SOLVING SYSYEM OF NONLINEAR EQUATIONS || PRECALCULUS

WOW MATH

27 Oct 202120:59

EducationalLearning

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TLDRThis video lesson dives into solving systems of non-linear equations using three primary methods: substitution, elimination, and graphing. The instructor begins with the substitution method, illustrating how to solve for variables in a system like y = -x + 2 and x^2 + y = 2. The elimination method is then showcased, demonstrating how to eliminate variables algebraically to find solutions, using an example system involving x and y squared terms. Lastly, the graphing method is introduced, explaining how to visualize and solve non-linear systems by plotting them on a graph, which is particularly useful for systems that represent geometric shapes like circles and ellipses. The video serves as a comprehensive guide for learners to understand and apply these mathematical techniques.

Takeaways

📚 The video lesson covers solving systems of non-linear equations using three methods: substitution, elimination, and graphing.
🔍 The substitution method involves expressing one variable in terms of another and substituting it into the other equation.
📐 In the elimination method, equations are manipulated to cancel out one of the variables, making it easier to solve for the other.
📈 The graphing method involves plotting the equations on a graph and finding the points of intersection, which are the solutions.
🔑 The first example uses the substitution method to solve the system: y = -x + 2 and x^2 + y = 2, resulting in solutions (0, 2) and (1, 1).
🧩 The second example demonstrates the substitution method with the system: x^2 + y^2 = 10 and x - 3y = -10, leading to the solution (-1, 3).
✂️ The elimination method is shown with the system: x^2 - 6x + y^2 + 2y = 15 and x^2 - 6x - 2y = 3, resulting in solutions (3, -6), (7, 2), and (-1, 2).
📉 An example of graphing involves transforming the equations into standard form to identify them as a circle and an ellipse, then finding their intersection points.
📝 The video emphasizes the importance of understanding the different methods for solving non-linear systems and when to apply each one.
🎓 The lesson aims to educate viewers on mathematical problem-solving techniques, encouraging them to practice and apply these methods.
👨‍🏫 The instructor guides the viewers through each step of the process, from understanding the equations to finding the solutions.

Q & A

What are the three methods discussed in the video for solving systems of non-linear equations?
-The video discusses the substitution method, elimination method, and graphing method for solving systems of non-linear equations.
How does the substitution method work in the context of the given example where y = -x + 2 and x^2 + y = 2?
-In the substitution method, you substitute the value of y from the first equation into the second equation. Here, substituting y = -x + 2 into x^2 + y = 2 gives x^2 - x + 2 = 2, which simplifies to x^2 - x = 0. Factoring out x gives x(x - 1) = 0, so the solutions for x are 0 and 1.
What are the solutions for x and y when using the substitution method with the equations y = -x + 2 and x^2 + y = 2?
-The solutions are x = 0, y = 2 and x = 1, y = 1. These are found by substituting the values of x back into the equation for y.
What is the process of solving the system of equations x^2 + y^2 = 10 and x - 3y = -10 using the substitution method?
-First, express x in terms of y from the second equation: x = 3y - 10. Then substitute this expression into the first equation to get (3y - 10)^2 + y^2 = 10. Expanding and simplifying this equation will help find the values of y, which can then be substituted back to find the corresponding x values.
How do you solve for y in the system x^2 + y^2 = 10 and x - 3y = -10 after substituting x with 3y - 10?
-After substituting, you get (3y - 10)^2 + y^2 = 10. Expanding this gives 9y^2 - 60y + 100 + y^2 = 10. Simplifying leads to 10y^2 - 60y + 90 = 0. Dividing by 10 gives y^2 - 6y + 9 = 0, which factors to (y - 3)(y - 3) = 0, so y = 3.
What are the solutions for x and y in the system x^2 + y^2 = 10 and x - 3y = -10?
-The solution is x = -1 and y = 3, found by substituting y = 3 into x = 3y - 10.
What is the elimination method and how does it apply to the system x^2 - 6x + y^2 + 2y = 15 and x^2 - 6x - 2y = 3?
-The elimination method involves adding or subtracting equations to eliminate one variable. For these equations, subtracting the second from the first eliminates x, resulting in y^2 + 4y + 12 = 0, which can be factored and solved for y.
What are the solutions for x and y in the system x^2 - 6x + y^2 + 2y = 15 and x^2 - 6x - 2y = 3 using the elimination method?
-The solutions are x = 3, y = -6; x = 7, y = 2; and x = -1, y = 2, found by solving the resulting equations after elimination.
How does the graphing method help in solving the system x^2 + y^2 - 6x - 2y = 6 and x^2 + 4y^2 - 6x - 8y = 51?
-The graphing method involves plotting the equations on a graph to find their intersection points. The first equation represents a circle and the second an ellipse. Their intersection points give the solutions to the system.
What are the standard forms of the equations representing a circle and an ellipse in the system x^2 + y^2 - 6x - 2y = 6 and x^2 + 4y^2 - 6x - 8y = 51?
-The standard form of the circle is (x - 3)^2 + (y - 1)^2 = 16, and the standard form of the ellipse is (x - 3)^2/64 + (y - 1)^2/16 = 1.
How can you find the intersection points of a circle and an ellipse represented by the equations x^2 + y^2 - 6x - 2y = 6 and x^2 + 4y^2 - 6x - 8y = 51?
-By graphing both the circle and the ellipse, you can visually identify the points where they intersect. These points are the solutions to the system of equations.

