Problem 6 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=4 \\ x+y=-2 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((0, -2)\) and \((-2, 0)\).

Step 1: Analyze the System of Equations

The system comprises two equations: \(x^2 + y^2 = 4\) (a circle with radius 2 centered at the origin) and \(x + y = -2\) (a straight line). We aim to find the points where both equations hold true simultaneously.

Step 2: Express y in terms of x

From the second equation, express \(y\) in terms of \(x\): \(y = -2 - x\). This allows us to substitute \(y\) in the first equation.

Step 3: Substitute y in the Circle Equation

Substitute \(y = -2 - x\) into the circle equation: \(x^2 + (-2 - x)^2 = 4\). This results in a single equation in terms of \(x\).

Step 4: Simplify the Resulting Equation

Expand and simplify the equation: \[x^2 + (-2 - x)^2 = x^2 + (4 + 4x + x^2) = 4\] leading to \[2x^2 + 4x + 4 = 4\].

Step 5: Solve the Quadratic Equation

Simplify to \(2x^2 + 4x = 0\). Factor out the common term to get \(2x(x + 2) = 0\). Solve for \(x\), yielding \(x = 0\) or \(x = -2\).

Step 6: Find Corresponding y Values

Substitute \(x = 0\) into \(y = -2 - x\) to get \(y = -2\). Substitute \(x = -2\) to find \(y = 0\). The solutions are the pairs \((0, -2)\) and \((-2, 0)\).

Key Concepts

Circle Equation

A circle equation is a fundamental concept in geometry, often represented in a format like \(x^2 + y^2 = r^2\). The expression \(x^2 + y^2 = 4\) describes a circle centered at the origin (0, 0) with a radius of 2. This means that any point fulfilling the equation is located exactly 2 units away from the origin.
Understanding this equation helps with visualizing relationships between coordinates. The value on the right side of the equation, 4 in this case, indicates the square of the radius.
- Radius is 2 because \(\sqrt{4} = 2\).- Center of the circle is at (0, 0) because the equation is not shifted.By exploring circle equations, students gain insight into how algebraic equations translate into geometric figures. This fosters skills like sketching and interpreting geometric forms.

Linear Equation

Linear equations like \(x + y = -2\) are essential in algebra, forming straight lines once graphed. The equation provided is already in the form that shows the relationship between x and y. This particular relationship sets a constraint whereby the sum of x and y must always equal negative two.
Linear equations are characterized by:- Straight-line graphs.- Variables typically to the first power.- Constant rate of change between variables.To graph \(x + y = -2\), you can easily identify points by choosing any value for x, then solving for y. For example:- If \(x = 0\), \(y = -2\).- If \(x = -2\), \(y = 0\).These points, when connected, illustrate the line that meets the circle in the solution of the system.

Quadratic Equation

Quadratic equations, such as \(2x^2 + 4x = 0\), are pivotal in solving nonlinear systems. These equations include variables raised to the second power, often resulting in parabolic graphs.
The transformation of our circle's equation into the form of \(x^2 + y^2 = 4\) to ultimately \(2x(x + 2) = 0\) allows us to apply factorization for a straightforward solution.- A quadratic equation may have 0, 1, or 2 solutions based on its discriminant.- Factoring such as \(2x(x + 2)= 0\) simplifies finding solutions.In this instance, by factoring, we pinpoint the values of \(x\) where these expressions equal zero:- Either \(2x = 0\) gives \(x = 0\), or- \(x + 2 = 0\) gives \(x = -2\).Both solutions translate into points on our original circle equation, interacting with the linear equation to define the intersection.

Simultaneous Equations

Simultaneous equations occurs when dealing with multiple equations and seeking their common solutions. The system may comprise different types of equations, like linear and quadratic ones, as seen in our problem. To solve them:- We manipulate one or both equations to allow them to be combined or substitute values.- In this problem, substituting \(y = -2 - x\) into the circle equation simplifies to one solvable equation.- This method reduces the complexity and isolates one of the variables, making solution strategies like factoring more efficient.Ultimately, simultaneous equations unveil the intersection points of the given curves. For our system, the results were:- \((0, -2)\) and - \((-2, 0)\).These solutions represent the point(s) that satisfy both the circle and linear line equations concurrently. Understanding simultaneous equations is invaluable for not only math but also real-world problem-solving.