document.write( "Question 634601: find three consecutive odd positive integers such that 5 times the sum of all three is 42 more than the product of the first and second interger \n" ); document.write( "

Algebra.Com's Answer #399726 by sachi(548) $\"\"$ $\"About$

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find three consecutive odd positive integers such that 5 times the sum of all three is 42 more than the product of the first and second interger\r
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\n" ); document.write( "\n" ); document.write( "ans:=
\n" ); document.write( "let the three consecutive odd positive integers are x,x+2,x+4
\n" ); document.write( "as per question
\n" ); document.write( "5(x+x+2+x+4)=x(x+2)+42
\n" ); document.write( "or 5(3x+6)=x^2+2x+42
\n" ); document.write( "or x^2-13x+12=0
\n" ); document.write( "or x^2-x-12x+12=0
\n" ); document.write( "or (x-1)(x-12)=0
\n" ); document.write( "or x=1or12
\n" ); document.write( "so the integers 1,3,5 as we have to select odd integers ans
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