document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=105185>Question 105185</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One number is seven more than a second number. Five times the first is 11 less than 4 times the second. Find the numbers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #76537 by </B><A HREF=/tutors/aboutme.mpl?userid=cheesey><B>cheesey(22)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=cheesey><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=76537\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you must create a system of linear equations to represent each problem.  So, x + 7 =  y (one number - y is seven more than a second number -x).  Next, 5y = 4x -11.\r<BR>\n" );
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document.write( "Since the first equation is already solved for y, let's use it in the second equation.\r<BR>\n" );
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document.write( "5(x+7) = 4x - 11<BR>\n" );
document.write( "5x + 35 = 4x -11 <=  combine like terms to get<BR>\n" );
document.write( "5x - 4x = -11 - 35<BR>\n" );
document.write( "x = -46... NOW\r<BR>\n" );
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document.write( "Use this value of x to solve for y in the first equation.  <BR>\n" );
document.write( "x + 7 = y<BR>\n" );
document.write( "-46 + 7 = -39 = y. \r<BR>\n" );
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document.write( "Thus x = -46, y = -39.  So to check our results plug the values into 5y = 4x - 11.  <BR>\n" );
document.write( "Thus 5(-39) = 4(-46) - 11 => -195 = -195.  Done! </SPAN>\n" );
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