- Define cost functions: \(y_H(x) = 25x + 500\) for HE and \(y_S(x) = 30x + 400\) for Standard.
- Equate functions: \(25x + 500 = 30x + 400\).
- Calculate break-even point: \(5x = 100 \implies x = 20\) years.
- Final cost at equality is \(\boxed{\$1000}\).
Explanation
- Analyzing the Data and Defining Variables
We are given two sets of data points representing the cost of two different washing machines over time. Let \(x\) represent the number of years and \(y\) represent the total cost in dollars. We define \(y_H(x)\) as the cost function for the high-efficiency (HE) washing machine and \(y_S(x)\) as the cost function for the standard washing machine. We assume both follow a linear model of the form \(y = mx + b\), where \(m\) is the annual rate of cost increase and \(b\) is the initial cost (or fixed costs). The data points are:
- For HE: \((1, 525), (2, 550), (3, 575)\)
- For Standard: \((1, 430), (2, 460), (3, 490)\)
- Determining the High-Efficiency Cost Function
For the high-efficiency washing machine, we calculate the slope \(m_H\) using the formula \(m_H = \frac{y_2 - y_1}{x_2 - x_1}\):
\[m_H = \frac{550 - 525}{2 - 1} = \frac{25}{1} = 25\]
Now, we find the y-intercept \(b_H\) using one of the points, such as \((1, 525)\):
\[525 = 25(1) + b_H \implies b_H = 525 - 25 = 500\]
Thus, the cost equation for the HE machine is \(y_H(x) = 25x + 500\).
- Determining the Standard Cost Function
For the standard washing machine, we calculate the slope \(m_S\) similarly:
\[m_S = \frac{460 - 430}{2 - 1} = \frac{30}{1} = 30\]
Next, we find the y-intercept \(b_S\) using the point \((1, 430)\):
\[430 = 30(1) + b_S \implies b_S = 430 - 30 = 400\]
Thus, the cost equation for the standard machine is \(y_S(x) = 30x + 400\).
- Solving the System and Concluding
To find when the costs are equal, we set the two equations equal to each other and solve for \(x\):
\[25x + 500 = 30x + 400\]
Subtract \(25x\) from both sides:
\[500 = 5x + 400\]
Subtract \(400\) from both sides:
\[100 = 5x\]
Divide by \(5\):
\[x = 20\]
The costs are equal after 20 years. At this time, the cost for both machines is \(y = 25(20) + 500 = \boxed{\$1000}\).
Examples
This linear modeling approach is essential for long-term financial decision-making. By comparing the cost functions of different products, such as energy-efficient appliances versus standard models, consumers can determine the 'break-even point' where a more expensive but efficient product becomes more cost-effective. For instance, if a solar panel setup costs \(\$10000\) to install but saves \(\$500\) a year in electricity bills, while traditional power costs \(\$200\) more annually, this same linear system technique identifies exactly when the investment pays off.