document.write( "Question 893562: find two consecutive even integers such that seven more than twice the second is equal to the first \n" ); document.write( "

Algebra.Com's Answer #541393 by Fombitz(32388) $\"\"$ $\"About$

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Two consecutive even integers : $\"N\"$ , $\"N%2B2\"$
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\n" ); document.write( " $\"7%2B2%28N%2B2%29=N\"$
\n" ); document.write( " $\"7%2B2N%2B4=N\"$
\n" ); document.write( " $\"2N%2B11=4\"$
\n" ); document.write( " $\"2N=-7\"$
\n" ); document.write( " $\"N=-7%2F2\"$
\n" ); document.write( "There is no solution.
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\n" ); document.write( "Looking at the problem setup, double an even integer will get you another even integer but then adding an odd number will produce an odd number not an even number. That's why there's no solution.\r
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