RS Aggarwal Solutions Chapter 3 Linear Equations in Two Variables Exercise - 3D Class 10 Maths

 Chapter Name RS Aggarwal Chapter 3 Linear Equations in Two Variables Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 3AExercise 3BExercise 3CExercise 3EExercise 3F Related Study NCERT Solutions for Class 10 Maths

Exercise 3D Solutions

1. Show that the following system of equations has a unique solution:

3x + 5y = 12,

5x + 3y = 4.

Also, find the solution of the given system of equations.

Solution

The given system of equations is:

3x + 5y = 12

5x + 3y = 4

These equations are of the forms:

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = 3, b1 = 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4

For a unique solution, we must have:

a1/a2 ≠ b1/b2 i.e., 3/5 ≠ 5/3

Hence, the given system of equations has a unique solution.

Again, the given equations are:

3x + 5y = 12 ...(i)

5x + 3y = 4 ...(ii)

On multiplying (i) by 3 and (ii) by 5, we get:

9x + 15y = 36 ….(iii)

25x + 15y = 20 …(iv)

On subtracting (iii) from (iv), we get:

16x = -16

⇒ x = -1

On substituting x = -1 in (i), we get:

3(-1) + 5y = 12

⇒ 5y = (12 + 3) = 15

⇒ y = 3

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

2. Show that the following system of equations has a unique solution:

2x - 3y = 17,

4x + y = 13.

Also, find the solution of the given system of equations.

Solution

The given system of equations is:

2x - 3y - 17 = 0 ….(i)

4x + y - 13 = 0 …..(ii)

The given equations are of the form

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13

Now,

a1/a2 = 2/4 = 1/2 and b1/b2 = -3/1 = - 3

Since, a1/a2 ≠ b1/b2, therefore the system of equations has unique solution.

Using cross multiplication method, we have

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

⇒ x/(-3(-13) – 1 × (-17) = y/(-17 × 4 – (-13) × 2 = 1/(2 × 1 – 4 × (-3))

⇒ x/(39 + 17) = y/(-68 + 26) = 1/(2 + 12)

⇒ x/56 = y/-42 = 1/14

⇒ x = 56/14, y = −42/14

⇒ x = 4, y = -3

Hence, x = 4 and y = -3.

3. Show that the following system of equations has a unique solution:

x/3 + y/2 = 3, x – 2y = 2.

Also, find the solution of the given system of equations.

Solution

The given system of equations is:

x/3 + y/2 = 3

⇒ (2x + 3y)/6 = 3

2x + 3y = 18

⇒ 2x + 3y – 18 = 0 ….(i)

and

x – 2y = 2

x – 2y – 2 = 0 ...(ii)

These equations are of the forms:

a1x + b1y + c1 = 0 and a2x + b2y + c= 0

where, a1 = 2, b1= 3, c1 = -18 and a= 1, b2 = -2, c = -2

For a unique solution, we must have:

a1/a2 ≠ b1/b2, i.e., 2/1 ≠ 3/-2

Hence, the given system of equations has a unique solution. Again, the given equations are:

2x + 3y – 18 = 0 ….(iii)

x – 2y – 2 = 0 ...(iv)

On multiplying (i) by 2 and (ii) by 3, we get:

4x + 6y – 36 = 0 ….(v)

3x - 6y – 6 = 0 …(vi)

On adding (v) from (vi), we get:

7x = 42

⇒ x = 6

On substituting x = 6 in (iii), we get:

2(6) + 3y = 18

⇒3y = (18 - 12) = 6

⇒ y = 2

Hence, x = 6 and y = 2 is the required solution.

4. Find the value of k for which the system of equations has a unique solution:

2x + 3y = 5,

kx - 6y = 8.

Solution

The given system of equations are

2x + 3y – 5 = 0

kx - 6y - 8 = 0

This system is of the form:

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = 2, b1 = 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8

Now, for the given system of equations to have a unique solution, we must have:

a1/a2 ≠ b1/b2

⇒ 2/k ≠ 3/-6

⇒ k ≠ -4

Hence, k ≠ -4

5. Find the value of k for which the system of equations has a unique solution:

x – ky = 2,

3x + 2y + 5 = 0

Solution

The given system of equations are

x - ky – 2 = 0

3x + 2y + 5 = 0

This system of equations is of the form:

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5

Now, for the given system of equations to have a unique solution, we must have:

a1/a2 ≠ b1/b2

⇒ 1/3 ≠ - k/2

⇒ k ≠ -2/3

Hence, k ≠ -2/3

6. Find the value of k for which the system of equations has a unique solution:

5x – 7y = 5,

2x + ky = 1.

