# H1.03: Example 2

**Example 2**: Ajax Manufacturing bought a machine for $48,000. It is expected to last 15 years and, at the end of that time, have a salvage value of $7,000. Set up a linear depreciation model for this machine and find the worth at the end of 10 years.

**Solution**:

**1. What are the two variables and their units?**Ans. Time and worth. Time is measured in years and worth in dollars.

**2. What will we predict? (Make it y.) What are some points?**Ans. We will predict worth. So y = worth, which must always be above $7000. Then x = years and x will go from 0 to 15. At zero years, the worth is 48000. So x = 0 and y = 48000. Use (0, 48000) At 15 years, the worth is 7000. So x = 15 and y = 7000. Use (15, 7000)

**3. Is a linear model appropriate?**Yes, because the problem asks for linear depreciation.

**4. Slope**Find the slope: m = rise / run Let (0,48000) be the first point and (15,7000) be the second point.

[latex]m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{48000-7000}{0-15}=\frac{41000}{-15}=-2733.333[/latex]

**Interpret the slope, using the units of the numbers in the problem:**(As x increases by 1, y increases by m.) For each year, the worth increases by -2733.33. But, of course, increasing by a negative amount means decreasing. So, for each year, the worth decreases by $2733.33.

**5. Write the formula of the linear relationship:**(Choose either point and use the point-slope form of the line. Then simplify it to the slope-intercept form of the line.) Choose (0, 48000) for the point and use the slope of -2733.333.

[latex]\begin{align}&y-{{y}_{0}}=m(x-{{x}_{0}})\\&y-48000=-2733.333(x-0)\\&y-48000=-2733.333x\\&y=-2733.333x+48000\\\end{align}[/latex]

**6. Interpret the y-intercept.**

*(The value of y is b when .)*At the beginning of the process, when time=0, the worth is $48,000.

**7. Use the equation to make the requested prediction. Write the result in a sentence, with units.**At 10 years, x = 10:

[latex]\begin{align}&y=-2733.333x+48000\\&y=-2733.333(10)+48000\\&y=-27333.33+48000\\&y=20666.67\end{align}[/latex]

So the worth of the machine at 10 years is $20,666.67.**8. Make a graph and use it to check the prediction.**

time | worth |

0 | 48000 |

15 | 7000 |

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- Mathematics for Modeling.
**Authored by:**Mary Parker and Hunter Ellinger.**License:**CC BY: Attribution.