document.write( "Question 323799: The cost of a new math textbook was $80 in 2005. It went up to $120 in 2010. Assuming the cost increases linearly, write an equation representing the cost as a function of time. \n" ); document.write( "

Algebra.Com's Answer #231758 by josmiceli(19441) $\"\"$ $\"About$

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You need a linear equation, which has the form
\n" ); document.write( " $\"C+=+m%2Ay+%2B+b\"$
\n" ); document.write( " $\"C\"$ = cost of a textbook during a particular year
\n" ); document.write( " $\"y\"$ is a particular year, calling
\n" ); document.write( " $\"2005+=+0\"$
\n" ); document.write( " $\"2006+=+1\"$
\n" ); document.write( " $\"2007+=+2\"$
\n" ); document.write( "etc.
\n" ); document.write( "All you have to do is find $\"b\"$ and $\"m\"$ in the equation
\n" ); document.write( " $\"m\"$ is the slope, or rate of increase
\n" ); document.write( "When $\"y\"$ increases from $\"0\"$ to $\"5\"$ (2005- 2010),
\n" ); document.write( " $\"C\"$ increases from $\"80\"$ to $\"120\"$
\n" ); document.write( "The slope is (change in C)/(change in y)
\n" ); document.write( " $\"m+=+%28120+-+80%29%2F5\"$
\n" ); document.write( " $\"m+=+40%2F5\"$
\n" ); document.write( " $\"m+=+8\"$
\n" ); document.write( "So far the equation is
\n" ); document.write( " $\"C+=+8y+%2B+b\"$
\n" ); document.write( "Now to find $\"b\"$ ,
\n" ); document.write( "Use $\"C+=+120\"$ , and $\"y+=+5\"$ (a point on the line)
\n" ); document.write( " $\"120+=+8y+%2B+b\"$
\n" ); document.write( " $\"120+=+8%2A5+%2B+b\"$
\n" ); document.write( " $\"b+=+120+-+40\"$
\n" ); document.write( " $\"b+=+80\"$
\n" ); document.write( "So, the equation is
\n" ); document.write( " $\"C+=+8y+%2B+80\"$
\n" ); document.write( "The plot of this equation is:
\n" ); document.write( " $\"+graph%28+700%2C+500%2C+-5%2C+15%2C+-50%2C+200%2C+8x+%2B+80%29+\"$ \n" ); document.write( "

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