RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3G(3.7)

RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3G(3.7) Solutions Free PDF

For the students of Class 10, RS Aggarwal Solutions Class 10 Chapter 3 – Linear Equations In Two Variables are highly effective means while preparing for the exam. We at our BYJU’S website made available all the exercise solutions to help students develop a better understanding of the concepts through proper and detailed explanations of the questions. Practicing these solutions daily will help the students to enhance their performance in the exam. All the solutions are solved in a step-wise manner so that students can refer to it while solving difficult questions.

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Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3G(3.7)

Question 1: kx + 2y = 5, 3x + y = 1

Solution:

kx + 2y – 5 = 0 ……..(1)

3x + y – 1 = 0 …….. (2)

a1 = k, b1 = 2, c1 = -5

a2 = 3, b2 = 1, c2 = -1

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{a_{1}}{a_{2}}  \neq \frac{b_{1}}{b_{2}} i.e \frac{k}{3} \neq \frac{2}{1} \)

\( k \neq 6 \)

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

Question 2: x – 2y = 3, 3x + ky = 1

Solution:

x – 2y – 3 = 0 …….. (1)

3x + ky – 1 = 0 …….. (2)

a1 = 1, b1 = -2, c1 = -3

a2 = 3, b2 = k, c2 = -1

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{1}{3}  \neq \frac{-2}{k} \)

\( k \neq -6 \)

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

Question 3: kx + 3y = (k – 3), 12x + ky = k

Solution:

kx + 3y – (k – 3) = 0 ……..(1)

12x + ky – k = 0 …….. (2)

a1 = k, b1 = 3, c1 = -(k – 3)

a2 = 12, b2 = k, c2 = -k

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{k}{12} \neq \frac{3}{k} \)

\( k^{2} \neq 36 \Rightarrow k \neq \pm 6 \)

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

Question 4: 4x – 5y = k, 2x – 3y = 12

Solution:

4x – 5y -k = 0 …….. (1)

2x – 3y – 12 = 0 …….. (2)

a1 = 4, b1 = -5, c1 = -k

a2 = 2, b2 = -3, c2 = -12

For unique solution, we must have, \( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \)

Now,

\( \frac{4}{2} \neq \frac{-5}{-3} \)

\( 2 \neq \frac{5}{3} \Rightarrow 6 \neq 5 \)

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

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

Question 5: 2x + 3y = 7 , (k – 1)x + (k + 2)y = 3k

Solution:

2x + 3y – 7 = 0 …….. (1)

(k – 1)x + (k + 2)y – 3k = 0 ……..(2)

a1 = 2, b1 = 3, c1 = -7

a2 = (k – 1), b2 = (k + 2), c2 = -3k

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{2}{k – 1} = \frac{3}{k + 2} = \frac{-7}{-3k} \)

\( \frac{2}{k – 1} = \frac{3}{k + 2} = \frac{7}{3k} \)

Now, the following cases arises

Case 1:

\( \frac{2}{k – 1} = \frac{3}{k + 2} \)

\( \Rightarrow 2(k + 2) = 3(k – 1) \Rightarrow 2k + 4 = 3k -3 \)

\( \Rightarrow k = 7 \)

Case 2:

\( \frac{3}{k + 2} = \frac{7}{3k} \)

\( \Rightarrow 7(k + 2) = 9k \Rightarrow 7k + 14 = 9k \)

\( \Rightarrow k = 7 \)

Case 3:

\( \frac{2}{k – 1} = \frac{7}{3k} \)

\( \Rightarrow 7k – 7 = 6k \)

\( \Rightarrow k = 7 \)

For k = 7, there are infinitely many solutions of the given system of equations.

Question 6: 2x + (k – 2)y = k, 6x + (2k – 1)y = (2k + 5)

Solution:

2x + (k – 2)y – k = 0 ……..(1)

6x + (2k – 1)y – (2k + 5) = 0 ……..(2)

a1 = 2, b1 = (k – 2), c1 = -k

a2 = 6, b2 = (2k – 1), c2 = -(2k + 5)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{2}{6} = \frac{k – 2}{2k – 1} = \frac{-k}{-(2k + 5)} \)

\( \frac{1}{3} = \frac{k – 2}{2k – 1} = \frac{k}{2k + 5} \)

Now, the following cases arises

Case 1:

\( \frac{1}{3} = \frac{k – 2}{2k – 1} \)

