 # RS Aggarwal Class 10 Solutions Chapter 3 - Linear Equations In Two Variables - Ex 3F(3.6)

## RS Aggarwal Class 10 Chapter 3 - Linear Equations In Two Variables - Ex 3F(3.6) Solutions Free PDF

The RS Aggarwal Class 10 solutions for Chapter 3 provided here show you how to solve the exercise questions included in the RS Aggarwal Maths textbook step by step. The solutions are designed according to the latest CBSE syllabus and follow all the CCE guidelines. The detailed  Class 10 solutions of RS Aggarwal will help you lay a strong foundation of the basic Mathematics concepts. Practicing the exercise questions will help you ace the board exam as well as other competitive exams you are appearing for.

You can use these solutions to understand answers to the important and relevant questions asked in the examination. They serve as a valuable aid to you while preparing for your Class 10 board exam. All the solutions of RS Aggarwal Solutions Class 10 Chapter 3 – Linear Equations In Two Variables are well-formated in a proper stepwise manner in an easy to understand way.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3F(3.6)

Question 1: $\frac{1}{x} + \frac{1}{y} – 7 = 0$, $\frac{2}{x} + \frac{3}{y} – 17 = 0$

Solution:

$\frac{1}{x} + \frac{1}{y} – 7 = 0$

$\frac{2}{x} + \frac{3}{y} – 17 = 0$

Taking (1/x) = u and (1/y) = 0

We have,

u + v – 7 = 0

2u + 3v – 17 = 0

By cross multiplication, we have:

$\frac{u}{[1 \times (-17) – 3 \times (-7)]} = \frac{v}{[(-7) \times 2 – 1 \times (-17)] } = \frac{1}{3 – 2}$

$\Rightarrow \frac{u}{-17 + 21} = \frac{v}{-14 + 17} = \frac{1}{1}$

$\Rightarrow \frac{u}{4} = \frac{v}{3} = \frac{1}{1}$

$\Rightarrow \frac{u}{4} = 1, \, \frac{v}{3} = 1$

$\Rightarrow$ u = 4, v = 3

$\Rightarrow \frac{1}{x} = 4, \, \frac{1}{y} = 3$

Hence the solution is:

$x = \frac{1}{4} \, and \, y = \frac{1}{3}$

Question 2: $\frac{5}{\left ( x+y \right )}-\frac{2}{\left ( x-y \right )}+1=0,$

$\frac{15}{\left ( x+y \right )}-\frac{7}{\left ( x-y \right )}-10=0\:\:\left ( x\neq y,x\neq -y \right )$

Solution:

Let, $\frac{1}{x + y} = u \, and \, \frac{1}{x – y} = v$

In the equation

5u – 2v + 1 = 0

15u + 7v – 10 = 0

$\frac{u}{[-2 \times (-10) – 1 \times 7] } = \frac{v}{1 \times 15 – ( -10) \times 5} = \frac{1}{35 + 30}$

$\frac{u}{20 – 7} = \frac{v}{15 + 50} = \frac{1}{65}$

$\frac{u}{13} = \frac{v}{65} = \frac{1}{65}$

$\frac{u}{13} = \frac{1}{65}, \, \frac{v}{65} = \frac{1}{65}$

$u = \frac{13}{65}, \, v = \frac{65}{65}$

Hence,

u = (1/5), v = 1

So, $\frac{1}{x + y} = \frac{1}{5}, \frac{1}{x – y} = 1$

x + y = 5, x – y = 1

By cross multiplication, we have

$\frac{x}{[1 \times (-1) – (-5) \times (-1)]} = \frac{y}{[(-5) \times 1 – (-1) \times 1]} = \frac{1}{1 \times (-1) – 1 \times 1] }$

$\Rightarrow \frac{x} {(-1 – 5)} = \frac{y}{-5 + 1} = \frac{1}{-1 -1}$

$\Rightarrow \frac{x}{-6} = \frac{y}{-4} \frac{1}{-2}$

$\Rightarrow \frac{x}{-6} = \frac{1}{2}, \, \frac{y}{-4} = \frac{1}{-2}$

$\Rightarrow x = \frac{-6}{-2} = 3, \, y = \frac{-4}{-2} = 2$

Hence, the solution is x = 3, y = 2.

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

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

Solution:

3x + 5y – 12 = 0, 5x = 3y – 4 = 0

a1 = 3, b1 = 5, c1 = -12

a2 = 5, b2 = 3, c2 = -4

Thus,

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

Hence, the given system of equations has a unique solution.

