Problem 4 - Precalculus Mathematics for Calculus

Question

Use the substitution method to find all solutions of the system of equations. $$\left\\{\begin{aligned} x^{2}+y^{2} &=25 \\ y &=2 x \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

Step 1: Identify the Equations

We are given a system of equations: \(x^2 + y^2 = 25\) and \(y = 2x\). The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Substitute for y in the First Equation

Since \(y = 2x\), substitute \(2x\) for \(y\) in the first equation \(x^2 + y^2 = 25\). This gives us \(x^2 + (2x)^2 = 25\).

Step 3: Simplify the Equation

Simplify the equation from Step 2: \(x^2 + 4x^2 = 25\). Combine like terms to get \(5x^2 = 25\).

Step 4: Solve for x

Divide both sides of the equation \(5x^2 = 25\) by 5 to isolate \(x^2\): \(x^2 = 5\). Take the square root of both sides to solve for \(x\): \(x = \sqrt{5}\) or \(x = -\sqrt{5}\).

Step 5: Find corresponding y values

Use \(y = 2x\) to find the corresponding \(y\) values for each \(x\). If \(x = \sqrt{5}\), \(y = 2 \cdot \sqrt{5} = 2\sqrt{5}\). If \(x = -\sqrt{5}\), \(y = 2(-\sqrt{5}) = -2\sqrt{5}\).

Step 6: Write the Solutions

The solutions to the system are \((\sqrt{5}, 2\sqrt{5})\) and \((-\sqrt{5}, -2\sqrt{5})\).

Key Concepts

System of Equations

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.

Solving Quadratic Equations

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 Solutions

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.