If 2x + y = 32 and 3x + 4y = 68, find the value of $\dfrac{x}{y}$.

Given,

Equations :

2x + y = 32     ………(1),

3x + 4y = 68     ………(2)

Multiplying equation (1) by 4, we get :

⇒ 4(2x + y) = 32 × 4

⇒ 8x + 4y = 128     ………(3)

Subtracting equation (2) from (3), we get :

⇒ 8x + 4y - (3x + 4y) = 128 - 68

⇒ 8x + 4y - 3x - 4y = 60

⇒ 5x = 60

⇒ x = $\dfrac{60}{5}$

⇒ x = 12.

Substituting value of x in equation (2), we get :

⇒ 3x + 4y = 68

⇒ 3 × 12 + 4y = 68

⇒ 36 + 4y = 68

⇒ 4y = 68 - 36

⇒ 4y = 32

⇒ y = $\dfrac{32}{4}$

⇒ y = 8.

Substituting value of x and y in $\dfrac{x}{y}$, we get :

$\Rightarrow \dfrac{x}{y} = \dfrac{12}{8} = \dfrac{3}{2}$.

Hence, $\dfrac{x}{y} = \dfrac{3}{2}$.