**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:

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

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

\( \Rightarrow \frac{u}{4} = \frac{v}{3} = \frac{1}{1} \)

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

\( \Rightarrow \)

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

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

a_{1 }= 3, b_{1} = 5, c_{1 }= -12

a_{2} = 5, b_{2} = 3, c_{2 }= -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)

a_{1 }= 2, b_{1} = 3, c_{1 }= -18

a_{2} = 1, b_{2} = -2, c_{2 }= -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)

a_{1 }= 3, b_{1} = -5, c_{1 }= -7

a_{2} = 6, b_{2} = -10, c_{2 }= -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)

a_{1 }= 2, b_{1} = -3, c_{1 }= -5

a_{2} = 6, b_{2} = -9, c_{2 }= -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)

a_{1 }= k, b_{1} = 2, c_{1 }= -5

a_{2} = 3, b_{2} = -4, c_{2 }= -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)

a_{1 }= 1, b_{1} = 2, c_{1 }= -5

a_{2} = 3, b_{2} = k, c_{2 }= 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)

a_{1 }= 1, b_{1} = 2, c_{1 }= -3

a_{2} = 5, b_{2} = k, c_{2 }= 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)

a_{1 }= 8, b_{1} = 5, c_{1 }= -9

a_{2} = k, b_{2} = 10, c_{2 }= -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)

a_{1 }= k, b_{1} = 3, c_{1 }= -3

a_{2} = 12, b_{2} = k, c_{2 }= -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)

a_{1 }= 3, b_{1} = 1, c_{1 }= -1

a_{2} = (2k – 1), b_{2} = (k – 1), c_{2 }= -(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,** **(k**^{2}** + 1)x + (k – 2)y = 5 **

**Solution:**

(3k + 1)x + 3y – 2 = 0 ……..(1)

(k^{2} + 1)x + (k – 2)y – 5 = 0 ……..(2)

a_{1 }= (3k + 1), b_{1} = 3, c_{1 }= -2

a_{2} = (k^{2} + 1), b_{2} = (k – 2), c_{2 }= -5

For no solution we must have

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

Now,

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

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

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

\( \Rightarrow \)^{2} + k – 6k – 2 = 3k^{2} + 3 and 2k – 4 \( \neq \)

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)

a_{1 }= 3, b_{1} = -1, c_{1 }= -5

a_{2} = 6, b_{2} = -2, c_{2 }= k

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

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