Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.5

Question: 1In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

x − 3y − 3 = 0, x − 3y − 3 = 0

3x − 9y − 2 = 0, 3x − 9y − 2 = 0

Solution:The given system may be written as

x − 3y − 3 = 0

3x − 9y − 2 = 0

The given system of equation is of the form

a

_{2}x + b_{2}y − c_{2}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 1, b_{1}= −3, c_{1}= −3a

_{2}= 3, b_{2}= −9, c_{2}= −2We have,

Therefore, the given equation has no solution.

Question: 2In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

2x + y − 5 = 0

4x + 2y − 10 = 0

Solution:The given system may be written as

2x + y − 5 = 0 4x + 2y − 10 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= 1, c_{1}= −5a

_{2}= 4, b_{2}= 2, c_{2}= −10We have,

Therefore, the given equation has infinitely many solution.

Question: 3In each of the following systems of equation determine whether the system has a unique solution, no solution or infinite solutions. In case there is a unique solution,

3x − 5y = 20

6x − 10y = 40

Solution:The given system may be written as

3x − 5y = 20 6x − 10y = 40

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 3, b_{1}= −5, c_{1}= − 20a

_{2}= 6, b_{2}= −10, c_{2 }= − 40We have,

Therefore, the given equation has infinitely many solution.

Question: 4x − 2y − 8 = 0

5x − 10y − 10 = 0

Solution:The given system may be written as

x − 2y − 8 = 0 5x − 10y − 10 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 1, b_{1}= −2, c_{1}= −8a

_{2}= 5, b_{2}= −10, c_{2}= −10We have,

Therefore, the given equation has no solution.

Question: 5Find the value of k for each of the following system of equations which have a unique solution

kx + 2y − 5 = 0

3x + y − 1 = 0

Solution:The given system may be written as

kx + 2y − 5 = 0

3x + y − 1 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0a

_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= k, b_{1 }= 2, c_{1}= −5a

_{2}= 3, b_{2}= 1, c_{2}= −1For unique solution, we have

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

Question: 6Find the value of k for each of the following system of equations which have a unique solution

4x + ky + 8 = 0

2x + 2y + 2 = 0

Solution:The given system may be written as

4x + ky + 8 = 0 2x + 2y + 2 = 0

The given system of equation is of the form

a

_{1}x +b_{1}y − c_{1 }= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 4, b_{1}= k, c_{1}= 8a

_{2 }= 2, b_{2}= 2, c_{2}= 2For unique solution, we have

Therefore, the given system will have unique solution for all real values of k other than 4.

Question: 7Find the value of k for each of the following system of equations which have a unique solution

4x − 5y = k

2x − 3y = 12

Solution:The given system may be written as

4x − 5y − k = 0

2x − 3y − 12 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= 4, b_{1}= −5, c_{1}= −ka

_{2}= 2, b_{2 }= -3, c_{2 }= -12For unique solution, we have

⇒ k can have any real values.

Therefore, the given system will have unique solution for all real values of k.

Question: 8Find the value of k for each of the following system of equations which have a unique solution

x + 2y = 3

5x + ky + 7 = 0

Solution:The given system may be written as

x + 2y = 3

5x + ky + 7 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where a

_{1 }= 1, b_{1}= 2, c_{1}= −3a

_{2 }= 5, b_{2}= k, c_{2}= 7For unique solution, we have

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

Question: 9Find the value of k for which each of the following system of equations having infinitely many solution:

2x + 3y − 5 = 0

6x − ky − 15 = 0

Solution:The given system may be written as

2x + 3y − 5 = 0 6x − ky − 15 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= 3, c_{1}= −5a

_{2}= 6, b_{2}= k, c_{2}= −15For unique solution, we have

Therefore, the given system of equation will have infinitely many solutions, if k = 9.

