# 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

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## Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 â€“Â Linear Equations In Two Variables â€“ Ex 3G(3.7)

Question 1:Â kx + 2y = 5,Â 3x + y = 1

Solution:

kx + 2y – 5 = 0 â€¦â€¦..(1)

3x + y – 1 = 0 â€¦â€¦.. (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,

$\frac{a_{1}}{a_{2}} Â \neq \frac{b_{1}}{b_{2}} i.e \frac{k}{3} \neq \frac{2}{1}$

$k \neq 6$

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

Question 2:Â x – 2y = 3,Â 3x + ky = 1

Solution:

x – 2y – 3 = 0 â€¦â€¦.. (1)

3x + ky – 1 = 0 â€¦â€¦.. (2)

a1 = 1,Â 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,

$\frac{1}{3} Â \neq \frac{-2}{k}$

$k \neq -6$

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

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

Solution:

kx + 3y – (k – 3) = 0 â€¦â€¦..(1)

12x + ky – k = 0 â€¦â€¦.. (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:Â 4x – 5y = k,Â 2x – 3y = 12

Solution:

4x – 5y -k = 0 â€¦â€¦.. (1)

2x – 3y – 12 = 0 â€¦â€¦.. (2)

a1 = 4, b1 = -5,Â 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:Â 2x + 3y = 7 ,Â (k – 1)x + (k + 2)y = 3k

Solution:

2x + 3y – 7 = 0 â€¦â€¦.. (1)

(k – 1)x + (k + 2)y – 3k = 0 â€¦â€¦..(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:Â 2x + (k – 2)y = k,Â 6x + (2k – 1)y = (2k + 5)

Solution:

2x + (k – 2)y – k = 0 â€¦â€¦..(1)

6x + (2k – 1)y – (2k + 5) = 0 â€¦â€¦..(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:Â kx + 3y = (2k + 1),Â 2(k + 1)x + 9y = (7k + 1)

Solution:

kx + 3y – (2k + 1) = 0 â€¦â€¦..(1)

2(k + 1)x + 9y – (7k + 1) = 0 â€¦â€¦..(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:Â 5x + 2y = 2k,Â 2(k + 1)x + ky = (3k + 4)

Solution:

5x + 2y – 2k = 0 â€¦â€¦.. (1)

2(k + 1)x + ky – (3k + 4) = 0 â€¦â€¦..(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:Â x+ (k – 1) y = 5,Â (k + 1)x + 9y = -(8k – 1)

Solution:

x + (k + 1)y – 5 = 0 â€¦â€¦..(1)

(k + 1)x + 9y – (8k – 1) = 0 â€¦â€¦..(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:

$\frac{1}{k + 1} = \frac{k + 1}{9}$

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

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

Solution:

(k – 1)x – y – 5 = 0 â€¦â€¦..(1)

(k + 1)x + (1 – k)y – (3k + 1) = 0 â€¦â€¦..(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:

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

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

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

Solution:

(2a – 1)x + 3y – 5 = 0 â€¦â€¦..(1)

3x + (b – 1)y – 2 = 0 â€¦â€¦..(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 $\frac{11}{5}$

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