PROBLEMS INVOLVING SYSTEM OF NONLINEAR EQUATIONS || PRECALCULUS

WOW MATH
2 Nov 202120:23
EducationalLearning
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TLDRThis video lesson delves into solving systems of non-linear equations, which are sets of two or more equations involving variables with at least one non-linear component. The presenter explains the concept, noting that solutions can be determined by the number of intersections. Three example problems are tackled, including finding the intersection of a circle and a parabola, solving a system involving x and y squared, and calculating the dimensions of a rectangle given its perimeter and area. The video aims to educate viewers on these mathematical concepts in a clear and engaging manner.

Takeaways
  • 📚 The video lesson discusses solving problems involving systems of non-linear equations, which are equations with at least one non-linear component in two or more variables.
  • 🔍 A system of non-linear equations can be made up of first and second-degree equations, and solutions are determined by the number of points of intersection.
  • ⚠️ If there are no intersections, there is no solution; one intersection indicates a single solution; and two intersections indicate two solutions.
  • 📈 The first problem involves finding the intersection point of a circle with a radius of √5 cm centered at the origin and a parabola defined by x = y^2 - 3.
  • 🧩 The circle's equation is derived from its radius and center, resulting in x^2 + y^2 = 5, and the intersection is found by substituting into the parabola's equation.
  • 🔢 The solution process for the first problem involves algebraic manipulation and solving a quartic equation to find the y-coordinates of the intersection points.
  • 📐 Substituting the y-values back into the parabola's equation yields the corresponding x-values, resulting in multiple intersection points.
  • 🔑 The second problem solves a system of non-linear equations: x + 2y = -4 and y = x^2 - 2x - 3, by substituting and simplifying to find a quadratic equation in terms of y.
  • 📉 The quadratic equation is solved to find the y-values, which are then substituted back into the first equation to find the corresponding x-values.
  • 📏 The third problem involves finding the dimensions of a rectangle with a perimeter of 40 cm and an area of 96 cm^2 using the formulas for perimeter and area of a rectangle.
  • 📐 By setting up and solving a quadratic equation, the possible dimensions of the rectangle are found to be 12 cm by 8 cm and vice versa.
Q & A

  • What is a system of non-linear equations?

    -A system of non-linear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear.

  • How can the solutions of a system of non-linear equations be determined graphically?

    -The solutions can be determined by the number of points of intersection of the graphs of the equations. If there are no intersections, there is no solution. If there is one intersection, there is one solution, and if there are two intersections, there are two solutions.

  • What is the given equation of the circle in Problem 1?

    -The equation of the circle is x^2 + y^2 = 5, with a radius of √5 and centered at the origin (0,0).

  • How is the parabola defined in Problem 1?

    -The parabola is defined by the equation x = y^2 - 3.

  • How do you find the points of intersection between the circle and the parabola in Problem 1?

    -To find the points of intersection, substitute the equation of the parabola (x = y^2 - 3) into the equation of the circle (x^2 + y^2 = 5) and solve for the values of y and x.

  • What are the points of intersection found in Problem 1?

    -The points of intersection are (1, 2), (1, -2), (-2, 1), and (-2, -1).

  • What is the system of equations given in Problem 2?

    -The system of equations is x + 2y = -4 and y = x^2 - 2x - 3.

  • How do you solve the system of equations in Problem 2?

    -Solve for x in the first equation to get x = -2y - 4, then substitute this into the second equation and solve for y. Substitute the found values of y back into the equation to find x.

  • What are the solutions of the system of equations in Problem 2?

    -The solutions are (-0.5, -1.75) and (3, -3).

  • What is the given information in Problem 3 about the rectangle?

    -The perimeter of the rectangle is 40 cm, and the area is 96 square cm.

  • How do you represent the perimeter and area equations for the rectangle in Problem 3?

    -The perimeter equation is 2L + 2W = 40, and the area equation is L * W = 96, where L is the length and W is the width.

  • How do you solve for the dimensions of the rectangle in Problem 3?

    -Express W in terms of L from the area equation (W = 96/L), substitute this into the perimeter equation, and solve the resulting quadratic equation to find the values of L and W.

  • What are the dimensions of the rectangle in Problem 3?

    -The dimensions of the rectangle are 12 cm by 8 cm.

Outlines
00:00
📚 Introduction to Non-Linear Equation Systems

This paragraph introduces the concept of a system of non-linear equations, which involves two or more equations with at least one non-linear component. It explains that the solutions to such systems can be determined by the number of points of intersection. The video will present problems involving circles and parabolas, and their intersection points, starting with a circle with a radius of √5 cm and a parabola defined by the equation x = y^2 - 3.

05:03
🔍 Solving Intersection Points of Circle and Parabola

The paragraph delves into the process of finding the intersection points of a circle centered at the origin with radius √5 cm and a parabola defined by x = y^2 - 3. The equation of the circle is given as x^2 + y^2 = 5. By substituting the parabola's equation into the circle's equation, a fourth-degree polynomial in terms of y is formed. After simplification, the roots for y are found to be ±2 and ±1. Substituting these values back into the equations yields the x-coordinates of the intersection points, resulting in the points (1, 2), (1, -2), (-1, 1), and (-1, -1).