Question: 10Find the value of k for which each of the following system of equations having infinitely many solution:

4x + 5y = 3

x + 15y = 9

Solution:The given system may be written as

4x + 5y = 3

kx +15y = 9

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 4, b_{1}= 5, c_{1}= 3a

_{2}= k, b_{2 }= 15, c_{2}= 9For unique solution, we have

Therefore, the given system will have infinitely many solutions if k = 12.

Question: 11Find the value of k for which each of the following system of equations having infinitely many solution:

kx − 2y + 6 = 0

4x + 3y + 9 = 0

Solution:The given system may be written as

kx − 2y + 6 = 0 4x + 3y + 9 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= k, b_{1}= −2, c_{1}= 6a

_{2}= 4, b_{2}= −3, c_{2}= 9For unique solution, we have

Therefore, the given system of equations will have infinitely many solutions, if k = 8/3.

Question: 128x + 5y = 9

kx + 10y = 19

Solution:The given system may be written as

8x + 5y = 9 kx + 10y = 19

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2 }=0Where, a

_{1}= 8, b_{1}= 5, c_{1 }= −9a

_{2}= k, b_{2}= 10, c_{2}= −18 a_{2}= k, b_{2}= 10,c_{2 }= −18For unique solution, we have

Therefore, the given system of equations will have infinitely many solutions, if k = 16.

Question: 132x − 3y = 7

(k + 2)x − (2k + 1)y = 3(2k − 1)

Solution:The given system may be written as

2x − 3y = 7 (k + 2)x − (2k + 1)y = 3(2k − 1)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= −3, c_{1}= −7a

_{2 }= k, b_{2}= − (2k + 1), c_{2}= −3(2k − 1)For unique solution, we have

⇒ 2(2k + 1) = 3(k + 2) and 3 × 3(2k − 1) = 7(2k + 1)

⇒ 4k + 2 = 3k + 6 and 18k − 9 = 14k + 7

⇒ k = 4 and 4k = 16

⇒ k = 4

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

Question: 142x + 3y = 2

(k + 2)x + (2k + 1)y = 2(k − 1)

Solution:The given system may be written as

2x + 3y = 2 (k + 2)x + (2k + 1)y = 2(k − 1)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= 3,c_{1}= −2a

_{2}= (k + 2), b_{2}= (2k + 1),c_{2}= −2(k − 1)For unique solution, we have

⇒ 2(2k + 1) = 3(k + 2) and 3(k − 1) = (2k + 1)

⇒ 4k + 2 = 3k + 6 and 3k − 3 = 2k + 1

⇒ k = 4 and k = 4

Therefore, the given system of equations will have infinitely many solutions, if k = 4.

Question: 15x + (k + 1)y = 4

(k + 1)x + 9y = (5k + 2)

Solution:The given system may be written as

x + (k + 1)y = 4 (k + 1)x + 9y = (5k + 2)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 1, b_{1}= (k + 1), c_{1}= −4a

_{2}= (k + 1), b_{2}= 9, c_{2}= − (5k + 2)For unique solution, we have

⇒ 9 = (k + 1)

^{2}and (k + 1)(5k + 2) = 36⇒ 9 = k

^{2}+ 2k + 1 and 5k^{2}+ 2k + 5k + 2 = 36⇒ k

^{2 }+ 2k − 8 = 0 and 5k^{2}+ 7k − 34 = 0⇒ k

^{2}+ 4k − 2k − 8 = 0 and 5k^{2}+ 17k − 10k − 34 = 0⇒ k(k + 4) −2 (k + 4) = 0 and (5k + 17) − 2 (5k + 17) = 0

⇒ (k + 4)(k − 2) = 0 and (5k + 17)(k − 2) = 0

⇒ k = - 4 or k = 2 and k = -17/5 or k = 2

Thus, k = 2 satisfies both the condition.

