In 1990, the cost of tuition at a large Midwestern university was $97 per credit hour. In 1997, tuition had risen to $188 per credit hour.

A linear equation is one that has the following form:
[tex]y=m\cdot x+b[/tex]Where "m" is the slope of the line, and "b" is the point at which it crosses the y-axis.
For our case "y" is actually represented by a "c", so we can rewrite the expression as:
[tex]c=m\cdot x+b[/tex]The first thing we need to do is determine the slope of the line, which is done by using the following formula:
[tex]m=\frac{c_2-c_1}{x_2-x_1}[/tex]Where (x1,c1) and (x2,c2) are two data points, which were provided to us by the problem. Applying these on the expression, we have:
[tex]\begin{gathered} m=\frac{188-97}{1997-1990} \\ m=\frac{91}{7}=13 \end{gathered}[/tex]We have the following expression for now:
[tex]c=13\cdot x+b[/tex]We need to determine the value for b, to do that we can apply a data point and solve for b.
[tex]\begin{gathered} 188=13\cdot1997+b \\ 188=25961+b \\ b=188-25961 \\ b=-25773 \end{gathered}[/tex]The full expression is:
[tex]c=13\cdot x-25773[/tex]The expression is c=13*x-25773.
Now we need to use this expression to predict the cost in 2005. This is done by replacing x with 2005 and solving for c.
[tex]\begin{gathered} c=13\cdot2005-25773 \\ c=292 \end{gathered}[/tex]The cost of the tuition in 2005 will be $292 per credit hour.
Now we need to use the expression to predict in which year the tuition will cost $331 per credit hour.
[tex]\begin{gathered} 331=13\cdot x-25773 \\ 13\cdot x=331+25773 \\ 13\cdot x=26104 \\ x=\frac{26104}{13} \\ x=2008 \end{gathered}[/tex]The tuition will cost $331 per credit hour in 2008.