We can use a trick to make the numbers easy to deal with.
We will let y be the number of students per computer, but
for x, we will let x = the number of years after 2000.
That means when x = 1, the year is 2001.
When x = 2, the year is 2002, etc.
We are told that in 2001, there were 20 computers per student.
That gives us the ordered pair (1, 20).
Then in 2008, there were 4.5 computers per student.
That gives us the ordered pair (8, 4.5).
Now we find the equation of the line that contains the points (1, 20) and (8, 4.5).
The equation of the line through points \((x_1, y_1)\) and \((x,_2, y_2) \) is:
\(y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)\)
Now we plug in the coordinates of our two points:
\(y - 20 = \dfrac{4.5 - 20}{8 - 1}(x - 1)\)
\(y - 20 = \dfrac{-15.5}{7}(x - 1)\)
Multiply both sides by 7 to get rid of the denominator 7:
\(7y - 140 = -15.5x + 15.5\)
\(15.5x + 7y = 155.5\)
Multiply both sides by 2 to get integer coefficients:
\(31x + 14y = 311\)
We now have the equation of the line.
Now we want to know the year that corresponds to 10 students per computer.
We need to find the value of x that corresponds to y = 10.
\(31x + 14(10) = 311\)
\(31x + 140 = 311\)
\(31x = 171\)
\(x = \dfrac{171}{31}\)
\(x \approx 5.52\)
Since x is approximately equal to 5.5 years, the year is 2005.5, or the middle of 2005. That means the school will have 10 computers per student approximately in the middle of 2005.