This is what I would do, and I believe it is the correct way to approach this problem.

First, let's rename the years such that 1995 = 1, 1996 = 2, etc.

Thus we have the following seven data points: (1, 130), (2, 140), (3, 152), (4, 160), (5, 169), (6, 182), (7, 194).

We wish to find a line that "best" fits the abovementioned seven data points. We will define "best" in a moment. For now, just note that we want to solve for and in the following linear system:

.

Notice what we did here: for each point we included the equation in our linear system.

Now, back to what we meant by "best."

"Best" in our case, means minimizing the sum of squares of "errors," i.e., the least squares approach. That is, we wish to find a line that fits our seven data points, such that the sum of squares of "errors" is minimized. We define "error" as follows: For each in our linear system, the error is given by . Note that each individual error can be positive or negative (hence why they end up squared).

So, we want find and that solve the equations in our linear system, such that is as small as possible, i.e., minimized. This is analogous to the following procedure:

First, we need to set up the sum of squares of errors function, which easily enough is

.

Next, we want to compute the minimum of , which is done by calculating the partial derivatives of with respect to and and setting them equal to zero. Finally, we solve for and . Then plug those values you get back into the original equation y = b1 + b2 x