The given equations are,

3x + 5y = 12 ……..(1)

5x + 3y = 4 ……..(2)

Multiplying (1) by 3, and (2) by 5, we get

9x + 15y = 36 ……..(3)

25x + 15y = 20 ……..(4)

Subtracting (3) from (4), we get

16x = -16 $x = \frac{-16}{16} = -1$

Putting x = -1 in (3), we get

(9)(-1) + 15y = 36

-9 + 15y = 36

15y = 36 + 9 $\Rightarrow y = \frac{45}{15} = 3$

Hence, the solution is x = -1, y = 3

Question 4: $\frac{x}{3} + \frac{y}{2} = 3$, $\Rightarrow \frac{2x + 3y}{6} = 3$

Solution:

$\frac{x}{3} + \frac{y}{2} = 3$

$\Rightarrow \frac{2x + 3y}{6} = 3$

2x + 3y – 18 = 0 …….. (1)

x – 2y – 2 = 0 …….. (2)

a1 = 2, b1 = 3, c1 = -18

a2 = 1, b2 = -2, c2 = -2

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

Hence, the given system of equations has a unique solution.

The given equations are,

2x + 3y = 18 …….. (1)

3x – 6y = 2 …….. (2)

Multiplying (1) by 2, and (2) by 3, we get

4x + 6y = 36 …….. (3)

3x – 6y = 6 …….. (4)

Adding (3) and (4), we get

7x = 42 $x = 6$

Putting x = 6 in (1), we get

(2)(6) + 3y = 18

3y = 18 – 12

3y = 6 $\Rightarrow y = \frac{6}{3} = 2$

Hence, the solution is x = 6, y = 2

Question 5: 3x – 5y – 7 = 0, 6x – 10y – 3 = 0

Solution:

3x – 5y – 7 = 0 …….. (1)

6x – 10y – 3 = 0 …….. (2)

a1 = 3, b1 = -5, c1 = -7

a2 = 6, b2 = -10, c2 = -3

Thus, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Hence, the given system of equations is inconsistent.

Question 6: 2x – 3y – 5 = 0, 6x – 9y – 15 = 0

Solution:

2x – 3y – 5 = 0 …….. (1)

6x – 9y – 15 = 0 …….. (2)

a1 = 2, b1 = -3, c1 = -5

a2 = 6, b2 = -9, c2 = -15

Thus, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Hence, the given system of equations has infinitely many solutions.

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

Question 7: kx + 2y – 5 = 0, 3x – 4y – 10 = 0

Solution:

kx + 2y – 5 = 0 ……..(1)

3x – 4y – 10 = 0 …….. (2)

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

a2 = 3, b2 = -4, c2 = -10

For unique solution, we must have

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

This happens when,

$k \neq \frac{-3}{2}$

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

(ii) for no solution we must have:

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

Now,

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

$\frac{k}{3} = \frac{2}{-4} \neq \frac{-5}{-10}$

$\frac{k}{3} = \frac{2}{-4} and \frac{k}{3} \neq \frac{1}{2}$

$k = \frac{-3}{2}, k \neq \frac{3}{2}$

Hence, the given system of equations is has no solution if k = (-3/2).

Question 8: x + 2y – 5 = 0, 3x + ky + 15 = 0

Solution:

x + 2y – 5 = 0 …….. (1)

3x + ky + 15 = 0 …….. (2)

a1 = 1, b1 = 2, c1 = -5

a2 = 3, b2 = k, c2 = 15

For unique solution, we must have

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} or \frac{1}{3} \neq \frac{2}{k} \Rightarrow k \neq 6$

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

(ii) for no solution we must have

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

Now,

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

$\frac{1}{3} = \frac{2}{k} \neq \frac{-5}{15}$

$\frac{1}{3} = \frac{2}{k} and \frac{2}{k} \neq \frac{-5}{15}$

Therefore, k = 6.

Hence, the given system of equations has no solution if k = 6.

Question 9: x + 2y = 3, 5x – ky = 7

Solution:

x + 2y – 3 = 0 …….. (1)

5x – ky – 7 = 0 …….. (2)

a1 = 1, b1 = 2, c1 = -3

a2 = 5, b2 = k, c2 = 7

For unique solution, we must have

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} or \frac{1}{5} \neq \frac{2}{k} \Rightarrow k \neq 10$

Thus, for all real value of k other than 10, the given system of equations will have a unique solution.

(ii) For no solution we must have

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

Now,

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

$\frac{1}{5} = \frac{2}{k} \neq \frac{-3}{7}$

$\frac{1}{5} = \frac{2}{k} or \frac{2}{k} \neq \frac{-3}{7}$

$k = 10, or k \neq \frac{-14}{3}$

Hence, the given system of equations is has no solution if k =10, $k \neq \frac{-14}{3}$.

