# 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.

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.