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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.
Solution
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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