Find the numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.

Let the first number be x and the second number be y. 

Then, we have:

3x + y = 142 ……….(i) 

4x - y = 138 ………(ii) 

On adding (i) and (ii), we get 

7x = 280 

⇒ x = 40 

On substituting x = 40 in (i), we get:

 3 × 40 + y = 142 

⇒ y = (142 – 120) = 22 

⇒ y = 22 

Hence, the first number is 40 and the second number is 22.

Let the first number be x and the second number be y. 

Then, we have:

3x + y = 142 ……….(i) 

4x - y = 138 ………(ii) 

On adding (i) and (ii), we get 

7x = 280 

⇒ x = 40 

On substituting x = 40 in (i), we get: 

3 × 40 + y = 142 

⇒ y = (142 – 120) = 22 

⇒ y = 22 

Hence, the first number is 40 and the second number is 22.