(i) Given,

0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 are the two equations.

From 1^st 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.

(ii) Given,

√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.

From 1^st 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.

(iii) Given,

(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.

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

\(\frac{\frac{-12 + 10y}{9}}{3}\) + \(\frac{y}{2}\) = \(\frac{13}{6}\)

⇒ \(\frac{-12 + 10y}{27}\) + \(\frac{y}{2}\) = \(\frac{13}{6}\)

⇒ \(\frac{y}{2}\) = \(\frac{13}{6}\) - \(\frac{-12 + 10y}{27}\)

⇒ \(\frac{y}{2}\) = \(\frac{117}{54}\) - \(\frac{-24 + 20y}{54}\)

⇒ \(\frac{y}{2}\) = \(\frac{117 + 24 - 20y}{54}\)

⇒ 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.