\( \Rightarrow 2k – 1 = 3k – 6 \)

\( \Rightarrow k = 5 \)

Case 2:

\( \frac{k – 2}{2k – 1} = \frac{k}{2k + 5} \)

\( \Rightarrow 2k^{2} + 5k – 4k -10 2k^{2} – k \Rightarrow k + k = 10 \)

\( \Rightarrow k = 5 \)

Case 3:

\( \frac{1}{3} = \frac{k}{2k + 5} \)

\( \Rightarrow 2k + 5 = 3k \)

\( \Rightarrow k = 5 \)

For k = 5, there are infinitely many solutions of the given system of equations.

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

Solution:

kx + 3y – (2k + 1) = 0 ……..(1)

2(k + 1)x + 9y – (7k + 1) = 0 ……..(2)

a1 = k, b1 = 3, c1 = -(2k + 1)

a2 = 2(k + 1), b2 = 9, c2 = -(7k + 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{k}{2(k + 1)} = \frac{3}{9} = \frac{-(2k + 1)}{-(7k + 1)} \)

\( \frac{k}{2(k + 1)} = \frac{1}{3} = \frac{2k +1 }{7k + 1} \)

Now, the following cases arises

Case 1:

\( \frac{k}{2(k + 1)} = \frac{1}{3} \)

\( \Rightarrow 2(k + 1) = 3k \Rightarrow 2k + 2 = 3k \)

\( \Rightarrow k = 2 \)

Case 2:

\( \frac{1}{3} = \frac{2k + 1}{7k + 1} \)

\( \Rightarrow 7k + 1 = 6k + 3 \Rightarrow 7k – 6k = 3 – 1 \)

\( \Rightarrow k = 2 \)

Case 3:

\( \frac{k}{2(k + 1)} = \frac{2k + 1}{7k + 1} \)

\( \Rightarrow k(7k + 1) = 2(2k + 1)(k + 1) \)

\( \Rightarrow 7k^{2} + k = 2(2k^{2} + 2k + k + 1) \)

7k2 + k = 4k2 + 6k + 2

7k2 – 4k2 + k – 6k – 2 = 0

3k2 – 5k – 2 = 0

3k2 – (6k – 1k) – 2 = 0

3k(k – 2) + 1(k – 2) = 0

(k – 2) (3k + 1) = 0

\( k = 2 or k = \frac{-1}{3} \)

Thus, for k = 2, there are infinitely many solutions of the given system of equations.

Question 8: 5x + 2y = 2k, 2(k + 1)x + ky = (3k + 4)

Solution:

5x + 2y – 2k = 0 …….. (1)

2(k + 1)x + ky – (3k + 4) = 0 ……..(2)

a1 = 5, b1 = 2, c1 = -2k

a2 = 2(k + 1), b2 = k, c2 = -(3k + 4)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{5}{2(k + 1)} = \frac{2}{k} = \frac{-2k}{-(3k + 4)} \)

\( \frac{5}{2(k + 1)} = \frac{2}{k} = \frac{2k}{(3k + 4)} \)

Now, the following cases arises

Case 1:

\( \frac{5}{2(k + 1)} = \frac{2}{k} \)

\( \Rightarrow 5k = 4(k + 1) \Rightarrow 5k = 4k + 4 \)

\( \Rightarrow k = 4 \)

Case 2:

\( \frac{2}{k} = \frac{2k}{(3k + 4)} \)

\( \Rightarrow 2(3k + 4) = 2k^{2} \Rightarrow 6k + 8 = 2k^{2} \)

\( \Rightarrow 2k^{2} – 6k – 8 = 0 \)

On solving the above quadratic equation, we get k = 4 or k = -1

Case 3:

\( \frac{5}{2(k + 1)} = \frac{2k}{(3k + 4)} \)

\( \Rightarrow 15k + 20 = 4k^{2} = 4k \)

\( \Rightarrow 4k^{2} + 4k – 15 – 20 = 0 \)

On solving the above quadratic equations we get, \( k = 4 or k = \frac{-5}{4} \)

Thus, k = 4 is a common value for which there are infinitely many solutions.