(iii) for infinite number of solutions we must have,

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{1}{5} = \frac{5}{k} = \frac{-3}{7}$

This is never possible since, $\frac{1}{5} \neq \frac{-3}{7}$

There is no value of k for which system of equations has infinitely many solutions.

Question 10: 8x + 5y = 9, kx – 10y = 15

Solution:

8x + 5y – 9 = 0 …….. (1)

kx – 10y – 15 = 0 ……..(2)

a1 = 8, b1 = 5, c1 = -9

a2 = k, b2 = 10, c2 = -15

For no solution, we must have

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

Now, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{8}{k} = \frac{5}{10 \neq \frac{-9}{-15}}$

$\Rightarrow \frac{8}{k} = \frac{1}{2} \neq \frac{3}{5}$

$\Rightarrow \frac{8}{k} = \frac{1}{2} and \frac{8}{k} \neq \frac{3}{5}$

$\Rightarrow k = 16 and k \neq \frac{40}{3}$

Clearly, K = 16 also satisfies the condition k $\neq \frac{40}{3}$

Hence, the given system will have no solution when k = 16.

Question 11: kx + 3y – 3 = 0, 12x – ky – 6 = 0

Solution:

kx + 3y – 3 = 0 ……..(1)

12x – ky – 6 = 0 …….. (2)

a1 = k, b1 = 3, c1 = -3

a2 = 12, b2 = k, c2 = -6

for no solution we must have:

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

Now,

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

$\frac{k}{12} = \frac{3}{k} \neq \frac{-3}{6}$

$\frac{k}{12} = \frac{3}{k} and \frac{3}{k} \neq \frac{1}{2}$

$k^{2} = 36, and k \neq 6$

Hence, k = -6

Hence, the given system of equations is has no solution if k =-6.

Question 12: 3x + y – 1 = 0, (2k – 1) x + (k – 1) y – (2k + 1) = 0

Solution:

3x + y – 1 = 0 …….. (1)

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

a1 = 3, b1 = 1, c1 = -1

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

For no solution we must have:

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

Now,

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

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

$\frac{3}{2k + 1} = \frac{1}{k – 1} and \frac{1}{k – 1} \neq \frac{1}{2k + 1}$

$3k – 3 = 2k – 1 and (2k + 1) \neq (k – 1)$

Hence,$k = 2 and k \neq -2$

Hence, the given system of equations is has no solution when k = 2

Question 13: (3k + 1)x + 3y = 2, (k2 + 1)x + (k – 2)y = 5

Solution:

(3k + 1)x + 3y – 2 = 0 ……..(1)

(k2 + 1)x + (k – 2)y – 5 = 0 ……..(2)

a1 = (3k + 1), b1 = 3,  c1 = -2

a2 = (k2 + 1), b2 = (k – 2), c2 = -5

For no solution we must have

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

Now,

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

$\frac{3k + 1}{(k^{2} + 1)} = \frac{3}{k – 2} \neq \frac{2}{-5}$

$\frac{3k + 1}{(k^{2} + 1)} = \frac{3}{k – 2} and \frac{3}{k – 2} \neq \frac{2}{5}$

$\Rightarrow$ 3k2 + k – 6k – 2 = 3k2 + 3 and 2k – 4 $\neq$ 15

Hence,$k = -1 and k \neq \frac{19}{2}$

Hence, the given system of equations is has no solution when $k = -1 and k \neq \frac{19}{2}$

Thus, k = -1 also satisfies the condition, $k \neq \frac{19}{2}$

Hence, the given system will have no solution when k = -1

Question 14: 3x – y = 5, 6x – 2y = – k

Solution:

3x – y – 5 = 0 …….. (1)

6x – 2y + k = 0 …….. (2)

a1 = 3, b1 = -1, c1 = -5

a2 = 6, b2 = -2, c2 = k

Equations (1) and (2) have no solution, if

$\frac{-5}{k} \neq \frac{1}{2} or k \neq -10$

### Key Features of RS Aggarwal Class 10 Solutions Chapter 3 – Linear Equations In Two Variables – Ex 3F(3.6)

• The solutions steps are explained in the simplest manner without missing out on important aspects of solving a question.
• These solutions teach different techniques to solve tricky questions.
• The RS Aggarwal Maths solution includes questions from each chapter and topic.
• The solutions are prepared by subject experts according to the CBSE syllabus.

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Find the volume of the figure. 