Solution

The given system of equations are

5x - 7y – 5 = 0 ….(i)

2x + ky - 1 = 0 …(ii)

This system is of the form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

where, a1 = 5, b1= -7, c1 = -5 and a2 = 2, b2 = k, c2 = -1

Now, for the given system of equations to have a unique solution, we must have:

a1/a2 ≠ b1/b2

⇒ 5/2 ≠ -7/k

⇒ k ≠ -(14/5)

Hence, k ≠ -14/5.

7. Find the value of k for which the system of equations has a unique solution:

4x + ky + 8 = 0,

x + y + 1 = 0.

Solution

The given system of equations are

4x + ky + 8 = 0

x + y + 1 = 0

This system is of the form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

where, a1 = 4, b1= k, c1 = 8 and a2 = 1, b2 = 1, c2 = 1

For the given system of equations to have a unique solution, we must have:

a1/a2 ≠ b1/b2

⇒ 4/1 ≠ k/1

⇒ k ≠ 4

Hence, k ≠ 4.

8. Find the value of k for which the system of equations has a unique solution:

4x - 5y = k,

2x - 3y = 12.

Solution

The given system of equations are

4x - 5y = k

⇒ 4x - 5y - k = 0 ….(i)

And, 2x - 3y = 12

⇒ 2x - 3y - 12 = 0 …(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Here, a1 = 4, b1= -5, c1 = -k and a2 = 2, b2 = -3, c2 = -12

For a unique solution, we must have:

a1/a2 ≠ b1/b2

i.e., 4/2 ≠ −5/−3

⇒ 2 ≠ 5/3 ⇒ 6 ≠ 5

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

9. Find the value of k for which the system of equations has a unique solution:

kx + 3y = (k – 3),

12x + ky = k

Solution

The given system of equations:

kx + 3y = (k – 3)

⇒ kx + 3y – (k - 3) = 0 ….(i)

And, 12x + ky = k

⇒ 12x + ky - k = 0 …(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Here, a1 = k, b1 = 3, c1 = -(k – 3) and a2 = 12, b2 = k, c2 = -k

For a unique solution, we must have:

a1/a2 ≠ b1/b2

i.e., k/12 ≠ 3/k

⇒ k2 ≠ 36

⇒ k ≠ ±6

Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.

10. Show that the system equations

2x - 3y = 5,

6x - 9y = 15 has an infinite number of solutions

Solution

The given system of equations:

2x - 3y = 5

⇒ 2x - 3y – 5 = 0 ….(i)

6x - 9y = 15

⇒ 6x - 9y - 15 = 0 …(ii)

These equations are of the following forms:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Here, a1 = 2, b1= -3, c1 = -5 and a2 = 6, b2 = -9, c2 = -15

∴ a1/a2 = 2/6 = 1/3, b1/b2 = -3/-9 = 1/3 and c1/c2 = -5/-15 = 1/3

Thus, a1/a2 = b1/b2 = c1/c2

Hence, the given system of equations has an infinite number of solutions.

11. Show that the system of equations

6x + 5y = 11,

9x + 15/2y = 21 has no solution.

Solution

The given system of equations can be written as

6x + 5y – 11 = 0 ….(i)

⇒ 9x + 15/2y - 21 = 0 …(ii)

This system is of the form

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Here, a1 = 6, b1 = 5, c1 = -11 and a= 9, b2 = 15/2, c2 = -21

Now,

a1/a2 = 6/9 = 2/3

b1/b2 = 5/(15/2) = 2/3

Thus, a1/a2 = b1/b2 ≠ c1/c2, therefore the given system has no solution.

12. For what value of k, the system of equations

kx + 2y = 5,

3x - 4y = 10 has (i) a unique solution, (ii) no solution?

Solution

The given system of equations:

kx + 2y = 5

⇒ kx + 2y - 5 = 0 ….(i)

3x - 4y = 10

⇒ 3x - 4y - 10 = 0 …(ii)

These equations are of the forms:

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = k, b1 = 2, c1 = -5 and a2 = 3, b2 = -4, c2 = -10

(i) For a unique solution, we must have:

∴ a1/a2 ≠ b1/b2 i.e., k/3 ≠ 2/-4 ⇒ k ≠ −3/2

Thus for all real values of k other than −3/2, the given system of equations will have a unique solution.

