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

R S Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables, contains solutions for exercise 3d questions. These solutions are prepared by subject matter experts at BYJUâ€™S, explaining about conditions for the solvability of linear equations. Students can download the R S Aggarwal Solutions of Class 10Â  and score well in their exams.

## Access other exercise solutions of Class 10 Maths Chapter 3 Linear Equations in Two Variables

Exercise 3A Solutions: 29 Questions (Long Answers)

Exercise 3B Solutions: 20 Questions (Short Answers)

Exercise 3C Solutions: 13 Questions (Long Answers)

Exercise 3E Solutions: 20 Questions (Long Answers)

Exercise 3F Solutions: 10 Questions (6 Short Answers and 4 Long Answers)

## Exercise 3D Page No: 122

Show that each of the following systems of equations has a unique solution and solve it:

Question 1:

3x + 5y = 12, 5x + 3y = 4

Solution:

3x + 5y = 12 â€¦â€¦(1)

5x + 3y = 4 â€¦â€¦(2)

Here,

a1 = 3, b1 = 5, c1 = 12

a2 = 5, b2 = 3, c2 = 4

$\frac{a_{1}}{a_{2}} = \frac{3}{5} , \frac{b_{1}}{b_{2}} = \frac{5}{3}$

Which shows,

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

The system of equations have unique solution.

Solve the equations using substitution method:

From (1): x = (12-5y)/3

From (2): 5x + 3y – 4 = 0

5((12-5y)/3) + 3y – 4 = 0

60 â€“ 25y + 9y = 12

60 â€“ 16y = 12

16y = 48

y = 3

Now, substitute y in (1)

3x + 5(3) = 12

3x + 15 = 12

3x = 12 â€“ 15

x = – 1

Answer: x = – 1 and y = 3

Question 2.

2x â€“ 3y = 17, 4x + y = 13.

Solution:

2x â€“ 3y = 17 â€¦â€¦..(1)

4x + y = 13 â€¦â€¦â€¦(2)

Here

a1 = 2, b1 = – 3, c1 = 17

a2 = 4, b2 = 1, c2 = 13

$\frac{a_{1}}{a_{2}} = \frac{2}{4} , \frac{b_{1}}{b_{2}} = \frac{-3}{1} = -3$

Which shows:

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

System has unique solutions.

Solve the equations using substitution method:

From (1): x = (17 + 3y)/2

Putting value of x in (2)

4((17 + 3y)/2) + y = 13

68 + 12y + 2y = 26

68 + 14y = 26

y = -3

Again, x = (17 + 3(-3))/2

2x = 8

x = 4

Answer: x = 4 and y = – 3

Question 3:

x/3 + y/2 = 3, x – 2y = 2

Solution:

2x + 3y = 18 …(1)

x – 2y = 2 …(2)

a1 = 2, b1 = 3, c1 = 18

a2 = 1, b2 = – 2, c2 = 2

$\frac{a_{1}}{a_{2}} = \frac{2}{1} = 2 , \frac{b_{1}}{b_{2}} = \frac{3}{-2}$

which shows ,

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

This system has a unique solution.

Solve the equations:

From (2), x = 2 + 2y

Substituting the value of x in (1),

2(2 + 2y) + 3y = 18

=> 4 + 4y + 3y = 18

=> 7y = 18 â€“ 4 = 14

=> y = 2

and x = 2 + 2 x 2 = 2 + 4 = 6

Answer: x = 6 and y = 2

Find the value of k for which each of the following systems of equations has a unique solution:

Question 4:

2x + 3y â€“ 5 = 0, kx â€“ 6y â€“ 8 = 0.

Solution:

2x + 3y â€“ 5 = 0 â€¦â€¦â€¦â€¦.(1)

kx â€“ 6y â€“ 8 = 0 â€¦â€¦â€¦..(2)

a1 = 2, b1 = 3, c1 = – 5

a2 = k, b2 = – 6, c2 = – 8

$\frac{a_{1}}{a_{2}} = \frac{2}{k} , \frac{b_{1}}{b_{2}} = \frac{3}{-6}$

Which shows:

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

System has a unique solution.

Now, Find the value of k:

2/k â‰  3/-6

k â‰  -4

Question 5.

x â€“ ky = 2, 3x + 2y + 5 = 0.