Outlines

00:00

📚 Introduction to Solving Non-Linear Equations

This paragraph introduces the topic of solving systems of non-linear equations. The video will cover three methods: substitution, elimination, and graphing. The first example provided is a substitution method where the equation y = -x + 2 is used to solve the system x^2 + y = 2. The process involves substituting the expression for y into the second equation, simplifying, and solving for x. The solutions are found to be x = 0, y = 2 and x = 1, y = 1.

05:00

🔍 Substitution Method in Detail

The second paragraph delves deeper into the substitution method by providing another example. The system x - 3y = -10 and x^2 + y^2 = 10 is solved by expressing x in terms of y and substituting it into the second equation. After simplification, the equation is reduced to a quadratic form in y, which is then solved to find y = 3. Subsequently, x is found to be -1, resulting in the solution (-1, 3) for the system.

10:02

🚫 Elimination Method for Non-Linear Systems

The third paragraph discusses the elimination method applied to a system of equations. The given system x^2 - 6x + y^2 + 2y = 15 and x^2 - 6x - 2y = 3 is manipulated to eliminate like terms, resulting in a new equation y^2 + 4y - 12 = 0. Solving this, y is found to be either -6 or 2. Substituting these values back into the original equations yields the corresponding x values, resulting in three possible solutions: (3, -6), (7, 2), and (-1, 2).

15:03

📈 Graphing Method and Standard Equations

In the fourth paragraph, the graphing method is introduced alongside the concept of standard equations. The system x^2 + y^2 - 6x - 2y = 6 and x^2 + 4y^2 - 6x - 8y = 51 is transformed into the standard form of a circle and an ellipse, respectively. The center and radius of the circle, as well as the center and axes of the ellipse, are identified. The intersection points of these geometric shapes represent the solutions to the system of equations.

20:05

👋 Conclusion and Call to Action

The final paragraph concludes the video lesson by summarizing the methods taught: substitution, elimination, and graphing for solving systems of non-linear equations. The presenter thanks the viewers for watching and encourages them to like, subscribe, and hit the bell for more video tutorials. The channel aims to be a guide for learning and mastering mathematical lessons.

Mindmap

Keywords

💡System of Non-linear Equations

A system of non-linear equations refers to a set of two or more equations where at least one equation is not a linear equation (i.e., it contains variables raised to a power higher than one or products of variables). In the video, this concept is central as the instructor discusses methods for solving such systems, which are more complex than those for linear systems due to their potential for multiple solutions or no solutions at all.

💡Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of another and then substituting this expression into the other equation(s). The video demonstrates this method by first isolating one variable in one equation and then substituting this value into the second equation to find a solution. For instance, the script shows solving 'y = -x + 2' and 'x^2 + y = 2' by substituting 'y' from the first equation into the second.

