(i) x + y = 14

x – y = 4

(ii) s – t = 3

s/3 + t/2 = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) √2x + √3y = 0

√3x – √8y = 0

(vi) 3x/2 – 5y/3 = -2

x/3 + y/2 = 13/6

**Solution:**

**(i) x + y = 14 and x – y = 4 **

From the first equation, we get,

x = 14 – y

Now, substitute the value of x in the second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

**(ii) s – t = 3 and (s/3) + (t/2) = 6**

From the first equation, we get,

s = 3 + t ….(1)

Now, substitute the value of s in the second equation to get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒ 5t = 30

⇒ t = 6

Now, substitute the value of t in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given, 3x – y = 3 and 9x – 3y = 9 are the two equations.

From the first equation, we get,

x = (3 + y)/3

Now, substitute the value of x in the given second equation to get,

9(3 + y)/3 – 3y = 9

⇒ 9 + 3y – 3y = 9

⇒ 9 = 9

Therefore, y has infinite values and since, x = (3 + y) /3, so x also has infinite values.

**(iv) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 **

From the first equation, we get,

x = (1.3 – 0.3y)/0.2 ….(1)

Now, substitute the value of x in the given second equation to get,

0.4(1.3 – 0.3y)/0.2 + 0.5y = 2.3

⇒ 2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

Now, substitute the value of y in equation (1), we get,

x = (1.3 – 0.3(3))/0.2 = (1.3 – 0.9)/0.2 = 0.4/0.2 = 2

Therefore, x = 2 and y = 3.

**(v) √2 x + √3 y = 0 and √3 x – √8 y = 0**

are the two equations.

From the first equation, we get,

x = – (√3/√2)y ….(1)

Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0

⇒ (-3/√2)y – √8 y = 0

⇒ y = 0

Now, substitute the value of y in equation (1), we get,

x = 0

Therefore, x = 0 and y = 0.

**(vi) (3x/2) – (5y/3) = -2 and (x/3) + (y/2) = 13/6 **

From 1^{st} equation, we get,

(3/2)x = -2 + (5y/3)

⇒ x = 2(-6 + 5y)/9 = (-12 + 10y)/9 ….(1)

Putting the value of x in the given second equation to get,

((-12 + 10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12 + 10y)/27 ) + y/2 = 13/6

Solving this equation, we get y = 3.

Now, substitute the value of y in equation (1), we get,

(3x/2) – 5(3)/3 = -2

⇒ (3x/2) – 5 = -2

⇒ x = 2

Therefore, x = 2 and y = 3.