document.write( "Question 846596: Set up an equation and solve the problem.
\n" ); document.write( "Find two integers whose product is 65 such that one of the integers is three more than one of the integers is three more than twice the other integer.\r
\n" ); document.write( "\n" ); document.write( "I have:
\n" ); document.write( "x(y)=65
\n" ); document.write( "x=2y+3\r
\n" ); document.write( "\n" ); document.write( "(2y+3) y=65
\n" ); document.write( "2y^2+3y-65=0
\n" ); document.write( "(2y^2+13y)+(-10y-65)=0
\n" ); document.write( "y(2y+13)-5(2y+13)
\n" ); document.write( "(2y+13) (y-5)
\n" ); document.write( "2y+13=0 =-13/2
\n" ); document.write( "y-5=0. =5\r
\n" ); document.write( "\n" ); document.write( "Webassign told me it was wrong and I just want to know how to do it. \r
\n" ); document.write( "\n" ); document.write( "Thank you so much!!
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Algebra.Com's Answer #509826 by josh_jordan(263)\"\" \"About 
You can put this solution on YOUR website!
Hello. You were correct up until you stopped at y = -13/2 and 5. The word problem says that both numbers have to be integers, and -13/2 is not an integer. So, 5 would be one of the integers. To find the other integer, substitute 5 for y in x = 2y + 3:\r
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\n" ); document.write( "\n" ); document.write( "x = 2(5) + 3 -----> x = 13\r
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\n" ); document.write( "\n" ); document.write( "So, your equation would be set up as \"xy=65\" AND \"x=2y%2B3\" and your two integers would be 5 and 13
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