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

Question: 16kx + 3y = 2k + 1

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

Solution:The given system may be written as

kx + 3y = 2k + 1 2(k + 1)x + 9y = (7k + 1)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= k, b_{1}= 3, c_{1}= −(2k + 1)a

_{2}= 2(k + 1), b_{2}= 9, c_{2}= −(7k + 1)For unique solution, we have

⇒ 9k = 3 × 2(k + 1) and 3(7k + 1) = 9(2k + 1)

⇒ 9k − 6k = 6 and 21k − 18k = 9 − 3

⇒ 3k = 6 ⇒ k = 2 and k = 2

Therefore, the given system of equations will have infinitely many solutions, if k = 2.

Question: 172x + (k − 2)y = k

6x + (2k − 1)y = (2k + 5)

Solution:The given system may be written as

2x +( k − 2)y = k 6x + (2k − 1)y = (2k + 5)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2,b_{1}= (k − 2),c_{1}= −ka

_{2}= 6,b_{2}= (2k − 1),c_{2}= −(2k + 5)For unique solution, we have

⇒ 2k − 3k = −6 + 1 and k + k = 10

⇒ −k = −5 and 2k = 10 ⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

Question: 182x + 3y = 72x + 3y = 7

(k + 1)x + (2k − 1)y = (4k + 1)

Solution:The given system may be written as

2x + 3y = 7 (k + 1)x + (2k − 1)y = (4k + 1)

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2 }= 0Where, a

_{1}= 2, b_{1}= 3, c_{1}= −7a

_{2 }= k + 1, b_{2}= 2k − 1, c_{2}= −(4k + 1)For unique solution, we have

Extra close brace or missing open brace

⇒ 4k − 2 = 3k + 3 and 12k + 3 = 14k − 7

⇒ k = 5 and 2k = 10 ⇒ k = 5 and k = 5

Therefore, the given system of equations will have infinitely many solutions, if k = 5.

Question: 192x + 3y = k

(k − 1)x + (k + 2)y = 3k

Solution:The given system may be written as

2x + 3y = k (k − 1)x + (k + 2)y = 3k

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= 2,b_{1 }= 3, c_{1}= −ka

_{2}= k − 1, b_{2 }= k + 2, c_{2}= −3kFor unique solution, we have

Extra close brace or missing open brace

⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ 2k + 4 = 3k − 3 and 9 = k + 2 ⇒ k = 7 and k = 7

Therefore, the given system of equations will have infinitely many solutions, if k = 7.

Question: 20Find the value of k for which the following system of equation has no solution:

kx − 5y = 2

6x + 2y = 7

Solution:The given system may be written as

kx − 5y = 2 6x + 2y = 7

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= k, b_{1}= −5, c_{1 }= −2a

_{2}= 6 b_{2 }= 2, c_{2}= −7For no solution, we have

⇒ 2k = -30 ⇒ k = -15

Therefore, the given system of equations will have no solutions, if k = −15.

Question: 21Find the value of k for which the following system of equation has no solution:

x + 2y = 0

2x + ky = 5

Solution:The given system may be written as

x

_{2}y = 0 2x + ky = 5The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 1, b_{1}= 2, c_{1}= 0a

_{2}= 2, b_{2}= k, c_{2}= −5For no solution, we have

⇒ k = 4

Therefore, the given system of equations will have no solutions, if k = 4.

Question: 22Find the value of k for which the following system of equation has no solution:

3x − 4y + 7 = 0

kx + 3y − 5 = 0

Solution:The given system may be written as

3x − 4y + 7 = 0 kx + 3y − 5 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 3, b_{1}= −4, c_{1}= 7a

_{2}= k, b_{2}= 3, c_{2}= −5For no solution, we have

Therefore, the given system of equations will have no solutions, if k = - 9 /4.

Question: 23Find the value of k for which the following system of equation has no solution:

2x − ky + 3 = 0

3x + 2y − 1 = 0

Solution:The given system may be written as

2x − ky + 3 = 0 3x + 2y − 1 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= −k, c_{1}= 3a

_{2}= 3, b_{2}= 2, c_{2}= −1For no solution, we have

Therefore, the given system of equations will have no solutions, if k = – 4/3.

Question: 24Find the value of k for which the following system of equation has no solution:

2x + ky − 11 = 0

5x − 7y − 5 = 0

Solution:The given system may be written as

2x + ky − 11 = 0 5x − 7y − 5 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= k, c_{1}= −11a

_{2}= 5, b_{2}= −7, c_{2}= −5a_{2}= 5, b_{2 }= −7, c_{2}= − 5For no solution, we have

Therefore, the given system of equations will have no solutions, if k = -14/5.

Question: 25Find the value of k for which the following system of equation has no solution:

kx + 3y = 3

12x + ky = 6

Solution:The given system may be written as

kx + 3y = 3 12x + ky = 6

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= k, b_{1}= 3, c_{1}= −3a

_{2 }= 12, b_{2}= k, c_{2}= − 6For no solution, we have

⇒ k

^{2}= 36 ⇒ k = + 6 or −6From (i)

Therefore, the given system of equations will have no solutions, if k = − 6.

Question: 26For what value of a, the following system of equation will be inconsistent?

4x + 6y − 11 = 0

2x + ay − 7 = 0

Solution:The given system may be written as

4x + 6y − 11 = 0 2x + ay − 7 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 4, b_{1}= 6, c_{1}= −11a

_{2}= 2, b_{2}= a, c_{2}= −7For unique solution, we have

Therefore, the given system of equations will be inconsistent, if a = 3.

Question: 27For what value of a, the following system of equation have no solution?

ax + 3y = a − 3

12x + ay = a

Solution:The given system may be written as

ax + 3y = a − 3 12x + ay = a

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= a, b_{1 }= 3, c_{1}= - (a − 3)a

_{2}= 12, b_{2}= a, c_{2 }= − aFor unique solution, we have

And,

⇒ a

^{2 }= 36⇒ a = + 6 or – 6?

a ≠ 6 ⇒ a = – 6

Therefore, the given system of equations will have no solution, if a = − 6.

Question: 28Find the value of a, for which the following system of equation have

(i) Unique solution

(ii) No solution

kx + 2y = 5

3x + y = 1

Solution:The given system may be written as

kx + 2y − 5 = 0 3x + y − 1 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= k, b_{1}= 2, c_{1}= −5a

_{2}= 3, b_{2}= 1, c_{2}= −1(i) For unique solution, we have

Therefore, the given system of equations will have unique solution, if k ≠ 6 k ≠ 6.

(ii) For no solution, we have

Therefore, the given system of equations will have no solution, if a = 6.

Question: 29For what value of c, the following system of equation have infinitely many solution (where c ≠ 0 c ≠ 0)?

6x + 3y = c − 3

12x + cy = c

Solution:The given system may be written as

6x + 3y − (c − 3) = 0 12x + cy − c = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 6, b_{1}= 3, c_{1}= −(c − 3)a

_{2 }= 12, b_{2}= c, c_{2}= - cFor infinitely many solution, we have

⇒ c = 6 and c – 3 = 3

⇒ c = 6 and c = 6

Therefore, the given system of equations will have infinitely many solution, if c = 6.

Question: 30Find the value of k, for which the following system of equation have

(i) Unique solution

(ii) No solution

(iii) Infinitely many solution

2x + ky = 1

3x − 5y = 7

Solution:The given system may be written as

2x + ky = 1 3x − 5y = 7

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= k, c_{1}= −1a_{1 }= 2, b_{1}= k, c_{1}= −1a

_{2 }= 3, b_{2}= −5, c_{2}= −7(i) For unique solution, we have

Therefore, the given system of equations will have unique solution, if k ≠ -10/3.

(ii) For no solution, we have

Therefore, the given system of equations will have no solution, if k = -10)/3.

(iii) For the given system to have infinitely many solution, we have

So there is no value of k for which the given system of equation has infinitely many solution.

