SOLUTION: Given two consecutive integers. The sum of four times the first integer and five times the second integer is 77. Find the integers

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Question 1158342: Given two consecutive integers. The sum of four times the first integer and five times the second integer is 77. Find the integers
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
Two consecutive integers, N and N+1;

Now just follow the description exactly...
Answer by ikleyn(52884)   (Show Source): You can put this solution on YOUR website!
.

The consecutive integers are n and (n+1).


The equation, according to the condition, is

    4n + 5*(n+1) = 77.


It gives

    4n + 5n + 5 = 77

    9n          = 77 - 5 = 72

    n                    = 72/9 = 8.


ANSWER.  the integers are 8 and 9.

. . . . . . . .

Actually, this formulation is provocative.

It is provocative, because the notion "one number" is not the same as the "smaller number".

Therefore, in parallel with the equation (1), we should consider the other case

     4*(n+1) + 5*n = 77. 

It gives

    4n + 4 + 5n = 77

    9n          = 77 - 4  = 73

     n                    = 73/9  is not an integer number, so we reject this case.

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Completed and solved.



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