(ii) For the given system of equations to have no solutions, we must have:

a1/a2 = b1/b2 ≠ c1/c2

⇒ k/3 = 2/-4 ≠ -5/-10

⇒ k/3 = 2/-4 and k/3 ≠ 1/2

⇒ k = −3/2, k ≠ 3/2

Hence, the required value of k is −3/2.

13. For what value of k, the system of equations

x + 2y = 5,

3x + ky + 15 = 0

has (i) a unique solution, (ii) no solution?

Solution

The given system of equations:

x + 2y = 5

⇒ x + 2y - 5 = 0 ….(i)

3x + ky + 15 = 0 …(ii)

These equations are of the forms:

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

where, a1 = 1, b1= 2, c1 = -5 and a2 = 3, b2 = k, c2 = 15

(i) For a unique solution, we must have:

∴ a1/a2 ≠ b1/b2 i.e., 1/3 ≠ 2/k ⇒ k ≠ 6

Thus for all real values of k other than 6, the given system of equations will have a unique solution.

(ii) For the given system of equations to have no solutions, we must have:

⇒ a1/a2 ≠ b1/b2 ≠ c1/c2

⇒ 1/3 = 2/k ≠ -5/15

⇒ 1/3 = 2/k and 2/k ≠ −5/15

⇒ k = 6, k ≠ -6

Hence, the required value of k is 6.

14. For what value of k, the system of equations

x + 2y = 3,

5x + ky + 7 = 0

Have (i) a unique solution, (ii) no solution?

Also, show that there is no value of k for which the given system of equation has infinitely namely solutions

Solution

The given system of equations:

x + 2y = 3

⇒ x + 2y - 3 = 0 ….(i)

And, 5x + ky + 7 = 0 …(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = 1, b1 = 2, c1 = -3 and a2 = 5, b2 = k, c2 = 7

(i) For a unique solution, we must have:

∴ a1/a2 ≠ b1/b2 i.e., 1/5 ≠ 2/k ⇒ k ≠ 10

Thus for all real values of k other than 10, the given system of equations will have a unique solution.

(ii) In order that the given system of equations has no solution, we must have:

a1/a2 = b1/b2 ≠ c1/c2

⇒ 1/5 ≠ 2/k ≠ −3/7

⇒ 1/5 ≠ 2/k and 2/k ≠ −3/7

⇒ k = 10, k ≠ 14/−3

Hence, the required value of k is 10.

There is no value of k for which the given system of equations has an infinite number of solutions.

15. Find the value of k for which the system of linear equations has an infinite number of solutions:

2x + 3y = 7,

(k – 1)x + (k + 2)y = 3k.

Solution

The given system of equations:

2x + 3y = 7,

⇒ 2x + 3y - 7 = 0 ….(i)

And, (k – 1)x + (k + 2)y = 3k

⇒ (k – 1)x + (k + 2)y - 3k = 0 …(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = 2, b1 = 3, c1 = -7 and a2 = (k – 1), b2 = (k + 2), c2 = -3k

For an infinite number of solutions, we must have:

⇒ a1/a2 = b1/b2 = c1/c2

2/(k – 1) = 3/(k + 2) = -7/-3k

⇒ 2/(k – 1) = 3/(k + 2) = 7/3k

Now, we have the following three cases:

Case I:

2/(k – 1) = 3/(k + 2)

⇒ 2(k + 2) = 3(k – 1) ⇒ 2k + 4 = 3k – 3 ⇒ k = 7

Case II:

3/(k + 2) = 7/3k

⇒ 7(k + 2) = 9k ⇒ 7k + 14 = 9k ⇒ 2k = 14 ⇒ k = 7

Case III:

2/(k – 1) = 7/3k

⇒ 7k – 7 = 6k ⇒ k = 7

Hence, the given system of equations has an infinite number of solutions when k is equal to 7.

