Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2. 

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1 Answers

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

Then, we have: 

2x + 3y = 92 ……….(i) 

4x - 7y = 2 ………(ii) 

On multiplying (i) by 7 and (ii) by 3, we get 

14x + 21y = 644 ………..(iii) 

12x - 21y = 6 ………..(iv) 

On adding (iii) and (iv), we get 

26x = 650 

⇒ x = 25 

On substituting x = 25 in (i), we get 

2 × 25 + 3y = 92 

⇒ 50 + 3y = 92 

⇒ 3y = (92 – 50) = 42 

⇒ y = 14 

Hence, the first number is 25 and the second number is 14

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