💡Elimination Method

The elimination method involves adding or subtracting equations from each other to eliminate one of the variables, thereby simplifying the system to a single-variable equation. The video script mentions this method as a way to simplify the system 'x^2 - 6x + y^2 + 2y = 15' and 'x^2 - 6x - 2y = 3' by eliminating 'x' and finding a solution for 'y'.

💡Graphing Method

The graphing method is a visual approach to solving systems of equations by plotting them on a graph and finding the point(s) of intersection, which represent the solution(s). The video mentions this method in the context of solving systems of non-linear equations, suggesting that it can be used to visually identify where the graphs of the equations intersect, indicating the solutions.

💡Factoring

Factoring is a mathematical process of breaking down a polynomial into a product of its factors. In the script, the instructor uses factoring to simplify equations after applying the substitution method, such as when transforming 'x^2 - x = 0' into 'x(x - 1) = 0' to find the values of 'x' that satisfy the equation.

💡Circle

A circle is a geometric shape defined as the set of all points in a plane that are at a given distance (the radius) from a given point (the center). The video discusses converting the equation 'x^2 + y^2 - 6x - 2y = 6' into the standard form of a circle's equation, which is used to graph the circle and find its center and radius.

💡Ellipse

An ellipse is a geometric shape similar to a circle but with two axes of different lengths, making it more elongated than a circle. The script describes converting the equation 'x^2 + 4y^2 - 6x - 8y = 51' into the standard form of an ellipse's equation, which helps in graphing the ellipse and identifying its center and axes lengths.

💡Center

In the context of circles and ellipses, the center refers to the point from which all points on the curve are equidistant (for a circle) or have a constant sum of distances to the two foci (for an ellipse). The video explains how to find the center of a circle and an ellipse by completing the square and using the standard form of their equations.

💡Radius

The radius of a circle is the distance from the center to any point on the circle. In the video, the radius is discussed in the context of converting the general equation of a circle into its standard form, which allows for the identification of the circle's radius and center.

💡Axis

In the context of an ellipse, the axes refer to the two lines that pass through the ellipse's center and are perpendicular to each other, with the ellipse being symmetrical about these lines. The video mentions the axes of an ellipse when converting the general equation into the standard form, which reveals the lengths of the major and minor axes.

Highlights

Introduction to solving systems of non-linear equations using substitution, elimination, and graphing methods.

Application of the substitution method with the given equations y = -x + 2 and x^2 + y = 2.

Finding the value of y when x = 0 and solving for x by substituting y in the equation.

Solving for x by factoring x^2 - x = 0 and finding x = 0 and x = 1 as solutions.

Determining corresponding y values for each x solution, resulting in (0, 2) and (1, 1).

Using the substitution method for the system x^2 + y^2 = 10 and x - 3y = -10.

Substituting x = 3y - 10 into the equation and simplifying to find y.

Solving the quadratic equation y^2 - 6y + 9 = 0 to find y = 3.

Finding x when y = 3 by substituting back into the equation, resulting in x = -1.

Introduction to the elimination method for solving systems of non-linear equations.

Eliminating variables by adding or subtracting equations to simplify the system.

Solving the system x^2 - 6x + y^2 + 2y = 15 and x^2 - 6x - 2y = 3 using elimination.

Finding y values by factoring and solving the simplified equation y^2 + 4y - 12 = 0.

Determining y = -6 and y = 2 from the factored form and substituting back to find x.

Solving for x when y = -6 and y = 2, resulting in x = 3, x = 7, and x = -1.

Introduction to the graphing method for visualizing and solving systems of non-linear equations.

Transforming the system x^2 + y^2 - 6x - 2y = 6 and x^2 + 4y^2 - 6x - 8y = 51 into standard form.

Identifying the equations as representing a circle and an ellipse and their respective centers and radii.

Graphical representation of the intersection points of the circle and ellipse.

Conclusion summarizing the three methods: substitution, elimination, and graphing for solving non-linear systems.

Transcripts

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