# Thread: Model the problem with a linear equation

1. ## Model the problem with a linear equation

Hello everyone. I have spent a long time trying to figure this problem out. I have an exam soon and I am at the stage of anxiety because I cannot move forward studying because I am still not understanding what appears to be a simple problem here. I think that the "where x=0" part is really throwing me off. Any help would be greatly appreciated! thank you.

Problem:
Model the problem with a linear equation
An investment is worth $2958 in 1995. By 2000 it has grown to$5628. Let y be the value of the investment in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the investment in the year x.

2. Originally Posted by lanipr
Hello everyone. I have spent a long time trying to figure this problem out. I have an exam soon and I am at the stage of anxiety because I cannot move forward studying because I am still not understanding what appears to be a simple problem here. I think that the "where x=0" part is really throwing me off. Any help would be greatly appreciated! thank you.

Problem:
Model the problem with a linear equation
An investment is worth $2958 in 1995. By 2000 it has grown to$5628. Let y be the value of the investment in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the investment in the year x.
A linear equation is of the form $y = mx + c$, where $m$ is the slope and $c$ is the $y$-intercept. Note, the $y$-intercept is given, it's 2958. So all you need is the slope.

It's found by $m = \frac{y_2 - y_1}{x_2 - x_1}$

$= \frac{5628 - 2958}{5 - 0}$ (the second $x$ value must be 5, because it's 5 years on...)

$= \frac{2670}{5}$

$= 534$.

So the equation of the line is $y = 534x + 2958$.