document.write( "Question 844234: two consecutive odd numbers are chosen so that one third of the smaller exceeds one seventh of the larger by 6. find the numbers.
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Algebra.Com's Answer #508639 by josmiceli(19441) $\"\"$ $\"About$

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Call the smaller number $\"+n+\"$
\n" ); document.write( "The larger number will be $\"+n+%2B+2+\"$
\n" ); document.write( " $\"+%281%2F3%29%2An+=+%281%2F7%29%2A%28+n+%2B+2+%29+%2B+6+\"$
\n" ); document.write( "-----------------------------
\n" ); document.write( " $\"+%281%2F3+%29%2An+=+%28+1%2F7+%29%2An+%2B+2%2F7+%2B+6+\"$
\n" ); document.write( "Multiply both sides by $\"+21+\"$
\n" ); document.write( " $\"+7n+=+3n+%2B+6+%2B+126+\"$
\n" ); document.write( " $\"+4n+=+132+\"$
\n" ); document.write( " $\"+n+=+33+\"$
\n" ); document.write( " $\"+n+%2B+2+=+35+\"$
\n" ); document.write( "The consecutive odd numbers are 33 and 35
\n" ); document.write( "check:
\n" ); document.write( " $\"+%281%2F3%29%2An+=+%281%2F7%29%2A%28+n+%2B+2+%29+%2B+6+\"$
\n" ); document.write( " $\"+%281%2F3%29%2A33+=+%281%2F7%29%2A35+%2B+6+\"$
\n" ); document.write( "Multiply both sides by $\"21\"$
\n" ); document.write( " $\"+7%2A33+=+3%2A35+%2B+126+\"$
\n" ); document.write( " $\"+231+=+105+%2B+126+\"$
\n" ); document.write( " $\"+231+=+231+\"$
\n" ); document.write( "OK \n" ); document.write( "

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