The sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by 2, then find the numbers.
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Answer:
let the first no. be x and second no. be y
2x+3y=92...1
4x=7y+2...2
so, 4x-7y=2
multiplying 1 by 2
4x+6y=184
subtracting 1 from 2
4x+7y=2
4x+6y=182
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