16. Find the value of k for which the system of linear equations has an infinite number of solutions:

2x + (k – 2)y = k,

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

Solution

The given system of equations:

2x + (k – 2)y = k

⇒ 2x + (k – 2)y – k = 0 ...(i)

And, 6x + (2k – 1)y = (2k + 5)

⇒ 6x + (2k – 1)y – (2k + 5) = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Where, a1 = 2, b1 = (k – 2), c1 = - k and a2 = 6, b2 = (2k – 1), c2 = -(2k + 5)

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

2/6 = (k – 2)/(2k – 1) = -k/-(2k + 5)

⇒ 1/3 = (k – 2)/(2k – 1) = k(2k + 5)

Now, we have the following three cases:

Case I:

1/3 = (k – 2)/(2k – 1)

⇒ (2k – 1) = 3(k – 2)

⇒ 2k – 1 = 3k – 6 ⇒ k = 5

Case II:

(k – 2)/(2k – 1) = k/(2k + 5)

⇒ (k – 2)(2k + 5) = k(2k – 1)

⇒ 2k2 + 5k – 4k – 10 = 2k2 – k

⇒ k + k = 10 ⇒ 2k = 10 ⇒ k = 5

Case III:

1/3 = k/(2k + 5)

⇒ 2k + 5 = 3k ⇒ k = 5

Hence the given system of equations has an infinite number of solutions when k is equal to 5.

17. Find the value of k for which the system of linear equations has an infinite number of solutions:

kx + 3y = (2k + 1),

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

Solution

The given system of equations:

kx + 3y = (2k + 1)

⇒ kx + 3y – (2k + 1) = 0 ...(i)

And, 2(k + 1)x + 9y = (7k + 1)

⇒ 2(k + 1)x + 9y – (7k + 1) = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Where, a1 = k, b1 = 3, c1 = -(2k + 1) and a2 = 2(k + 1), b2 = 9, c2 = -(7k + 1)

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

i.e., k/2(k + 1) = 3/9 = -(2k + 1)/-(7k + 1)

⇒ k/2(k + 1) = 1/3 = (2k + 1)/(7k + 1)

Now, we have the following three cases:

Case I:

k/2(k + 1) = 1/3

⇒ 2(k + 1) = 3k

⇒ 2k + 2 = 3k

⇒ k = 2

Case II:

1/3 = (2k + 1)/(7k + 1)

⇒ (7k + 1) = 6k + 3

⇒ k = 2

Case III:

k/2(k + 1) = (2k + 1)/(7k + 1)

⇒ k(7k + 1) = (2k + 1) × 2|(k + 1)

⇒ 7k2 + k = (2k + 1)(2k + 2)

⇒ 7k2 + k = 4k2 + 4k + 2k + 2

⇒ 3k2 – 5k – 2 = 0

⇒ 3k2 – 6k + k – 2 = 0

⇒ 3k(k – 2) + 1(k – 2) = 0

⇒ (3k + 1)(k – 2) = 0

⇒ k = 2 or k = -1/3

Hence, the given system of equations has an infinite number of solutions when k is equal to 2.

18. Find the value of k for which the system of linear equations has an infinite number of solutions.

5x + 2y = 2k,

2(k + 1)x + ky = (3k + 4).

Solution

The given system of equations:

5x + 2y = 2k

⇒ 5x + 2y – 2k = 0 ...(i)

And, 2(k + 1)x + ky = (3k + 4)

⇒ 2(k = 1)x + ky – (3k + 4) = 0 ...(ii)

These equations are of the following form:

a1x + b2y + c1 = 0, a2x + b2y + c2 = 0

Where, a1 = 5, b1 = 2, c1 = - 2k and a2 = 2(k + 1), b2 = k, c2 = -(3k + 4)

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

5/2(k + 1) = 2/k = -2k/-(3k + 4)

⇒ 5/2(k + 1) = 2/k = 2k/(3k + 4)

Now, we have the following three cases:

Case I:

5/2(k + 1) = 2/k

⇒ 2 × 2(k + 1) = 5k

⇒ 4(k + 1) = 5k

⇒ 4k + 4 = 5k

⇒ k = 4

Case II:

2/k = 2k/(3k + 4)

⇒ 2k2 = 2 × (3k + 4)

⇒ 2k2 = 6k + 8 ⇒ 2k2 – 6k – 8 = 0

⇒ 2(k2 – 3k – 4) = 0

⇒ k2 – 4k + k – 4 = 0

⇒ k(k – 4) + 1(k – 4) = 0

⇒ (k + 1)(k – 4) = 0

⇒ (k + 1) = 0 or (k – 4) = 0

⇒ k = - 1 or k = 4

Case III:

5/2(k + 1) = 2k/(3k + 4)

⇒ 15k + 20 = 4k2 + 4k

⇒ 4k2 – 11k – 20 = 0

⇒ 4k2 – 16k + 5k – 20 = 0

⇒ 4k(k – 4) + 5(k – 4) = 0

⇒ (k – 4)(4k + 5) = 0

⇒ k = 4 or k = -5/4

Hence, the given system of equations has an infinite number of solutions when k is equal to 4.

