document.write( "Question 1093203: Find three consecutive odd integers such tha three times the middle integer is 11 more than the sum of the first and third integers \n" ); document.write( "

Algebra.Com's Answer #707823 by greenestamps(13203) $\"\"$ $\"About$

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To help follow the reasoning I use for my method of solving the problem, let's call the three numbers A, B, and C.

\n" ); document.write( "For three consecutive odd integers, the middle number is equal to the average of the first and last numbers.
\n" ); document.write( " $\"B+=+%28A%2BC%29%2F2\"$

\n" ); document.write( "That means that twice the middle number is equal to the sum of the first and last numbers.
\n" ); document.write( " $\"2B+=+A%2BC\"$

\n" ); document.write( "So if three times the middle number is 11 more than the sum of the first and last, then the middle term must be 11.
\n" ); document.write( " $\"3B+=+%28A%2BC%29%2B11\"$
\n" ); document.write( " $\"2B%2BB+=+%28A%2BC%29%2B11\"$
\n" ); document.write( " $\"%28A%2BC%29%2BB+=+%28A%2BC%29%2B11\"$
\n" ); document.write( " $\"B+=+11\"$

\n" ); document.write( "And that makes the three integers 9, 11, and 13. \n" ); document.write( "

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