SOLUTION: find two consecutive integers such that 4 times the lesser is 8 more than 3 times the greater

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Question 670326: find two consecutive integers such that 4 times the lesser is 8 more than 3 times the greater
Answer by VirtualMathTutor(26) About Me  (Show Source):
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let x be the first number
let x + 1 be the second number
4 times the lesser number (which is x) is 8 more than 3 times the greater (which is x + 1)
4x = 8 + 3(x + 1)
4x = 8 + 3x + 3
4x = 11 + 3x
subtract 3x from both sides
x = 11
The second number is x + 1 = 11 + 1 = 12
So the consecutive numbers are 11 and 12