Question 1203707
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Find five consecutive integers such that:
“The sum of the first and 4 times the third is equal to 56 less than 3 times 
the sum of the second, fourth, and fifth.”
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<pre>
Let the numbers n, (n+1), (n+2), (n+3) and (n+4) be five consecutive integer numbers.


Write equation as you read the problem

    n + 4*(n+2) = 3*((n+1) + (n+3) + (n+4)) - 56.


Simplify step by step

    n + 4n + 8 = 3*(3n+8) - 56

     5n    + 8 = 9n + 24 - 56

     8 - 24  + 56 = 9n - 5n

        40     =    4n

         n     =   40/4 = 10.


<U>ANSWER</U>.  The numbers are  10, 11, 12, 13, 14.
</pre>

Solved.