19. Find the value of k for which the system of linear equations has an infinite number of solutions:

(k – 1)x – y = 5,

(k + 1)x + (1 – k)y = (3k + 1).

Solution

The given system of equations:

(k – 1)x – y = 5

⇒ (k – 1)x – y – 5 = 0 ...(i)

And, (k + 1)x + (1 – k)y = (3k + 1)

⇒ (k + 1)x + (1 – k)y – (3k + 1) = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = (k – 1), b1 = -1, c1 = - 5, a2 = (k + 1), b2 = (1 – k), c2 = -(3k + 1)

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

i.e., (k – 1)/(k + 1) = -1/(k – 1) = -5/(-3k + 1)

⇒ (k – 1)/(k + 1) = 1/(k – 1) = 5/(3k + 1)

Now, we have the following three cases:

Case I:

(k – 1)/(k + 1) = 1/(k – 1)

⇒ (k – 1)2 = (k + 1)

⇒ k2 + 1 – 2k = k + 1

⇒ k2 – 3k = 0 ⇒ k(k – 3) = 0

⇒ k = 0 or k = 3

Case II:

1/(k – 1) = 5/(3k + 1)

⇒ 3k + 1 = 5k – 5

⇒ 2k = 6 ⇒ k = 3

Case III:

(k – 1)/(k + 1) = 5/(3k + 1)

⇒ (3k + 1)(k – 1) = 5(k + 1)

⇒ 3k2 + k – 3k – 1 = 5k + 5

⇒ 3k2 – 2k – 5k – 1 – 5 = 0

⇒ 3k2 - 7k – 6 = 0

⇒ 3k2 – 9k + 2k – 6 =0

⇒ 3k(k – 3) = 2(k – 3) = 0

⇒ (k – 3)(3k + 2) = 0

⇒ (k – 3) = 0 or (3k + 2) = 0

⇒ k = 3 or k = -2/3

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.

20. Find the value of k for which the system of linear equations has a unique solution:

(k – 3)x + 3y – k, kx + ky – 12 = 0

Solution

The given system of equations can be written as

(k – 3)x + 3y – k = 0

kx + ky – 12 = 0

This system is of the form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Where, a1 = k, b1 = 3, c1 = -k and a2 = k, b2 = k, c2 = -12

For the given system of equations to have a unique solution, we must have:

a1/a2 = b1/b2 = c1/c2

⇒ (k – 3)/k = 3/k = -k/-12

⇒ k – 3 = 3 and k2 = 36

⇒ k = 6 and k ± 6

⇒ k = 6

Hence, k = 6

21. Find the values of a and b for which the system of linear equations has an infinite number of solutions:

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

Solution

The given system of equations can be written as

(a – 1) x + 3y = 2

⇒ (a – 1) x + 3y – 2 = 0 ...(i)

and 6x + (1 – 2b) y = 6

6x + (1 – 2b) y - 6 = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

where, a1 = (a – 1), b1 = 3, c1 = - 2 and a2 = 6, b2 = (1 – 2b), c2 = - 6

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

⇒ (a – 1)/6 = 3/(1 – 2b) = -2/-6

⇒ (a – 1)/6 = 3/(1 – 2b) = 1/3

⇒ (a – 1)/6 = 1/3 and 3/(1 – 2b) = 1/3

⇒ 3a – 3 = 6 and 9 = 1 – 2b

⇒ 3a = 9 and 2b = - 8

⇒ a = 3 and b = -4

∴ a = 3 and b = - 4

22. Find the values of a and b for which the system of linear equations has an infinite number of solutions:

(2a – 1)x + 3y = 5, 3x + (b – 1)y = 2

Solution

The given system of equations can be written as

(2a – 1)x + 3y = 5

⇒ (2a – 1)x + 3y – 5 = 0 ...(i)

and 3x + (b – 1)y – 2 = 0

⇒ 3x + (b – 1)y – 2 = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = (2a – 1), b1 = 3, c1 = - 5 and a2 = 3, b2 = (b – 1), c2 = - 2