10:05
📘 Solving a System of Non-Linear Equations Involving a Rectangle

This section presents a problem involving a system of non-linear equations that describe a rectangle with a perimeter of 40 cm and an area of 96 cm². The equations are 2l + 2w = 40 for the perimeter and xy = 96 for the area, where l is the length and w is the width. By expressing w in terms of x and substituting into the perimeter equation, a quadratic equation in x is obtained. Solving this equation yields x = 12 or x = 8. Subsequently, the corresponding values for y (width) are calculated as 8 and 12, respectively, indicating the dimensions of the rectangle are 12 cm by 8 cm and vice versa.

15:06
🎓 Conclusion and Invitation to Learn More

The final paragraph serves as a conclusion to the video lesson, summarizing the topics covered and inviting viewers to continue learning by subscribing to the channel, hitting the bell button for updates, and watching more video tutorials. It emphasizes the educational value of the content and encourages viewers to engage with the channel for further lessons.

Mindmap
Keywords
💡System of Non-linear Equations
A system of non-linear equations refers to a set of two or more equations involving two or more variables, where at least one of the equations is not linear. In the context of the video, this concept is central as it forms the basis for discussing and solving complex mathematical problems. The script mentions that the system can be composed of first and second-degree equations, and solutions can be determined by the number of points of intersection, which is a key aspect of solving these systems.
💡Intersection Points
Intersection points are the values of the variables where two or more equations meet or cross each other on a graph. In the video, the concept of intersection points is used to determine the solutions to the system of non-linear equations. For instance, if there are no intersections, there is no solution; if there is one intersection, there is one solution; and if there are two intersections, there are two solutions.
💡Circle
A circle is a geometric shape defined by all points that are equidistant from a central point, known as the center. In the script, a circle with a radius of √5 centimeters and centered at the origin (0,0) is used as part of a problem involving the intersection with a parabola. The equation of the circle is given as x^2 + y^2 = 5, which is a key part of finding the intersection points with the parabola.
💡Parabola
A parabola is a U-shaped curve that is defined by a quadratic equation. In the video, a parabola is defined by the equation x = y^2 - 3. The script discusses finding the intersection points of this parabola with a circle, which is essential for solving the problem presented in the video.
💡Perimeter
The perimeter of a geometric shape, such as a rectangle, is the total length around the shape. In the video, the perimeter is used in the formula 2l + 2w to represent the total distance around a rectangle, where 'l' is the length and 'w' is the width. The script uses the perimeter to set up an equation that, along with the area, helps find the dimensions of the rectangle.
💡Area
Area refers to the amount of space inside a two-dimensional shape. In the context of the video, the area of a rectangle is calculated using the formula length times width (l * w). The script presents a problem where the area is given as 96 square centimeters, and this information is used alongside the perimeter to find the rectangle's dimensions.
💡Quadratic Equation
A quadratic equation is a polynomial equation of degree two, typically in the form of ax^2 + bx + c = 0. In the script, quadratic equations are used to model various problems, such as finding the intersection points of geometric shapes and determining the dimensions of a rectangle based on its perimeter and area.
💡Factoring
Factoring is a mathematical method used to break down a polynomial into a product of its factors. In the video, factoring is used as a technique to solve quadratic equations, such as when finding the y-values for the intersection points of a circle and a parabola. The script demonstrates factoring to simplify equations and find solutions.
💡Substitution
Substitution is a method used in algebra to solve systems of equations by replacing one variable with an expression from another equation. In the video, substitution is used to solve for x and y in the system of non-linear equations, such as when finding the intersection points of a circle and a parabola, and when solving for the dimensions of a rectangle.
💡Solving Systems of Equations
Solving systems of equations involves finding the values of the variables that satisfy all equations in the system simultaneously. The video demonstrates various methods for solving systems of non-linear equations, including factoring, substitution, and using geometric concepts like intersection points.
Highlights

Introduction to systems of non-linear equations, which are sets of two or more equations with at least one non-linear equation.

Explanation of how the solution of a system of non-linear equations can be determined by the number of points of intersection.

Problem 1 involves finding the intersection point of a circle with a radius of √5 cm and a parabola defined by x = y^2 - 3.

The equation of the circle is derived from its center at the origin and radius √5, resulting in x^2 + y^2 = 5.

Substitution of x = y^2 - 3 into the circle's equation leads to a fourth-degree equation in terms of y.

Solving the fourth-degree equation yields potential y-values for the points of intersection.

Substituting the y-values back into the original equations to find corresponding x-values.

Problem 2 presents a system of non-linear equations involving x + 2y = -4 and y = x^2 - 2x - 3.

Method to solve the system by substituting y from the second equation into the first.

Resulting in a quadratic equation in terms of x, which is solved to find the values of x.

Using the value of x to find the corresponding value of y in the system.

Problem 3 focuses on finding the dimensions of a rectangle given its perimeter and area.

The perimeter equation 2l + 2w = 40 and area equation x*y = 96 are set up, where l is length and w is width.

Solving the system by expressing one variable in terms of the other and substituting into the perimeter equation.

Resulting in a quadratic equation in x, which is factored to find the possible values of x.

Using the values of x to determine the corresponding values of y (width).

Conclusion that the dimensions of the rectangle are 12 cm by 8 cm and vice versa.

Encouragement for viewers to like, subscribe, and hit the bell button for more video tutorials.

Transcripts

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