Solution:

x â€“ ky = 2 â€¦.(1)

3x + 2y + 5 = 0 â€¦.(2)

Here,

a1 = 1, b1 = – k, c1 = – 2

a2 = 3, b2 = 2, c2 = 5

$\frac{a_{1}}{a_{2}} = \frac{1}{3} = 2 , \frac{b_{1}}{b_{2}} = \frac{-k}{2} , \frac{c_{1}}{c_{2}} = \frac{2}{-5}$

Systems has a unique solution.

Now,

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{3} \neq \frac{-k}{2}$

k â‰  -2/3

Question 6.

5x â€“ 7y â€“ 5 = 0, 2x + ky â€“ 1 = 0.

Solution:

5x â€“ 7y â€“ 5 = 0 â€¦.(1)

2x + ky â€“ 1 = 0 â€¦..(2)

Here,

a1 = 5, b1 = – 7, c1 = – 5

a2 = 2, b2 = k, c2 = – 1

$\frac{a_{1}}{a_{2}} = \frac{5}{2} , \frac{b_{1}}{b_{2}} = \frac{-7}{+k} = \frac{-7}{k}$

Systems has a unique solution.

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{5}{2} \neq \frac{-7}{k}$

k â‰  -14/5

Question 7.

4x + ky + 8 = 0, x + y + 1 = 0.

Solution:

4x + ky + 8 = 0 â€¦â€¦â€¦(1)

x + y + 1 = 0 â€¦â€¦â€¦â€¦.(2)

Here,

a1 = 4, b1 = k, c1 = 8

a2 = 1, b2 = 1, c2 = 1

$\frac{a_{1}}{a_{2}} = \frac{4}{1} , \frac{b_{1}}{b_{2}} = \frac{k}{1}$

Systems has a unique solution.

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{4}{1} \neq \frac{k}{1}$

k â‰  4

Question 8.

4x â€“ 5y = k, 2x â€“ 3y = 12.

Solution:

4x â€“ 5y = k â€¦â€¦.(1)

2x â€“ 3y = 12 â€¦â€¦â€¦â€¦(2)

Here,

a1 = 4, b1 = – 5, c1 = – k

a2 = 2, b2 = – 3, c2 = – 12

Which shows that,

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

2 â‰  $\frac{5}{3}$

Therefore, k is any real number.

Question 9.

kx + 3y = (k â€“ 3), 12x + ky = k.

Solution:

kx + 3y = (k â€“ 3), 12x + ky = k

Here,

a1 = k, b1 = 3, c1 = k â€“ 3

a2 = 12, b2 = k, c2 = k

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{k}{12} \neq \frac{3}{4}$

Systems has a unique solution.

Question 10:

Show that the system of equations 2x â€“ 3y = 5, 6x â€“ 9y = 15 has an infinite number of solutions.

Solution:

2x â€“ 3y = 5, 6x â€“ 9y = 15

Question 11.

Show that the system of equations 6x + 5y = 11, 9x + 15 /2 y = 21 has no solution.

Solution:

6x + 5y = 11

9x + 15/2 y = 21

Here,

a1 = 6, b1 = 5, c1 = – 11

a2 = 9, b2 = 15/2, c2 = -21

Question 12: For what value of k does the system of equations kx + 2y = 5, 3x â€“ 4y = 10 have

(i) a unique solution,

(ii) no solution?

Solution:

System of equations kx + 2y = 5 and 3x â€“ 4y = 10

Here,

a1 = k, b1 = 2, c1 = 5

a2 = 3, b2 = – 4, c2 = 10

Question 13:

For what value of k does the system of equations x + 2y = 5, 3x + ky + 15 = 0 have

(i) a unique solution,

(ii) no solution?

Solution:

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

Here,

a1 = 1, b1 = 2, c1 = – 5

a2 = 3, b2 = k, c2 = 15

Question 14.

For what value of k does 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 equations has infinitely many solutions.

Solution:

x + 2y = 3, 5x + ky + 7 = 0

Here

a1 = 1, b1 = 2, c1 = – 5

a2 = 3, b2 = k, c2 = 15

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

Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3d is based on the topic and subtopics:

• Conditions for solvability of linear equations
• Consistent and Inconsistent systems of linear equations
• Conditions for solvability of linear equations