# RS Aggarwal Solutions Class 10 Ex 3G

## RS Aggarwal Class 10 Ex 3G Chapter 3

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 $\frac{11}{5}$

#### Practise This Question

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig.). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m3, and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed (in m3)? (Take π=227)