document.write( "Question 700129: Find three consecutive even integers such that twice the sum of the first and third integers is twenty-one more than the second integer \n" ); document.write( "

Algebra.Com's Answer #431662 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let the consecutive even integers =
\n" ); document.write( " $\"+n+\"$ , $\"+n+%2B+2+\"$ , and $\"+n+%2B+4+\"$
\n" ); document.write( "given:
\n" ); document.write( " $\"+2%2A%28+n+%2B+n+%2B+4+%29+=+n+%2B+2+%2B+21+\"$
\n" ); document.write( " $\"+2%2A%28+2n+%2B+4+%29+=+n+%2B+23+\"$
\n" ); document.write( " $\"+4n+%2B+8+=+n+%2B+23+\"$
\n" ); document.write( " $\"+3n+=+23+-+8+\"$
\n" ); document.write( " $\"+3n+=+15+\"$
\n" ); document.write( " $\"+n+=+5+\"$
\n" ); document.write( "This didn't work because I', going to get
\n" ); document.write( "5,7, and 9 which are odd numbers.
\n" ); document.write( "I need to call the smallest number
\n" ); document.write( " $\"+2n+\"$ to make sure it will be even
\n" ); document.write( "My number are $\"+2n+\"$ , $\"+2n+%2B+2+\"$ , $\"+2n+%2B+4+\"$
\n" ); document.write( "given:
\n" ); document.write( " $\"+2%2A%28+2n+%2B+2n+%2B+4+%29+=+2n+%2B+2+%2B+21+\"$
\n" ); document.write( " $\"+2%2A%28+4n+%2B+4+%29+=+2n+%2B+23+\"$
\n" ); document.write( " $\"+8n+%2B+8++=+2n+%2B+23+\"$
\n" ); document.write( " $\"+6n+=+15+\"$
\n" ); document.write( " $\"+n+=+15%2F6+\"$
\n" ); document.write( " $\"+2n+=+5+\"$
\n" ); document.write( "This didn't work either.
\n" ); document.write( "--------------------
\n" ); document.write( "Now that I think about it, the problem has no solution.
\n" ); document.write( "It is saying that twice the sum of 2 numbers ( which is even ) is
\n" ); document.write( "equal to 21 plus an even number ( which is odd ), and
\n" ); document.write( "even cannot equal odd. \n" ); document.write( "

\n" );