Week 4 Systems Of Equations By Terminstructor Na
Week 4 Systems Of Equations By Terminstructor Na
Create a Voiceover Presentation where you:
Your instructor will provide you with a system of equations. Please reach out on Monday of Week 4 if you did not receive the system you are supposed to use on this presentation.
Pick TWO of the following ways to solve the systems of equations that you were assigned:
- A. Substitution Method
- B. Elimination Method
- C. Graphing
Using a graphing calculator (like the one at) will REALLY help with your work moving forward, and showing how to graph a system and solve it is not only fast and robust (100% as high-quality as with an algebraic method) but it teaches you new ways to work with equations that will help immensely in the second half of the course.
Paper For Above instruction
The focus of this presentation is to demonstrate the process of solving a system of linear equations using two different methods: substitution and graphing. The chosen system of equations for this exercise is:
4x + 10y = 2
x - 10y = 13
First, I will solve the system using the substitution method. This method involves solving one of the equations for one variable and then substituting that expression into the other equation. Here, I will solve the second equation for x:
From x - 10y = 13, solving for x gives x = 13 + 10y.
Next, I substitute this expression for x into the first equation:
4(13 + 10y) + 10y = 2
Expanding the equation yields: 52 + 40y + 10y = 2
Combining like terms results in: 52 + 50y = 2
Subtracting 52 from both sides gives: 50y = 2 - 52
Simplifying, we find: 50y = -50
Dividing both sides by 50 results in: y = -1
Now, substituting y = -1 back into x = 13 + 10y, we get:
x = 13 + 10(-1) = 13 - 10 = 3
Thus, the solution to the system using substitution is x = 3, y = -1.
Next, I will solve the same system using the graphing method. First, I express both equations in slope-intercept form (y = mx + b) to facilitate graphing:
4x + 10y = 2 → 10y = -4x + 2 → y = (-2/5)x + 0.2
x - 10y = 13 → -10y = -x + 13 → y = (1/10)x - 1.3
Plotting these two lines on a graph, their point of intersection corresponds to the solution of the system. By using graphing tools or calculator, I observe the intersection occurs at the point (3, -1), confirming the algebraic solution obtained earlier.
In conclusion, both methods yield the same solution: x = 3, y = -1. The substitution method involved algebraic manipulation and substitution, whereas graphing visually demonstrated the solution point as the intersection of the two lines. I prefer the algebraic method of substitution for its precision and clear step-by-step solution, especially when solving systems that are complex or do not lend themselves easily to graphing. However, I also appreciate the graphing method because it provides a visual understanding of the system's solution and is useful for verifying solutions quickly.
After completing this presentation, I learned the importance of multiple approaches to solving systems of equations. Understanding both algebraic and graphical methods enhances our problem-solving toolkit and improves comprehension of how equations relate geometrically and algebraically. Moreover, practicing both methods prepares me for more complex systems and helps me identify the most efficient approach depending on the problem's context.
References
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