Question 9: x+ (k – 1) y = 5, (k + 1)x + 9y = -(8k – 1)

Solution:

x + (k + 1)y – 5 = 0 ……..(1)

(k + 1)x + 9y – (8k – 1) = 0 ……..(2)

a1 = 1, b1 = (k + 1), c1 = -5

a2 = (k + 1), b2 = 9, c2 = -(8k – 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{1}{k + 1} = \frac{k + 1}{9} = \frac{-5}{-(8k – 1)} \)

\( \frac{1}{k + 1} = \frac{k + 1}{9} = \frac{5}{(8k – 1)} \)

Now, the following cases arises:

Case 1:

\( \frac{1}{k + 1} = \frac{k + 1}{9} \)

\( \Rightarrow (k + 1)^{2} = 9  \Rightarrow (k + 1) = \pm 3 \)

K + 1 = 3 or k + 1 = -3

K = 2 or k = -4

Case 2:

\( \frac{k + 1}{9} = \frac{-5}{-(8k – 1)} \)

\( \Rightarrow (k + 1)(8k – 1) = 45 \Rightarrow 8k^{2} + 7k – 46 = 0 \)

On solving the above equation,

\( \Rightarrow k = \frac{-23}{8} or k = 2 \)

Case 3:

\( \frac{1}{k + 1} = \frac{-5}{-(8k – 1)} \)

\( \Rightarrow 8k – 1 = 5(k + 1) \)

\( \Rightarrow 3k = 6 \Rightarrow k = 2 \)

Thus, k = 2 is the common value for which there are infinitely many solutions

Question 10: (k – 1)x – y = 5, (k + 1)x + (1 – k)y = (3k + 1)

Solution:

(k – 1)x – y – 5 = 0 ……..(1)

(k + 1)x + (1 – k)y – (3k + 1) = 0 ……..(2)

a1 = (k – 1), b1 = -1), c1 = -5

a2 = (k + 1), b2 = (1 – k), c2 = -(3k + 1)

For infinite number of solutions we must have,

\( \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \)

This holds only when

\( \frac{k – 1}{k + 1} = \frac{-1}{-(k – 1)} = \frac{-5}{-(3k + 1)} \)

\( \frac{k – 1}{k + 1} = \frac{1}{(k – 1)} = \frac{5}{(3k + 1)} \)

Now, the following cases arise:

Case 1:

\( \frac{k – 1}{k + 1} = \frac{1}{(k – 1)} \)

\( \Rightarrow (k – 1)^{2} = k + 1  \Rightarrow k^{2} = 3k \)

k = 3

Case 2:

\( \frac{k – 1}{k + 1} = \frac{5}{(3k + 1)} \)

\( \Rightarrow (k + 1)(3k + 1) = 5(k + 1) \Rightarrow 3k^{2} – 7k – 6 = 0 \)

On solving the above equation,

\( \Rightarrow k = 3 \)

Case 3:

\( \frac{k – 1}{k + 1} = \frac{5}{(3k + 1)} \)

\( \Rightarrow (k – 1)(3k + 1) = 5(k + 1) \)

\( \Rightarrow 3k^{2} – 7k – 6 = 0 \Rightarrow k = \frac{-2}{3} or k = 3 \)

K = 3 is the common value for which the number of solution is infinitely many.

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:

Question 11: (2a – 1)x + 3y = 5, 3x + (b – 1)y = 2

Solution:

(2a – 1)x + 3y – 5 = 0 ……..(1)

3x + (b – 1)y – 2 = 0 ……..(2)

a1 = (2a – 1), b1 = 3, c1 = -5

a2 = 3, b2 = (b – 1), c2 = -2

These holds only when,

\(\frac{2a – 1}{3}\) = \(\frac{3}{b – 1}\) = \(\frac{-5}{-2} \)

\(\frac{2a – 1}{3}\) = \( \frac{3}{b – 1}\) = \( \frac{5}{2} \)

= \( \frac{2a – 1}{3} = \frac{5}{2} \;\;and \;\; \frac{3}{b – 1} = \frac{5}{2} \)

4a – 2 = 15 and 5(b – 1) = 6

4a = 17 and 5b – 5 = 6

\( a = \frac{17}{4}\) and 5b = 11

Therefore, \( a = \frac{17}{4}\) and \( \frac{11}{5} \)

Key Features of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3G(3.7)

  • The solutions are prepared by subject experts in sync with latest CBSE syllabus.
  • It teaches different methods to solve difficult and tricky questions.
  • These solutions provide different types of questions to practice.
  • It boosts the self-confidence of the students.

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