document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=677179>Question 677179</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find two consecutive intergers such that the sum of five times the smaller and three times the larger is 67. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #420733 by </B><A HREF=/tutors/aboutme.mpl?userid=fcabanski><B>fcabanski(1391)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fcabanski><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=420733\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call these integers x and y where y is the larger so y=x+1.  Three times the larger is 3y.  Five times the smaller is 5x.  Their sum is 3y + 5x and it equals 67.<P><BR>\n" );
document.write( "We now have two equations and two unknowns.<P><BR>\n" );
document.write( "y=x+1 and 3y + 5x = 67.<P><BR>\n" );
document.write( "Substitute x+1 for y in the second equation and solve for x.  3(x+1) + 5x =67---> 3x + 3 +5x = 67 ---> 8x = 64 ---> x=8.<P><BR>\n" );
document.write( "y=x+1=9.  The integers are 8 and 9.\r<BR>\n" );
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