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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? \n" );
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Algebra.Com's Answer #444438 by checkley79(3341)![]() ![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " |