Since two points define a line, we can take two points through which the line passes and find its equation.
From the word problem, we know that when t = 0, the original value is equal to $43,000. Then, we have a first ordered pair.
[tex](0,43000)[/tex]Since the value is reduced by $4300 each year, we have a second ordered pair.
[tex]\begin{gathered} 43000-4300=38700 \\ (1,38700) \end{gathered}[/tex]Now, we find the slope of the line using the following formula.
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are two points through which the line passes} \end{gathered}[/tex][tex]\begin{gathered} (x_1,y_1)=(0,43000) \\ (x_2,y_2)=(1,38700) \\ m=\frac{38700-43000}{1-0} \\ m=\frac{-4300}{1} \\ m=-4300 \end{gathered}[/tex]Finally, we can use and solve for y the point-slope formula.
[tex]y-y_1=m(x-x_1)\Rightarrow\text{ Point-slope formula}[/tex]Since the situation described depends on the time t, we replace x with t in the formula.
[tex]\begin{gathered} y-43000=-4300(t-0) \\ y-43000=-4300t \\ \text{ Add 43000 from both sides} \\ y-43000+43000=-4300t+43000 \\ y=-4300t+43000 \end{gathered}[/tex]Therefore, the linear equation that models the value y of the machinery at the end of year t is:
[tex]y=-4300t+43000[/tex]