Question 1193708
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Three times the least of three consecutive odd integers exceeds twice the greatest by 19. Find the three integers.
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<pre>
Let the three consecutive integer numbers be n, (n+2) and (n+4).


Then your equation to find "n" is

    3*n = 2*(n+4) + 19.


Now simplify it and find "n"

    3n = 2n + 8 + 19

    3n - 2n = 8 + 19

       n    =   27.


<U>ANSWER</U>.  The numbers are  27, 29 and 31.
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Solved.