document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=995862>Question 995862</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In 2002, Bill bought a truck for $30,000. The value of the truck has been decreasing by $2600 every three years. In the same year, Bill also bought a painting from a starving artist at a farmer's market for $50. The artist then became famous and the value of the painting has been increasing at a rate of $800 per year since. Let x be the number of years after 2002.\r<BR>\n" );
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document.write( "a)Write the value of the truck as a linear function of x<BR>\n" );
document.write( "b)Write the value of the painting as a linear function of x<BR>\n" );
document.write( "c)In what year will the painting be worth as much as the truck? What is the value of each item in that year? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #614533 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=614533\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> truck = 30,000<BR>\n" );
document.write( "depreciation = 2600 every 3 years.<BR>\n" );
document.write( "painting = 50.<BR>\n" );
document.write( "appreciation = 800 every year.<BR>\n" );
document.write( "x = number of years since 2002.<BR>\n" );
document.write( "x = 0 = 2002<BR>\n" );
document.write( "x = 1 = 2003<BR>\n" );
document.write( "etc.<BR>\n" );
document.write( "y1 = value of the truck in year x.<BR>\n" );
document.write( "value of the truck in year x is given by the formula:<BR>\n" );
document.write( "y1 = -2600/3 * x + 30000.<BR>\n" );
document.write( "y2 = value of the painting in year x.<BR>\n" );
document.write( "value of the painting in year x is given by the formula:<BR>\n" );
document.write( "y2 = 800 * x + 50.\r<BR>\n" );
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document.write( "your break even point is when y1 = y2.<BR>\n" );
document.write( "y1 = y2 is the same as -2600/3 * x + 30000 = 800 * x + 50 because:<BR>\n" );
document.write( "y1 can be replaced with its equivalent value of -2600/3 * x + 30000.<BR>\n" );
document.write( "y2 can be replaced with its equivalent value of 800 * x + 50.\r<BR>\n" );
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document.write( "start with:\r<BR>\n" );
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document.write( "-2600/3 * x + 30000 = 800 * x + 50<BR>\n" );
document.write( "add 2600/3 * x to both sides of the equation and subtract 50 from both sides of the equation to get:<BR>\n" );
document.write( "30000 - 50 = 800 * x + 2600/3 * x<BR>\n" );
document.write( "combine like terms on the left side of the equation and multiply 800 by 3/3 on the right side of the equation to get:<BR>\n" );
document.write( "29950 = 2400/3 * x + 2600/3 * x<BR>\n" );
document.write( "combine like terms on the right side of the equation to get:<BR>\n" );
document.write( "29950 = 5000/3 * x<BR>\n" );
document.write( "divide both sides of the equation by (5000/3) to get:<BR>\n" );
document.write( "29950 / (5000/3) = x<BR>\n" );
document.write( "this is the same as 29950 * 3 / 5000 = x<BR>\n" );
document.write( "solve for x to get:<BR>\n" );
document.write( "x = 17.97<BR>\n" );
document.write( "your break even point hould be when x = 17.97.<BR>\n" );
document.write( "that would be somewhere between x = 17 and x = 18.<BR>\n" );
document.write( "when x = 17, the year is 2002 + 17 = 2019.<BR>\n" );
document.write( "when x = 18, the year is 2002 + 18 = 2020.<BR>\n" );
document.write( "the breakeven point is in the year 2019.\r<BR>\n" );
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document.write( "both equations can be graphed and their intersection is the break even point.<BR>\n" );
document.write( "the graph looks like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2015/101203.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "you can see that, at the 3 year point, the truck has depreciated by 2600 because 30000 - 2600 = 27400.\r<BR>\n" );
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document.write( "you can also see on the graph that, at the 3 year point, the painting has appreciated by 2400 because 50 + 2400 = 2450.\r<BR>\n" );
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document.write( "2400 is the result of 3 years appreciation at 800 per year because 800 * 3 = 2400.\r<BR>\n" );
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document.write( "the value of the painting and the truck at the breakeven point is 14426 as shown on the graph.\r<BR>\n" );
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document.write( "this point is found by calculating the value of the truck and the value of the painting when x = 17.97.\r<BR>\n" );
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document.write( "800 * 17.97 + 50 = 14426.\r<BR>\n" );
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document.write( "-2600/3 * 17.97 + 30000 = 14426.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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