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.

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## 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,

a_1 = 3, b_1 = 5, c_1 = 12

a_2 = 5, b_2 = 3, c_2 = 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

a_1 = 2, b_1 = – 3, c_1 = 17

a_2 = 4, b_2 = 1, c_2 = 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/2 + y/2 = 3, x – 2y = 2**

**Solution:**

2x + 3y = 18 …(1)

x – 2y = 2 …(2)

a_1 = 2, b_1 = 3, c_1 = 18

a_2 = 1, b_2 = – 2, c_2 = 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)

a_1 = 2, b_1 = 3, c_1 = – 5

a_2 = k, b_2 = – 6, c_2 = – 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,

a_1 = 1, b_1 = – k, c_1 = – 2

a_2 = 3, b_2 = 2, c_2 = 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,

a_1 = 5, b_1 = – 7, c_1 = – 5

a_2 = 2, b_2 = k, c_2 = – 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,

a_1 = 4, b_1 = k, c_1 = 8

a_2 = 1, b_2 = 1, c_2 = 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,

a_1 = 4, b_1 = – 5, c_1 = – k

a_2 = 2, b_2 = – 3, c_2 = – 12

Which shows that,

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)2 â‰ \(\frac{5}{7}\)

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,

a_1 = k, b_1 = 3, c_1 = k â€“ 3

a_2 = 12, b_2 = k, c_2 = 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,

a_1 = 6, b_1 = 5, c_1 = – 11

a_2 = 9, b_2 = 15/2, c_2 = -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,

a_1 = k, b_1 = 2, c_1 = 5

a_2 = 3, b_2 = – 4, c_2 = 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,

a_1 = 1, b_1 = 2, c_1 = – 5

a_2 = 3, b_2 = k, c_2 = 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

a_1 = 1, b_1 = 2, c_1 = – 5

a_2 = 3, b_2 = k, c_2 = 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