document.write( "Question 725993: \r
\n" ); document.write( "\n" ); document.write( "Find 3 consecutive odd integers such that the sum of the 1st, 2nd and 3 times the 3rd is 58 more than the smaller number?
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Algebra.Com's Answer #444438 by checkley79(3341)\"\" \"About 
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LET X, X+2 & X+4 BE THE 3 INTEGERS.
\n" ); document.write( "X+(X+2)+3(X+4)=X+58
\n" ); document.write( "X+X+2+3X+12=X=58
\n" ); document.write( "5X-X=58-2-12
\n" ); document.write( "4X=44
\n" ); document.write( "X=44/4
\n" ); document.write( "X=11 ANS. FOR THE SMALLER INTEGER.
\n" ); document.write( "11+2=13 ANS. FOR THE MIDDLE NUMBER.
\n" ); document.write( "11+4=15 ANS. FOR THE LARGEST INTEGER.
\n" ); document.write( "PROOF:
\n" ); document.write( "11+13+3*15=11+58
\n" ); document.write( "24+45=69
\n" ); document.write( "69=69
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