Question

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.

Answer 1

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)

⇒ 3y = 42

⇒ y = 14

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