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

⇒ (2a – 1)/3 = 3/(b – 1) = - 5/-2

⇒ (2a – 1)/6 = 3/(b – 1) = 5/2

⇒ (2a – 1)/6 = 5/2 and 3/(b – 1) = 5/2

⇒ 2(2a – 1) = 15 and 6 = 5(b – 1)

⇒ 4a – 2 = 15 and 6 = 5b – 5

⇒ 4a = 17 and 5b = 11

∴ a = 17/4 and b = 11/5

23. Find the values of a and b for which the system of linear equations has an infinite number of solutions:

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

Solution

The given system of equations can be written as

2x – 3y = 7

⇒ 2x – 3y – 7 = 0 ...(i)

and (a + b)x – (a + b – 3)y = 4a + b

⇒ (a + b)x – (a + b – 3)y – 4a + b = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c= 0, a2x + b2y + c2 = 0

Here, a1 = 2, b1 = - 3, c1 = - 7 and a2 = (a + b), b2 = -(a + b – 3), c2 = -(4a + b)

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

2/(a + b) = -3/-(a + b – 3) = -7/-(4a + b)

⇒ 2/(a + b) = 3/(a + b – 3) = 7/(4a + b)

⇒ 2/(a + b) = 7/(4a + b) and 3/(a + b - 3) = 7/(4a + b)

⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b – 3)

⇒ 8a + 2b = 7a + 5b and 12a + 3b = 7a + 7b – 21

⇒ 4a = 17 and 5b = 11

∴ a = 5b ...(iii)

And 5a = 4b – 21 ....(iv)

On substituting a = 5b in (iv), we get;

25b = 4b – 21

⇒ 21b = - 21

⇒ b = - 1

On substituting b = - 1 in (iii), we get;

a = 5(-1) = - 5

∴ a = - 5 and b = - 1.

24. Find the values of a and b for which the system of linear equations has an infinite number of solutions:

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

Solution

The given system of equations can be written as

2x + 3y = 7

⇒ 2x + 3y – 7 = 0 ...(i)

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

(a + b + 1)x – (a + 2b + 2)y – [4(a + b) + 1] = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = 2, b1 = 3, c1 = -7 and a2 = (a + b + 1), b2 = (a + 2b + 2), c2 = -[4(a + b) + 1]

For an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

2/(a + b + 1) = 3/(a + 2b + 2) = -7/-[4(a + b) + 1]

⇒ 2/(a + b + 1) = 3/(a + 2b + 2) = 7/[4(a + b) + 1]

⇒ 2/(a + b + 1) = 3/(a + 2b + 2) and 3/(a + 2b + 2) = 7/[4(a + b) + 1]

⇒ 2(a + 2b + 2) = 3(a + b + 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)

⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14

⇒ a – b – 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14

⇒ a – b = 1 and 5a - 2b = 11

a = (b + 1) ...(iii)

5a – 2b = 11 ...(iv)

On substituting a = (b + 1) in (iv), we get:

5(b + 1) – 2b = 11

⇒ 5b + 5 – 2b = 11

⇒ 3b = 6

⇒ b = 2

On substituting b = 2 in (iii), we get:

a = 3

∴ a = 3 and b = 2

25. Find the values of a and b for which the system of linear equations has infinite number of solutions:

2x +3y = 7, (a + b)x + (2a – b)y = 21.

Solution

The given system of equations can be written as

2x + 3y – 7 = 0 ...(i)

(a + b)x + (2a – b)y – 21 = 0 ...(ii)

This system is of the form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where, a1 = 2, b1 = 3, c1 = - 7 and a2 = a + b, b2 = 2a – b, c2 = - 21

For the given system of linear equations to have an infinite number of solutions, we must have:

a1/a2 = b1/b2 = c1/c2

⇒ 2/(a + b) = 3/(2a – b) = -7/-21

⇒ 2/(a + b) = -7/-21 = 1/3 and 3/(2a – b) = -7/-21 = 1/3

⇒ a + b = 6 and 2a – b = 9

Adding a + b = 6 and 2a – b = 9, we get

3a = 15 ⇒ a = 15/3 = 3

Now substituting a = 5 in a + b = 6, we have

5 + b = 6 ⇒ b = 6 – 5 = 1

Hence, a = 5 and b = 1.

26. Find the values of a and b for which the system of linear equations has an infinite number of solutions:

2x + 3y = 7, 2ax + (a + b)y = 28.