Question: 31For what value of k, the following system of equation will represent the coincident lines?

x + 2y + 7 = 0

2x + ky + 14 = 0

Solution:The given system may be written as

x + 2y + 7 = 0 2x + ky + 14 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 1, b_{1}= 2, c_{1}= 7a

_{2}= 2, b_{2}= k, c_{2}= 14The given system of equation will represent the coincident lines if they have infinitely many solution.

Therefore, the given system of equations will have infinitely many solution, if k = 4.

Question: 32Find the value of k, for which the following system of equation have unique solution.

ax + by = c

lx + my = n

Solution:The given system may be written as

ax + by − c = 0 lx + my − n = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2 }= 0Where, a

_{1}= a, b_{1}= b, c_{1 }= − ca

_{2 }= l, b_{2}= m, c_{2}= − nFor unique solution, we have

Therefore, the given system of equations will have unique solution, if am ≠ bl.

Question: 33Find the value of a and b such that the following system of linear equation have infinitely many solution:

(2a − 1)x + 3y − 5 = 0

3x + (b − 1)y − 2 = 0

Solution:The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0 3x + (b − 1)y − 2 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= (2a − 1), b_{1}= 3, c_{1}= −5a

_{2}= 3, b_{2}= b − 1, c_{2}= −2The given system of equation will have infinitely many solution, if

⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)

⇒ 4a − 2 = 15 and 6 = 5b − 5 ⇒ 4a = 17 and 5b = 11

Question: 34Find the value of a and b such that the following system of linear equation have infinitely many solution:

2x − 3y = 7

(a + b)x − (a + b − 3)y = 4a + b

Solution:The given system of equation may be written as,

2x − 3y − 7 = 0 (a + b)x − (a + b − 3)y − (4a + b) = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= −3, c_{1}= −7a

_{2}= (a + b), b_{2}= −(a + b − 3), c_{2 }= −(4a + b)The given system of equation will have infinitely many solution, if

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

⇒ 2a + 2b − 6 = 3a + 3b and 12a + 3b = 7a + 7b − 21

⇒ a + b = −6 and 5a − 4b = −21

⇒ a = − 6 − b

Substituting the value of a in 5a − 4b = −21 we have

5( – b – 6) – 4b = – 21

⇒ − 5b − 30 − 4b = − 21

⇒ 9b = − 9 ⇒ b = −1

As a = – 6 – b

⇒ a = − 6 + 1 = − 5

Hence the given system of equation will have infinitely many solution if

a = – 5 and b = –1.

Question: 35Find the value of p and q such that the following system of linear equation have infinitely many solution:

2x − 3y = 9

(p + q)x + (2p − q)y = 3(p + q + 1)

Solution:The given system of equation may be written as,

2x − 3y − 9 = 0 (p + q)x + (2p − q)y − 3(p + q + 1) = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}=2, b_{1}= 3, c_{1 }= −9a

_{2}= (p + q), b_{2}= (2p − q), c_{2}= -3(p + q + 1)The given system of equation will have infinitely many solution, if

2(2p - q) = 3(p + q) and (p + q + 1) = 2p - q

⇒ 4p - 2q = 3p + 3q and -p + 2q = -1

⇒ p = 5q and p - 2q = 1

Substituting the value of p in p - 2q = 1, we have

3q = 1

⇒ q = 1/3

Substituting the value of p in p = 5q we have

p = 5/3

Hence the given system of equation will have infinitely many solution if

p = 5/3 and q = 1/3.

