Problem 2 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{r} x^{2}+y^{2}=25 \\ 3 x+4 y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((-4, 3)\) and \((4, -3)\).

Step 1: Identify Equations

The problem gives us a system of nonlinear equations: 1. \( x^2 + y^2 = 25 \) (Equation 1)2. \( 3x + 4y = 0 \) (Equation 2). The first equation represents a circle with radius 5, and the second equation is a linear equation.

Step 2: Solve Equation 2 for One Variable

From Equation 2, solve for one variable, say \( x \): \[ 3x + 4y = 0 \]\[ 3x = -4y \]\[ x = -\frac{4}{3}y \]. This expresses \( x \) as a function of \( y \).

Step 3: Substitute into Equation 1

Substitute \( x = -\frac{4}{3}y \) into Equation 1:\[ (-\frac{4}{3}y)^2 + y^2 = 25 \]\[ \frac{16}{9}y^2 + y^2 = 25 \]Combine the terms:\[ \frac{25}{9}y^2 = 25 \].

Step 4: Solve for \( y \)

To solve for \( y \), first eliminate the fraction:\[ 25y^2 = 225 \]Divide both sides by 25:\[ y^2 = 9 \]Take the square root of both sides, remembering to consider both positive and negative roots:\[ y = 3 ext{ or } y = -3 \].

Step 5: Find Corresponding \( x \) Values

Use the values of \( y \) to find \( x \):1. If \( y = 3 \):\[ x = -\frac{4}{3}(3) = -4 \]2. If \( y = -3 \):\[ x = -\frac{4}{3}(-3) = 4 \].

Step 6: Write Solutions

The solutions to the system are the pairs \((x, y)\):1. \((-4, 3)\)2. \((4, -3)\). These pairs satisfy both equations in the system.

Key Concepts

Solving Algebraic Equations

When we talk about solving algebraic equations, we mean finding the value of the variable(s) that make the equation true. In this exercise, you're working with a system of equations, which means you're trying to find a solution that satisfies more than one equation at the same time. This usually involves finding values for variables like \( x \) and \( y \) that work for all given equations in the system.

To solve such systems, there are several methods you can use. Common strategies include substitution, elimination, and graphing, among others. Each method offers a different approach, but they all aim to isolate the variable and eventually solve for it.

The first step in solving any algebraic equation is understanding what's being asked. Pay attention to the type of equation—whether it's linear, quadratic, or another form—since this affects the solving method you choose. In our given problem, one equation is linear, while the other represents a circle, making the system nonlinear.

• Double check the equations for any errors in your initial read-through.
• Identify key characteristics (like degree and type of equation).
• Choose the most effective solving method based on equation types you’re dealing with.

Circle Equations

In the world of algebra, a circle equation is a specific type of equation representing all the points that lie on a circle. The general form of a circle's equation is \( x^2 + y^2 = r^2 \), where \( r \) is the radius of the circle.

In our exercise, \( x^2 + y^2 = 25 \) symbolizes a circle centered at the origin with a radius of 5. This equation helps us identify points \((x, y)\) that are at exactly 5 units distance from the origin.

• The term \( x^2 + y^2 \) ensures that all solutions lie at a consistent distance from the center.
• The value 25 is the square of the circle’s radius.

To solve this equation as part of the system, it's crucial to find other conditions (from other equations) that these solutions must also satisfy, like in the case of a linear equation intersecting the circle.

Linear Equations

Linear equations are your straightforward algebraic expressions where each variable is raised only to the first power. Typically, they appear in the form \( ax + by = c \).

In this exercise, we have \( 3x + 4y = 0 \). This equation represents a straight line in a two-dimensional plane.

• The coefficients of \( x \) and \( y \) (in this instance 3 and 4) determine the slope and intercept of the line when graphically represented.
• The zero on the right-hand side tells us this line passes through the origin.

Linear equations often serve as constraints in systems of equations where they intersect other geometric figures, like circles. Understanding how to manipulate them—solving for one variable in terms of another, as done here—becomes crucial in finding these intersection points.

Algebraic Substitution Method

The substitution method is a powerful technique used to solve systems of equations, especially when the system involves one or more nonlinear equations. The idea is to solve one of the equations for one variable and then substitute this expression into the other equations in the system.

In our exercise, we first solved the linear equation \( 3x + 4y = 0 \) for \( x \), resulting in \( x = -\frac{4}{3}y \).

• This expression for \( x \) was then substituted into the circle equation \( x^2 + y^2 = 25 \), reducing it to a single-variable quadratic equation in terms of \( y \).
• The expression \( (-\frac{4}{3}y)^2 + y^2 = 25 \) allowed us to simplify and solve for \( y \).
• Finally, we used the found \( y \) values to determine corresponding \( x \) solutions.

This method simplifies solving complex nonlinear systems by reducing the number of variables you deal with simultaneously, effectively making the process more manageable.