Problem 5 - Precalculus: Mathematics for Calculus

Question

Use the substitution method to find all solutions of the system of equations. $$\left\\{\begin{array}{l} x^{2}+y^{2}=8 \\ x+y=0 \end{array}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Solve for One Variable in the Second Equation

From the equation \( x + y = 0 \), isolate one of the variables. Let's solve for \( y \):\[ y = -x \].

Step 2: Substitute the Expression into the First Equation

Replace \( y \) in the first equation \( x^2 + y^2 = 8 \) with the expression from Step 1:\[ x^2 + (-x)^2 = 8 \].

Step 3: Simplify the Equation

Simplify \( x^2 + (-x)^2 = 8 \) to get:\[ x^2 + x^2 = 8 \].Combine like terms:\[ 2x^2 = 8 \].

Step 4: Solve for \( x \)

Divide both sides of the equation by 2:\[ x^2 = 4 \].Take the square root of both sides:\[ x = \pm 2 \].

Step 5: Find Corresponding \( y \) Values

Use \( y = -x \) to find \( y \):- If \( x = 2 \), then \( y = -2 \).- If \( x = -2 \), then \( y = 2 \).

Step 6: Write the Solution Pairs

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

Key Concepts

System of Equations

A system of equations consists of multiple equations that involve the same set of variables. In this exercise, we have a system composed of two equations:

• The first equation is a quadratic: \(x^2 + y^2 = 8\).
• The second equation is linear: \(x + y = 0\).

The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Understanding how to interpret each equation and identifying the right methods to solve them is crucial in systems of equations.

Systems of equations can either have a unique solution, infinitely many solutions, or no solution at all, depending on the nature of the equations and their intersections.

Solving Equations

To solve a system of equations, especially when dealing with one linear and one non-linear equation like in our example, a common approach is the substitution method. Here’s how it works in simple steps:

• Goal: First, we want to solve for one variable in terms of the other. This simplifies tackling the system step by step.
• Start with the simpler equation: From the equation \(x + y = 0\), you can easily express \(y\) in terms of \(x\) by isolating one of the variables: \(y = -x\).

This step simplifies one of the equations significantly, paving the way to substitute this relationship into the other equation, thereby reducing the number of variables you have to deal with in one equation.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make equations easier to solve. It plays a crucial role in both isolating variables and reducing equations in a system. Let's break it down:

• Substitute and Simplify: Once you have \(y = -x\), substitute \( y \) in the first equation: \(x^2 + (-x)^2 = 8\). Simplify this to \(2x^2 = 8\) by combining like terms.
• Solve for the variable: Divide both sides by 2 to isolate \(x^2\): \(x^2 = 4\). Then take the square root of both sides to solve for \(x\), giving \(x = \pm 2\).
• Find corresponding values: Use the relationship \(y = -x\) to find the pairs: if \(x = 2\), then \(y = -2\); if \(x = -2\), then \(y = 2\).

By carefully applying algebraic manipulations, one can systematically uncover all possible solutions to the system of equations.