Answer:

The given system of equations can be written as

2x + 3y – 7 = 0 ...(i)

2ax + (a + b)y – 28 = 0 ...(ii)

This system is of the form:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

where, a1 = 2, b1 = 3, c1 = - 7 and a2 = 2a, b= a + b, c2 = - 28

For the given system of linear equations to have an infinite number of solutions, we must have:

a1/a2 = b1/b= c1/c2

⇒ 2/2a = 3/(a + b) = -7/-28

⇒ 2/2a = -7/-28 = 1/4 and 3/(a + b) = -7/-28 = 1/4

⇒ a = 4 and a + b = 12

Substituting a = 4 in a + b = 12, we get

4 + b = 12 ⇒ b = 12 – 4 = 8

Hence, a = 4 and b = 8.

27. Find the value of k for which the systems of equations

8x + 5y = 9, kx + 10y = 15 has a non-zero solution.

Solution

The given system of equations:

8x + 5y = 9

8x + 5y – 9 = 0 ...(i)

kx + 10y = 15

kx + 10y – 15 = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

where a1 = 8, b1 = 5, c1 = - 9 and a2 = k, b2 = 10, c2 = - 15

In order that the given system has no solution, we must have:

a1/a2 = b1/b2 ≠ c1/c2

⇒ 8/k = 5/10 ≠ -9/-15

⇒ 8/k = 1/2 ≠ 3/5

⇒ 8/k = 1/2 and 8/k ≠ 3/5

⇒ k = 16 and k ≠ 40/3

Hence, the given system of equations has no solutions when K is equal to 16.

28. Find the value of k for which the system of equations kx + 3y = 3, 12x + ky = 6 has no solution.

Solution

The given system of equations:

kx + 3y = 3

⇒ kx + 3y – 3 = 0 ...(i)

12x + ky = 6

⇒ 12x + ky – 6 = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Where a1 = k, b1 = 3, c1 = - 3 and a2 = 12, b2 = k, c2 = - 6

In order that the given system has no solution, we must have:

a1/a2 = b1/b2 ≠ c1/c2

i.e., k/12 = 3/k ≠ -3/-6

k/12 = 3/k and 3/k ≠ 1/2

⇒ k2 = 36 and k ≠ 6

⇒ k = ± 6 and k ≠ 6

Hence, the given system of equations has no solution when k is equal to – 6.

29. Find the value of k for which the system of equations

3x – y = 5, 6x – 2y = k has no solution.

Solution

The given system of equations:

3x – y – 5 = 0 ...(i)

And, 6x – 2y + k = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Where, a1 = 3, b1 = - 1, c1 = - 5 and a2 = 6, b2 = - 2, c2 = k

In order that the given system has no solution, we must have:

a1/a2 = b1/b2 ≠ c1/c2

i.e., 3/6 = -1/-2 ≠ -5/k

⇒ -1/-2 ≠ -5/k ⇒ k ≠ - 10

Hence, equations (i) and (ii) will have no solution if k ≠ - 10.

30. Find the value of k for which the system of equations

kx + 3y + 3 – k = 0, 12x + ky – k = 0 has no solution.

Solution

The given system of equations can be written as

kx + 3y + 3 – k = 0 ...(i)

12x + ky – k = 0 ...(ii)

This system of the form:

a1x + b1y + c1 = 0,

a2x + b2y + c2 = 0

where, a1 = k, b1 = 3, c1 = 3 – k and a2 = 12, b2 = k, c2 = - k

For the given system of linear equations to have no solution, we must have:

a1/a2 = b1/b2 ≠ c1/c2

⇒ k/12 = 3/k ≠ (3 – k)/-k

⇒ k/12 = 3/k and 3/k ≠ (3 – k)/-k

⇒ k2 = 36 and -3 ≠ 3 – k

⇒ k ± 6 and k ≠ 6

⇒ k = - 6

Hence, k = - 6.

31. Find the value of k for which the system of equations 5x – 3y = 0, 2x + ky = 0

has a non-zero solution.

Solution

The given system of equations:

5x – 3y = 0 ...(i)

2x + ky = 0 ...(ii)

These equations are of the following form:

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Where, a1 = 5, b1 = - 3, c1 = 0 and a2 = 2, b2 = k, c2 = 0

For a non-zero solution, we must have:

a1/a2 = b1/b2

⇒ 5/2 = -3/k

⇒ 5k = - 6 ⇒ k = -6/5

Hence, the required value of k is -6/5.