Use method of cross-multiplication to solve :

(i) 2x + y = 8 and 3x - 2y = 5

(ii) x + 4y = 3 and 2x + 9y = 5

(i) Given, equations :

⇒ 2x + y = 8 and 3x - 2y = 5

⇒ 2x + y - 8 = 0 ……………(1)

⇒ 3x - 2y - 5 = 0 ……………(2)

If the two equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

By cross-multiplication method :

$\dfrac{x}{b$1c2 - b2c1} = \dfrac{y}{c1a2 - c2a1} = \dfrac{1}{b2a1 - b1a2}

Now substitute the values, we get:

⇒ $\dfrac{x}{1 \times (-5) - (-2) \times (-8)} = \dfrac{y}{(-8) \times 3 - (-5) \times 2} = \dfrac{1}{(-2) \times 2 - 1 \times 3}$

⇒ $\dfrac{x}{-5 - 16} = \dfrac{y}{-24 + 10} = \dfrac{1}{-4 - 3}$

⇒ $\dfrac{x}{-21} = \dfrac{y}{-14} = \dfrac{1}{-7}$

⇒ x = $\dfrac{-21}{-7}$ and y = $\dfrac{-14}{-7}$

⇒ x = 3 and y = 2

Hence, the value of x = 3 and y = 2.

(ii) Given, equations :

⇒ x + 4y = 3 and 2x + 9y = 5

⇒ x + 4y - 3 = 0 ……………(1)

⇒ 2x + 9y - 5 = 0 ……………(2)

If the two equation are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

By using the method of cross - multiplication

$\dfrac{x}{b$1c2 - b2c1} = \dfrac{y}{c1a2 - c2a1} = \dfrac{1}{b2a1 - b1a2}

Now substitute the values, we get:

⇒ $\dfrac{x}{4 \times (-5) - 9 \times (-3)} = \dfrac{y}{(-3) \times 2 - (-5) \times 1} = \dfrac{1}{9 \times 1 - 4 \times 2}$

⇒ $\dfrac{x}{-20 + 27} = \dfrac{y}{-6 + 5} = \dfrac{1}{9 - 8}$

⇒ $\dfrac{x}{7} = \dfrac{y}{-1} = \dfrac{1}{1}$

⇒ x = $\dfrac{7}{1}$ and y = $\dfrac{-1}{1}$

⇒ x = 7 and y = -1

Hence, the value of x = 7 and y = -1.