26. x + 2y = 1 x — 2y = —7 Solve the following system of equations graphically.…
27. 2x — y = 4 3y — x = 3 Solve the following system of equations graphically.…
28. 2x + 3y — 12 = 0 2x — y — 4 = 0 Solve the following system of equations…
29. 2x — y — 5 = 0 x — y — 3 = 0 Solve the following system of equations…
30. 2x — y — 4 = 0 x + y + 1 = 0 Solve the following system of equations…
31. 3x + y — 5 = 0 2x — y — 5 = 0 Solve the following system of equations…
32. 3x + 2y — 4 = 0 2x — 3y — 7 = 0 Shade the region bounded by the lines and the…
33. 3x — 2y - 1 = 0 2x — 3y + 6 = 0 Shade the region bounded by the lines and the…
34. 2x + y = 6 2x — y = 0 Solve the following pair of linear equations…
35. 2x + 3y = —5 3x — 2y = 12 Solve the following pair of linear equations…
36. 4x — 3y + 4 = 0 4x + 3y — 20 = 0 Solve the following pair of linear equations…
37. 2x + y = 6 2x — y + 2 = 0 Solve the following pair of linear equations…
38. x — y = 1 2x + y = 8 Solve the following pair of linear equations graphically…
39. 3x + y — 11 = 0 x — y — 1 = 0 Solve the following pair of linear equations…
40. x + 2y — 7 = 0 2x — y — 4 = 0 Solve the following pair of linear equations…
41. 4x — y = 4 3x + 2y = 14 Solve the following system of linear equations…
42. x — y = 1 2x + y = 8 Solve the following system of linear equations…
43. 5x — 6y + 30 = 0; 5x + 4y — 20 = 0 Also find the vertices of the triangle…
44. Draw the graphs of the equations 3x — y + 9 = 0 and 3x + 4y — 6 = 0. Also…
45. Draw the graphs of the following equations 3x — 4y + 6 = 0; 3x + y — 9 = 0.…
46. 2y - x = 8 5y - x = 14 y - 2x = 1 Use a single graph paper and draw the graph…
47. y = x Y = 2x x + y = 6 Use a single graph paper and draw the graph of the…
48. y = x 3y = x x + y = 8 Use a single graph paper and draw the graph of the…
49. (2a - 1) x - 3y = 5, 3x + (b - 2) y = 3 Find the values of a and b for which the following…
50. kx + 3y - (k - 3) = 0, 12x + ky - k = 0 Find the values of a and b for which the following…
51. 3x + 4y = 12, (a + b) x + 2 (a - b) y = 5a - 1 Find the values of a and b for which the…
52. (a - 1) x + 3y = 2, 6x + (1 - 2b) y = 6 Find the values of a and b for which the following…
53. 2x + 3y = 7, (a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1 Find the values of a and b for…
54. ax + 3y = a - 2, 12x + ay = a For what value of a, the following system of linear…
55. x + 2y = 5, 3x + ay + 15 = 0 For what value of a, the following system of linear equations…
56. 3x + y = 1, (2a - 1) x + (a - 1) y = 2a + 1 For what value of a, the following system of…
57. (3a + 1) x + 3y - 2 = 0, (a2 + 1) x + (a - 2) y - 5 = 0 For what value of a, the following…
58. cx + 3y - (c - 3) = 0, 12x + cy - c = 0 For what value of c, the following system of…
59. 2x + 3y = 2, (c + 2) x + (2c + 1) y = 2 (c - 1) For what value of c, the following system…
60. x + (c + 1) y = 5, (c + 1) x + 9y = 8c - 1 For what value of c, the following system of…
61. (c - 1) x - y = 5, (c + 1) x + (1 - c) y = 3c + 1 For what value of c, the following…
62. Solve the following system of equations graphically. Also determine the…
63. ax + 2y = 5, 3x + y = 1 Find the value of a for which the following system of…
64. 9x + py - 1 = 0, 3x + 4y - 2 = 0 Find the value of a for which the following…
65. 3x + 2y = 4, ax - y = 3 Find the value of a for which the following system of…
66. 4x + py + 8 = 0, 2x + 2y + 2 = 0 Find the value of a for which the following…
67. 10 students of class X took part in mathematics quiz. If the number of girls is…
68. Form the pair of linear equations in the following problems and find their…
69. Champa went to a 'sale' to purchase some pants and skirts. When her friends…
70. Priyanka purchased 2 pencils and 3 erasers for Rs. 9. Sayeeda purchased 1…

Exercise 3.1

Question 1.

