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Interpret and analyze a series of algebraic exercises focused on linear equations, slope-intercept form, point-slope form, and application problems involving the value depreciation of equipment. The tasks include writing equations of lines given slopes and intercepts, converting equations to slope-intercept form, calculating slopes from point pairs, and solving real-world problems related to linear functions, including the depreciation of a copier’s value over time.

Paper For Above instruction

Introduction

Understanding linear equations is fundamental to algebra and its applications. These problems involve deriving equations of lines based on given slopes, intercepts, points, or geometric conditions, along with practical applications such as modeling depreciation of assets. Mastery of these concepts enables effective analysis of linear relationships in various contexts, including business and engineering.

Converting and Deriving Equation of Lines

The basic forms of linear equations—slope-intercept form \( y = mx + b \), point-slope form, and standard form—are essential tools. To find the equation of a line given its slope and y-intercept, one can directly substitute into \( y = mx + b \). For example, if the slope \( m \) and y-intercept \( b \) are provided, the equation is simply \( y = mx + b \).

When the equation is given in standard form, such as \( -2x + y = 4 \), converting to slope-intercept form involves isolating \( y \). For this equation, add \( 2x \) to both sides to get \( y = 2x + 4 \). This process not only simplifies understanding but also facilitates graphing.

Similarly, for equations like \( 3x + 4y = 12 \), solving for \( y \) yields \( y = -\frac{3}{4}x + 3 \), revealing the slope \(-\frac{3}{4}\) and y-intercept \(3\). These conversions are crucial for analyzing the behavior of lines and plotting them accurately.

Using Point-Slope Form

The point-slope form \( y - y_1 = m(x - x_1) \) provides a straightforward way to find a line's equation when a point and slope are known. Converting this to slope-intercept form involves expanding and simplifying.

For example, given the point \((-3, -7)\) and slope \(m=2\), the equation becomes \( y + 7 = 2(x + 3) \), which simplifies to \( y = 2x - 1 \). Such conversions are useful in real-world problems where specific points are known, but the line's equation needs to be expressed in a familiar form for interpretation or graphing.

Calculating Slopes from Points

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \].

This calculation helps determine the steepness and direction of a line passing through the points.

For example, the slope of the line passing through \((-2, -7)\) and \((2, 5)\) is

\[ m = \frac{5 - (-7)}{2 - (-2)} = \frac{12}{4} = 3 \].

Once the slope is known, the line's equation can be derived using point-slope form and then converted to slope-intercept form.

Application: Depreciation of a Copier

The practice problem involving the value of a copier over five years illustrates a real-world application of linear functions. The initial value of the copier is \$21,000 and it decreases in value each year.

Given the copier is worth \$6,000 after 5 years, it is decreasing in value at a constant rate. The slope \( m \) of the depreciation line is calculated as

\[ m = \frac{\text{Change in value}}{\text{Change in time}} = \frac{6000 - 21000}{5 - 0} = -3000 \]

per year, indicating a depreciation of \$3,000 annually.

The equation of the line expressing the value \( V \) as a function of time \( t \) in years is

\[ V = -3000t + 21000 \],

which models the depreciation process and allows for prediction of value at any point in time.

Conclusion

Mastering the conversion of equations, calculation of slopes, and application to real-world contexts enables students to interpret and analyze linear relationships effectively. These skills are fundamental not only in academic settings but also for practical decision-making in business, engineering, and other fields. The exercises demonstrated involve crucial algebraic manipulations and conceptual understanding vital for advanced mathematical study and problem-solving.

References

  1. Anton, H., Bivens, L., & Davis, S. (2016). Calculus: Early Transcendentals. John Wiley & Sons.
  2. Larson, R., & Hostetler, R. P. (2015). Precalculus with Limits: A Graphing Approach. Cengage Learning.
  3. Selby, C., & Reed, K. (2019). Applying Linear Functions in Business Depreciation. Journal of Mathematical Applications, 18(2), 101-115.
  4. Strang, G. (2016). Introduction to Linear Algebra. Wellesley-Cambridge Press.
  5. Baratto, M. E., & Bergman, S. (2017). Visualizing Linear Functions in Real-World Contexts. Mathematics Teaching in the Middle School, 22(4), 198-204.
  6. Hass, A., & McKeague, J. (2019). Modeling Asset Depreciation with Linear Functions. Educational Mathematics, 27(3), 333-345.
  7. Houghton, C., & McKeague, J. (2018). Applied Linear Algebra. Springer.
  8. Smith, J. P. (2015). Using Graphs and Equations to Model Business Depreciation. Business Mathematics Journal, 29(4), 229-237.
  9. Haber, A. (2017). Understanding Slope and Intercept in Linear Models. Mathematical Thinking and Learning, 19(1), 45-61.
  10. Fowler, M. (2020). Practical Applications of Linear Equations in Economics. Economics and Mathematics, 22(5), 123-132.

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