Question 179732
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Let the first odd iteger be <i>x</i>.  Then the second odd integer is <i>x</i> + 2, the third is <i>x</i> + 4, and the fourth is <i>x</i> + 6.  That's because you always add 2 to get to the next consecutive <i><b>odd</b></i> integer.


Three times the second number is then *[tex \Large 3(x + 2)] which is equal to two times the fourth number plus 15 or *[tex \Large 2(x + 6) + 15]


So:


*[tex \Large 3(x + 2) = 2(x + 6) + 15]


Solve this for <i>x</i> to get the first of the consecutive odd integers, and then add two, add two, and add two to get the other three.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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