Problem 70 - Algebra for College Students

Question

An accountant has a schedule of depreciation for some business equipment. The schedule shows that after 12 months the equipment is worth \(\$ 7600\) and that after 20 months it is worth \(\$ 6000\). Let \(y\) represent the worth and \(x\) represent the time in months.

Step-by-Step Solution

Answer

The depreciation equation is \( y = -200x + 10000 \).

Step 1: Understanding the Exercise

We are given two points on a depreciation schedule: (12, 7600) and (20, 6000). Each point is in the form (months, worth). We need to find a linear equation relating the worth and the time.

Step 2: Finding the Slope

To find the slope of the line, we use two given points. The slope formula is \( m = \frac{y_2-y_1}{x_2-x_1} \). By substituting \((x_1, y_1) = (12, 7600)\) and \((x_2, y_2) = (20, 6000)\), we get:\[ m = \frac{6000 - 7600}{20 - 12} = \frac{-1600}{8} = -200 \]

Step 3: Using the Point-Slope Form

Now that we have the slope, we can use the point-slope form of a line to find the equation. The form is \( y - y_1 = m(x - x_1) \). Substituting \( m = -200 \) and \( (x_1, y_1) = (12, 7600) \), we get:\[ y - 7600 = -200(x - 12) \]

Step 4: Simplifying to Slope-Intercept Form

We simplify the equation found in the previous step to get the y-intercept form (\(y = mx + b\)). \[ y - 7600 = -200x + 2400 \]\[ y = -200x + 2400 + 7600 \]\[ y = -200x + 10000 \]

Step 5: Interpreting the Equation

The equation \( y = -200x + 10000 \) shows that for each month, the equipment's worth decreases by \(200. Initially (at x = 0 months), the equipment is worth \)10,000.

Key Concepts

Depreciation

Depreciation refers to the reduction in the value of an asset over time. In the context of business equipment, it means the equipment loses its financial worth as time passes. This process is often due to wear and tear, technological obsolescence, or market conditions.
In our exercise, depreciation is demonstrated by the equipment's decreasing value, which was initially $10,000 then valued at $7,600 after 12 months and $6,000 after 20 months. This constant drop in value can be represented by a straight line when graphed, indicating a linear relationship between time and value.

Understanding depreciation is crucial for businesses to manage their assets well. Knowing how fast an asset loses value helps in planning for reinvestment or selling it before it becomes a liability. Therefore, establishing a clear depreciation schedule as shown in the example assists in recognizing predictable financial changes over time.

Slope

The slope of a line in mathematics is a measure of its steepness or direction. In linear equations, it's represented as 'm'. Slope tells us how much the dependent variable, often 'y', will change with every single unit increase in the independent variable, 'x'.
In our depreciation example, the slope is calculated using two known points: (12, 7600) and (20, 6000). The formula used is:

• The change in y divided by the change in x: \[ m = \frac{\Delta y}{\Delta x} = \frac{6000 - 7600}{20 - 12} = -200 \]

The slope is -200, indicating that for every month, the worth of the equipment decreases by $200. This negative slope clearly signifies depreciation. Understanding slope provides insight into the rate of change and helps in predicting future values as time progresses.

Point-Slope Form

The point-slope form is a way of expressing the equation of a straight line, which utilizes the slope and a single point on the line. The formula is:

• \(y - y_1 = m(x - x_1)\)

Here,

• \(m\) is the slope, and
• \((x_1, y_1)\) is a known point on the line.

Using this form, calculation becomes simpler when inputting known values to derive a linear equation.
In our example exercise, we substituted

• \(m = -200\)
• (\x_1, y_1\ = (12, 7600)

to construct the equation \(y - 7600 = -200(x - 12)\). This form highlights the relationship between the initial value at identified points and allows for conversion to other forms, like slope-intercept form, which can be easier to interpret and use.

Slope-Intercept Form

The slope-intercept form is commonly used to express linear equations because it readily shows both the slope and the y-intercept in a straightforward manner. This form of a linear equation is given by:

• \(y = mx + b\)

where:

• \(m\) is the slope of the line, and
• \(b\) is the y-intercept (the value of \(y\) when \(x\) equals zero).

From the exercise example, the equation derived from the point-slope form was further simplified to slope-intercept form: \(y = -200x + 10000\). This tells us that the equipment's value decreases at a rate of \(200 per month, and the initial value of the equipment, when time, \(x\), is zero, is \)10,000.
This form is very useful for quickly understanding the relationship between variables in a linear context, and easily graphing the line or predicting values beyond the known points.