document.write( "Question 205487: Find three consecutive odd numbers such that 1/7 of the first plus 1/3 of the second plus 1/5 of the third is 63. \n" ); document.write( "
Algebra.Com's Answer #155096 by checkley77(12844)\"\" \"About 
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Let x, x+2, x+4 be the 3 odd numbers.
\n" ); document.write( "x/7+(x+2)/3+(x+4)/5=63
\n" ); document.write( "[(15x+35(x+2)+21(x+4)]/105=63
\n" ); document.write( "[15x+35x+70+21x+84]/105=63
\n" ); document.write( "[71x+154)/105=63
\n" ); document.write( "71x+154=63*105
\n" ); document.write( "71x+154=6,615
\n" ); document.write( "71x=6,615-154
\n" ); document.write( "71x=6,461
\n" ); document.write( "x=6,461/71
\n" ); document.write( "x=91 the first number
\n" ); document.write( "91+2=93 the second number
\n" ); document.write( "91+4=95 the third number
\n" ); document.write( "Proof:
\n" ); document.write( "91/7+93/3+95/5=63
\n" ); document.write( "13+31+19=63
\n" ); document.write( "63=63 \r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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