Sudha went to market with her friends. They wanted to eat `gol - gappa' as well as `dahi - bhalla'. The number of plates of gol - gappa taken by them is half that of dahi - bhalla. The cost of one plate of gol - gappa was Rs. 10 and cost of one plate of dahi - bhalla was Rs. 5. She spent Rs. 60. Represent the situation algebraically and graphically.

Answer:

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).

Question 2.

Romila went to a stationary shop and purchased 2 pencils and 3 erasers for Rs. 9. Her friend Sonali saw the new variety of pencils and erasers with Romila and she also bought 4 pencils and 6 erasers of the same kind for Rs. 18. Represent this situation algebraically and graphically.

Answer:

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.

Question 3.

Present age of father is 30 years more than twice that of his son. After 10 years, the age of father will be thrice the age of his son. Represent this situation algebraically and geometrically.

Answer:

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.

Question 4.

The path of a wheel of train A is given by the equation x + 2y — 4 = 0 and the path of a wheel of another train B is given by the equation 2x + 4y — 12 = O. Represent this situation geometrically.

Answer:

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.

Question 5.

The path of highway number 1 and 2 are given by the equations x — y = 1 and 2x + 3y = 12 respectively. Represent these equations geometrically.

Answer:

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)

Question 6.

Person A walks along the path joining points (0, 3) and (1, 3) and person B walks along the path joining points (0, 4) and (1, 5). Represent this situation geometrically.

Answer:

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.

Question 7.

Examine which of the following pair of values of x and y is a solution of equation 4x — 3y + 24 = 0.

(i) x = 0, y = 8 (ii) x = — 6, y = 0

(iii) x = 1, y = — 2 (iv) x = – 3,y = 4

(v) x = 1, y = — 2 (vi) x = — 4, y = 2

Answer:

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

Question 8.

Examine which of the following points lie on the graph of the linear equation 5x — 3y + 30 = 0.

(i) A (— 6, 0) (ii) B (0, 10)

(iii) C (3, — 5) (iv) D (4, 2)

(v) E (— 9, 5) (vi) F (— 3, 5)

(vii) G (— 9, — 5)

Answer:

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

Question 9.

Solve graphically the following system of linear equations if it has unique solution:

3x + y = 2

6x + 2y = 1

Answer:

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 a2 = 6, b2 = 2 and c2 = – 1

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

Question 10.

Solve graphically the following system of linear equations if it has unique solution:

2x — 3y + 13 = 0

3x — 2y + 12 = 0

Answer:

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 a2 = 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.

Question 11.

Solve graphically the following system of linear equations if it has unique solution:

3x + 2y = 14

x — 4y = — 14

Answer:

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 a2 = 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.

Question 12.

Solve graphically the following system of linear equations if it has unique solution:

2x — 3y = 1

3x — 4y = 1

Answer:

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 a2 = 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.

Question 13.

Solve graphically the following system of linear equations if it has unique solution:

2x — y = 9

5x + 2y = 27

Answer:

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 a2 = 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.

Question 14.

Solve graphically the following system of linear equations if it has unique solution:

3y = 5 — x

2x = y + 3

Answer:

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 a2 = 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.

Question 15.

Solve graphically the following system of linear equations if it has unique solution:

3x — 5y = —1

2x — y = —3

Answer:

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 a2 = 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.

Question 16.

Solve graphically the following system of linear equations if it has unique solution:

2x — 6y + 10 = 0

3x — 9y + 15 = 0

Answer:

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 a2 = 3, b2 = – 9 and c2 = 15

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

Question 17.

Solve graphically the following system of linear equations if it has unique solution:

3x + y — 11 = 0

x — y — 1 = 0

Answer:

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 a2 = 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.

Question 18.

Solve the following system of linear equations graphically:

3x — 5y = 19, 3y — 7x + 1 = 0

Does the point (4, 9) lie on any of the lines? Write its equation.

Answer:

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.

Question 19.

Solve the following system of linear equations graphically: 2x — 3y = 1, 3x — 4y = 1 Does the point (3, 2) lie on any of the lines? Write its equation.

Answer:

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

Question 20.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

2x + 3y = 7, (a + b) x + (2a – b) y = 3(a + b + 1)

Answer:

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 a2 = (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

Question 21.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x — 2y = – 3

2x + y = 4

Answer:

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)

Question 22.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

2x + 3y = 8

x — 2y = — 3

Answer:

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).

Question 23.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 5

2x — 3y = — 4

Answer:

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)

Question 24.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x — y + 1 = 0

4x + 3y = 24

Answer:

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)

Question 25.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 1

x— 2y = 7

Answer:

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)

Question 26.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 1

x — 2y = —7

Answer:

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)

Question 27.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y = 4

3y — x = 3

Answer:

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)

Question 28.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x + 3y — 12 = 0

2x — y — 4 = 0

Answer:

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)

Question 29.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y — 5 = 0

x — y — 3 = 0

Answer:

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)

Question 30.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y — 4 = 0

x + y + 1 = 0

Answer:

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)

Question 31.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

3x + y — 5 = 0

2x — y — 5 = 0

Answer:

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)

Question 32.

Solve the following system of linear equations graphically.

3x + 2y — 4 = 0

2x — 3y — 7 = 0

Shade the region bounded by the lines and the x – axis.

Answer:

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).

Question 33.

Solve the following system of linear equations graphically.

3x — 2y – 1 = 0

2x — 3y + 6 = 0

Shade the region bounded by the lines and the x – axis.

Answer:

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).

Question 34.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

2x + y = 6

2x — y = 0

Answer:

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 = 

Question 35.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

2x + 3y = —5

3x — 2y = 12

Answer:

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)

 ()

Question 36.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

4x — 3y + 4 = 0

4x + 3y — 20 = 0

Answer:

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

Question 37.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

2x + y = 6

2x — y + 2 = 0

Answer:

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

Question 38.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and y – axis. Also find the area of the shaded region.

x — y = 1

2x + y = 8

Answer:

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)

Question 39.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and y – axis. Also find the area of the shaded region.

3x + y — 11 = 0

x — y — 1 = 0

Answer:

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

Question 40.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and y – axis. Also find the area of the shaded region.

x + 2y — 7 = 0

2x — y — 4 = 0

Answer:

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)

 ()

Question 41.

Solve the following system of linear equations graphically. Also shade the region bounded by the lines and y – axis.

4x — y = 4

3x + 2y = 14

Answer:

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).

Question 42.

Solve the following system of linear equations graphically. Also shade the region bounded by the lines and y – axis.

x — y = 1

2x + y = 8

Answer:

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).

Question 43.

Solve the following system of linear equations graphically:

5x — 6y + 30 = 0; 5x + 4y — 20 = 0

Also find the vertices of the triangle formed by the two lines and x – axis.

Answer:

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)

Question 44.

Draw the graphs of the equations 3x — y + 9 = 0 and 3x + 4y — 6 = 0.

Also determine the vertices of the triangle formed by the lines and the x – axis.

Answer:

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)

Question 45.

Draw the graphs of the following equations 3x — 4y + 6 = 0; 3x + y — 9 = 0.

Also, determine the coordinates of the vertices of the triangle formed by these lines and the x – axis.

Answer:

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)

Question 46.

Use a single graph paper and draw the graph of the following equations. Obtain the vertices of the triangle so formed:

2y – x = 8

5y – x = 14

y – 2x = 1

Answer:

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)

Question 47.

Use a single graph paper and draw the graph of the following equations. Obtain the vertices of the triangle so formed:

y = x

Y = 2x

x + y = 6

Answer:

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)

Question 48.

