document.write( "Question 148035: An item costs $1300 has a scrap value of $100, and a useful life of six years. The linear equation relating book value and number of years is: \n" ); document.write( "

Algebra.Com's Answer #108439 by josmiceli(19441) $\"\"$ $\"About$

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The data given tells me two points on the linear relation
\n" ); document.write( " $\"P%5B1%5D\"$ ( $\"0\"$ , $\"1300\"$ )
\n" ); document.write( " $\"P%5B2%5D\"$ ( $\"6\"$ , $\"100\"$ )
\n" ); document.write( "This is with $\"B\"$ ,the book value plotted on the y-axis
\n" ); document.write( "and $\"n\"$ , the number of years on the x-axis
\n" ); document.write( " $\"P%5B1%5D\"$ says that when it's $\"0\"$ years old (brand new),
\n" ); document.write( "it's worth $\"1300\"$ dollars
\n" ); document.write( " $\"P%5B2%5D\"$ says that when it's $\"6\"$ years old, it's worth
\n" ); document.write( " $\"100\"$ dollars (scrap value)
\n" ); document.write( "The formula to use, in terms of B, and n is
\n" ); document.write( " $\"%281300+-+B%29+%2F+%280+-+n%29+=+%281300+-+100%29+%2F+%280+-+6%29\"$
\n" ); document.write( " $\"%281300+-+B%29+%2F+%28-n%29+=+1200+%2F+%28-6%29\"$
\n" ); document.write( "multiply both sides by $\"%28-1%29%2F-1\"$
\n" ); document.write( " $\"%28B+-+1300%29+%2F+n+=+-1200+%2F+6\"$
\n" ); document.write( "multiply both sides by $\"6n\"$
\n" ); document.write( " $\"6B+-+7800+=+-1200n\"$
\n" ); document.write( " $\"B+=+-200n+%2B+1300\"$ answer
\n" ); document.write( "Try values of n = 1,2,3,4,5,6 to check this
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