Question: 36Find the values of a and b for which the following system of equation has infinitely many solution:

(i) (2a − 1)x + 3y = 5

3x + (b − 2)y = 3

(ii) 2x − (2a + 5)y = 5

(2b + 1)x − 9y = 15

(iii) (a − 1)x + 3y = 2

6x + (1 − 2b)y = 6

(iv) 3x + 4y = 12

(a + b)x + 2(a − b)y = 5a – 1

(v) 2x + 3y = 7

(a − 1)x + (a + 1)y = 3a − 1

(vi) 2x + 3y = 7

(a − 1)x + (a + 2)y = 3a

Solution:(i) The given system of equation may be written as,

(2a − 1)x + 3y − 5 = 0 3x + (b − 2)y − 3 = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2a − 1, b_{1 }= 3, c_{1}= −5a

_{2}= 3, b_{2}= b − 2, c_{2 }= -3(p + q + 1)The given system of equation will have infinitely many solution, if

2a - 1 = 5 and - 9 = 5(b - 2)

⇒ a = 3 and -9 = 5b - 10

a = 3 and b = 1/5

Hence the given system of equation will have infinitely many solution if

a = 3 and b = 1/5.

(ii) The given system of equation may be written as,

2x − (2a + 5)y = 5 (2b + 1)x − 9y = 15

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= - (2a + 5), c_{1}= −5a

_{2}= (2b + 1), b_{2}= −9, c_{2}= −15The given system of equation will have infinitely many solution, if

Hence the given system of equation will have infinitely many solution if

a = - 1 and b = 5/2.

(iii) The given system of equation may be written as,

(a − 1)x + 3y = 2 6x + (1 − 2b)y = 6

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= a-1, b_{1}= 3, c_{1}= −2a

_{2}= 6, b_{2}= 1 − 2b, c_{2}= −6The given system of equation will have infinitely many solution, if

⇒ a – 1 = 2 and 1 – 2b = 9

⇒ a - 1 = 2 and 1 - 2b = 9

⇒ a = 3 and b = -4

⇒ a = 3 and b = -4

Hence the given system of equation will have infinitely many solution if

a = 3 and b = −4.

(iv) The given system of equation may be written as,

3x + 4y − 12 = 0 (a + b)x + 2(a − b)y − (5a − 1) = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 3, b_{1}= 4, c_{1}= −12a

_{2 }= (a + b), b_{2}= 2(a − b), c_{2}= – (5a − 1)The given system of equation will have infinitely many solution, if

⇒ 3(a - b) = 2a + 2b and 2(5a - 1) = 12(a - b)

⇒ a = 5b and -2a = -12b + 2

Substituting a = 5b in -2a = -12b + 2, we have

-2(5b) = -12b + 2

⇒ −10b = −12b + 2 ⇒ b = 1

Thus a = 5

Hence the given system of equation will have infinitely many solution if

a = 5 and b = 1.

(v) The given system of equation may be written as,

2x + 3y − 7 = 0 (a − 1)x + (a + 1)y − (3a − 1) = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1}= 2, b_{1}= 3, c_{1}= −7a

_{2}= (a − 1), b_{2}= (a + 1), c_{2}= - (3a − 1)The given system of equation will have infinitely many solution, if

⇒ 2(a + 1) = 3(a - 1) and 3(3a - 1) = 7(a + 1)

⇒ 2a - 3a = -3 - 2 and 9a - 3 = 7a + 7

⇒ a = 5 and a = 5

Hence the given system of equation will have infinitely many solution if

a = 5 and b = 1.

(vi) The given system of equation may be written as,

2x + 3y − 7 = 0 (a − 1)x + (a + 2)y − 3a = 0

The given system of equation is of the form

a

_{1}x + b_{1}y − c_{1}= 0 a_{2}x + b_{2}y − c_{2}= 0Where, a

_{1 }= 2, b_{1}= 3, c_{1}= −7a

_{2}= (a − 1), b_{2 }= (a + 2), c_{2}= −3aThe given system of equation will have infinitely many solution, if

⇒ 2(a + 2) = 3(a - 1) and 3(3a) = 7(a + 2)

⇒ 2a + 4 = 3a - 3 and 9a = 7a + 14

⇒ a = 7 and a = 7

Hence the given system of equation will have infinitely many solution if

a = 7and b = 1.