Find two numbers such that the sum of twice the first and thrice the second is 103 and four times the first exceeds seven times the second by 11.

Let two numbers be x and y.

Given,

Sum of twice the first and thrice the second = 103

⇒ 2x + 3y = 103

⇒ 2x = 103 - 3y

⇒ x = $\Big(\dfrac{103 - 3y}{2}\Big)$     ……(1)

Given,

Four times the first number exceeds seven times the second by 11.

⇒ 4x - 7y = 11

⇒ 4x = 7y + 11     …..(2)

Substituting value of x from equation (1) in (2), we get :

⇒ $4\Big(\dfrac{103 - 3y}{2}\Big)$ = 7y + 11

⇒ 2(103 - 3y) = 7y + 11

⇒ 206 - 6y = 7y + 11

⇒ 206 - 11 = 7y + 6y

⇒ 195 = 13y

⇒ y = $\dfrac{195}{13} = 15$.

Substituting value of y in equation (1), we get :

$\Rightarrow x = \Big(\dfrac{103 - 3y}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 3 \times 15}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 45}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{58}{2}\Big) = 29.$

Hence, the numbers are 29 and 15.