Use a single graph paper and draw the graph of the following equations. Obtain the vertices of the triangle so formed:

y = x

3y = x

x + y = 8

Answer:

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)

Question 49.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

(2a – 1) x – 3y = 5, 3x + (b – 2) y = 3

Answer:

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 a2 = 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

Question 50.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

kx + 3y – (k – 3) = 0, 12x + ky – k = 0

Answer:

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 a2 = 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 = k2 – 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

Question 51.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

3x + 4y = 12, (a + b) x + 2 (a – b) y = 5a – 1

Answer:

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 a2 = (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

Question 52.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

(a – 1) x + 3y = 2, 6x + (1 – 2b) y = 6

Answer:

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 a2 = 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

Question 53.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

2x + 3y = 7, (a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1

Answer:

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 a2 = (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

Question 54.

For what value of a, the following system of linear equations has no solutions:

ax + 3y = a – 2, 12x + ay = a

Answer:

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 a2 = 12, b2 = a and c2 = – a

For no solutions,

∴ 

On taking I and II terms, we get

⇒ a2 = 36

⇒ a = √36

⇒a =±6

Question 55.

For what value of a, the following system of linear equations has no solutions:

x + 2y = 5, 3x + ay + 15 = 0

Answer:

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 a2 = 3, b2 = a and c2 = 15

For no solutions,

∴ 

On taking I and II terms, we get

Question 56.

For what value of a, the following system of linear equations has no solutions:

3x + y = 1, (2a – 1) x + (a – 1) y = 2a + 1

Answer:

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 a2 = (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

Question 57.

For what value of a, the following system of linear equations has no solutions:

(3a + 1) x + 3y – 2 = 0, (a2 + 1) x + (a – 2) y – 5 = 0

Answer:

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 a2 = 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

Question 58.

For what value of c, the following system of linear equations has infinite number of solutions:

cx + 3y – (c – 3) = 0, 12x + cy – c = 0

Answer:

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 a2 = 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.

Question 59.

For what value of c, the following system of linear equations has infinite number of solutions:

2x + 3y = 2, (c + 2) x + (2c + 1) y = 2 (c – 1)

Answer:

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 a2 = 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.

Question 60.

For what value of c, the following system of linear equations has infinite number of solutions:

x + (c + 1) y = 5, (c + 1) x + 9y = 8c – 1

Answer:

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.

Question 61.

For what value of c, the following system of linear equations has infinite number of solutions:

(c – 1) x – y = 5, (c + 1) x + (1 – c) y = 3c + 1

Answer:

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.

Question 62.

Solve the following system of equations graphically. Also determine the vertices of the triangle formed by the lines and y – axis.

4x – 5y – 20 = 0, 3x + 5y – 15 = 0

Answer:

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)

Question 63.

Find the value of a for which the following system of equations has unique solution:

ax + 2y = 5, 3x + y = 1

Answer:

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 a2 = 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.

Question 64.

Find the value of a for which the following system of equations has unique solution:

9x + py – 1 = 0, 3x + 4y – 2 = 0

Answer:

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 a2 = 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.

Question 65.

Find the value of a for which the following system of equations has unique solution:

3x + 2y = 4, ax – y = 3

Answer:

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 a2 = 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 .

Question 66.

Find the value of a for which the following system of equations has unique solution:

4x + py + 8 = 0, 2x + 2y + 2 = 0

Answer:

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 a2 = 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.

Question 67.

10 students of class X took part in mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Answer:

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.

Question 68.

Form the pair of linear equations in the following problems and find their solutions graphically. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son.

Answer:

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.

Question 69.

Champa went to a 'sale' to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, "The number of skirts is two less than twice the number of pants purchased. Also the number of skirts is four less than tbur times the number of pants purchased". Find how many pants and skirts Champa bought?

Answer:

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.

Question 70.

Priyanka purchased 2 pencils and 3 erasers for Rs. 9. Sayeeda purchased 1 pencil and two erasers for Rs. 5. Find the cost of one pencil and one eraser.

Answer:

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.