document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=28627>Question 28627</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am taking a college algebra class and am having some problems with linear equations and graphs.  Here is a problem that I would desperatley like to see solved with an explaination of how you did it.  Thanks so much, you folks are fantastic.\r<BR>\n" );
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document.write( "Straight line depreciation is a linear equation.  The current value, \"y\" of a certain company's piece of equipment after \"x\" years is y = 30,000 - 1000x<BR>\n" );
document.write( "a.  Find the intial value of the equipment<BR>\n" );
document.write( "b.  Find the value after 4 years<BR>\n" );
document.write( "c.  Find the value after 6 years<BR>\n" );
document.write( "d.  Graph the depreciation line </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #15696 by </B><A HREF=/tutors/aboutme.mpl?userid=emmychan><B>emmychan(11)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=emmychan><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=15696\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To begin this problem, keep in mind that \"x\" is the number of years.\r<BR>\n" );
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document.write( "A) The initial value of the equipment means that they have had it for 0 years so sub in 0 for \"x\".\r<BR>\n" );
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document.write( "y = 30,000 - 1,000x<BR>\n" );
document.write( "y = 30,000 - 1,000(0)<BR>\n" );
document.write( "y = 30,000 - 0<BR>\n" );
document.write( "y = 30,000\r<BR>\n" );
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document.write( "B)Number of years goes to 4 so sub in 4 for \"x.\"\r<BR>\n" );
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document.write( "y = 30,000 - 1,000x<BR>\n" );
document.write( "y = 30,000 - 1,000(4)<BR>\n" );
document.write( "y = 30,000 - 4,000<BR>\n" );
document.write( "y = 26,000\r<BR>\n" );
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document.write( "C)Number of years goes to 6 so again, sub in for \"x.\"\r<BR>\n" );
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document.write( "y = 30,000 - 1,000x<BR>\n" );
document.write( "y = 30,000 - 1,000(6)<BR>\n" );
document.write( "y = 30,000 - 6,000<BR>\n" );
document.write( "y = 24,000\r<BR>\n" );
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document.write( "D) Graph.  All you need to do is use your year values as \"x\" coordinates and the \"y\"'s you solved for as \"y\" coordinates.\r<BR>\n" );
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document.write( "Hope it helps.  :) </SPAN>\n" );
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