When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\dfrac{3}{2}$. When the difference of these numbers is divided by the smaller, the result is $\dfrac{1}{2}$ . Find the numbers.

Let two numbers be x and y, where x is the larger and y is the smaller number.

Given,

When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\dfrac{3}{2}$.

$\therefore \dfrac{x + y}{x + 1} = \dfrac{3}{2} \\[1em] \Rightarrow 2(x + y) = 3(x + 1) \\[1em] \Rightarrow 2x + 2y = 3x + 3 \\[1em] \Rightarrow 3x - 2x = 2y - 3 \\[1em] \Rightarrow x = 2y - 3 ………..(1)$

Given,

When the difference of these numbers is divided by the smaller, the result is $\dfrac{1}{2}$.

$\therefore \dfrac{x - y}{y} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x - y) = y \\[1em] \Rightarrow 2x - 2y = y \\[1em] \Rightarrow 2y + y = 2x \\[1em] \Rightarrow 2x = 3y \\[1em] \Rightarrow x = \dfrac{3y}{2} …….(2)$

From (1) and (2), we get :

$\Rightarrow 2y - 3 = \dfrac{3y}{2} \\[1em] \Rightarrow 2(2y - 3) = 3y \\[1em] \Rightarrow 4y - 6 = 3y \\[1em] \Rightarrow 4y - 3y = 6 \\[1em] \Rightarrow y = 6.$

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

$\Rightarrow x = \dfrac{3 \times 6}{2} \\[1em] = \dfrac{18}{2} \\[1em] = 9.$

Hence, numbers are 6 and 9.