Question 159071
Let's do this.
We'll setup a linear equation in the form y = mx +b that uses the given info.
y is the cost of tuition
x is the number of years after 1980

You are given : $1979 in 1980 to $4895 in 1997. So you know 2 points on the line. Thise points are {0, 1979) and (17, 4985). We find the x values as {{{x = year - 1980}}}

First let's find the slope of the line
m = (y[2] - y[1])/(x[2] - x(1])
{{{m = (4985 - 1979)/(17 -0)}}}
{{{m = 3006/17}}}
{{{m = 176.8}}}

b is the y intercept. That is the point where the x value is 0. So for us {{{b = 1979}}}

We have what we need. Just plug it all in
{{{y = mx + b}}}
{{{y = 176.8 * x + 1979}}}

Check the eqaution using the two points you know. At year 1980, x =0, so y = 1979. That is correct

At year 1997, x = 17. So {{{y = 176.8 * 17 + 1797}}} = {{{4984.6}}} close enough

So for part A) {{{E = 176.8 * x + 1979}}}  (same equation, just use the variables the problem states.

Finally, you are asked to estimate the cost in the year 2010.
In 2010 {{{x = 2010 - 1980}}} = {{{30}}}
Plugging that into {{{y = 176.8 * x + 1979}}} yeilds
{{{y = 176.8 * 30 + 1979}}}
{{{y = 5304 + 1979}}}
{{{y = 7283}}}