College Algebra word problem

Question

An investment is worth $3197 in 1991. by 1996 it has grown to $4907. Let y be the value of the investment in the year x, where x =0 represents 1991. write a linear equation that models the value of the investment in the year x.al

Answer 1

Hello, thank you for taking the time to post your question!

Essentially you are given two points here …. (0, 3197) and (5, 4907) and we want to find an equation of the line between those in the form y = mx + b. What’s nice about this is that the y-intercept is given to us directly as the 3197 so we just need to find the slope using the formula

m = (4907 – 3197) / (1996 – 1991) = 1710/5 = 342

That makes the full equation of the line here y = 342x + 3197, meaning that the value of the investment grew by $342 each year

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

A linear model is required, so the first step is to find the slope of the line. There are two pieces of data:

(0,3197) and (5,4907).  Where the X-coordinate represents the year # and the Y-coordinate is the value of the investment. The problem specifically states that X=0 represents year 1991, which means 1996 MUST be represented by X=5. 

 

The slope of the line is then (4907-3197)/(5-0) = 1710/5 = 342.

 

The point (0,3197) however also serves as the y-intercept where X=0.

 

Therefore, the linear model is y = 342X + 3197.