# RS Aggarwal Solutions Class 10 Ex 3F

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

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

#### Practise This Question

Silicon is also tetravalent. But why can’t it form long chain compounds like carbon?