91Ó°ÊÓ

A system of equations contains two or more equations with the same set of variables. The objective is to find the values of these variables that satisfy all equations at the same time.
In our exercise, the system of equations given is:

• Equation 1: \(x^2 + y^2 = 25\)
• Equation 2: \(y = 2x\)

This means we are looking for values of \(x\) and \(y\) that make both equations true simultaneously.
To approach solving them, particularly when dealing with linear and non-linear equations, methods such as substitution are very useful.
The substitution method involves replacing one variable in one equation with its equivalent from another equation. In this instance, we substituted \(2x\) for \(y\) in the first equation. This approach effectively reduces the system to one equation with one unknown, simplifying the process to find a solution.

Quadratic equations are equations that can be written in the form \(ax^2 + bx + c = 0\). They have a degree of two, indicated by the exponent on the variable.
In the system we solved, after substituting \(y = 2x\) into \(x^2 + y^2 = 25\), the equation simplifies to a quadratic form:

• \(5x^2 = 25\)

To solve this quadratic equation, you can follow these steps:

• Isolate \(x^2\) by dividing both sides by 5, giving \(x^2 = 5\).
• Take the square root of both sides to solve for \(x\). Remember, the square root of a number can be positive or negative, hence, \(x = \sqrt{5}\) or \(x = -\sqrt{5}\).

This solution process highlights that quadratic equations can have more than one solution, making it important to consider both the positive and negative square roots.

Graphical methods offer a visual approach to solving systems of equations. You can graph each equation on the same coordinate plane and find the intersections, which represent the solutions.
For instance:

• The equation \(x^2 + y^2 = 25\) represents a circle centered at the origin with a radius of 5.
• The equation \(y = 2x\) is a line through the origin with a slope of 2.

The solutions to the system are where the circle and the line intersect. In this case, the intersection points are \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).
This graphical representation is not only intuitive but also confirms the accuracy of the algebraic solutions found through methods like substitution. It illustrates how different types of equations interact on a graph, providing a deeper understanding of their relationships.