Problem 12 - College Algebra

Question

For the following exercises, solve the system of nonlinear equations using elimination. $$x^{2}+y^{2}=25$$ $$x^{2}-y^{2}=1$$

Step-by-Step Solution

Answer

The solution pairs are (\(\sqrt{13}, 2\sqrt{3}\)), (\(\sqrt{13}, -2\sqrt{3}\)), (\(-\sqrt{13}, 2\sqrt{3}\)), and (\(-\sqrt{13}, -2\sqrt{3}\)).

Step 1: Understand the Equations

We have two nonlinear equations:1. \( x^2 + y^2 = 25 \) 2. \( x^2 - y^2 = 1 \).We are to use the method of elimination to find values for \(x\) and \(y\).

Step 2: Set Up Elimination

To eliminate \(y^2\), we add the equations:First equation: \( x^2 + y^2 = 25 \) Second equation: \( x^2 - y^2 = 1 \).By adding them, we have:\[ (x^2 + y^2) + (x^2 - y^2) = 25 + 1 \].

Step 3: Simplify the Equation

Simplify the equation from Step 2:\[ x^2 + y^2 + x^2 - y^2 = 26 \] simplifies to:\[ 2x^2 = 26 \].

Step 4: Solve for x

Solve the equation \( 2x^2 = 26 \) for \(x\):Divide both sides by 2:\[ x^2 = 13 \].Take the square root of both sides:\[ x = \pm \sqrt{13} \].

Step 5: Solve for y

Use one of the original equations to solve for \(y\). Let's use \( x^2 + y^2 = 25 \):Substitute \( x^2 = 13 \) into it:\[ 13 + y^2 = 25 \].Solve for \(y^2\):\[ y^2 = 25 - 13 \]\[ y^2 = 12 \].Take the square root:\[ y = \pm \sqrt{12} \], which simplifies to \[ y = \pm 2\sqrt{3} \].

Step 6: State Solution Pairs

The solution pairs \((x, y)\) are:1. \( x = \sqrt{13}, y = 2\sqrt{3} \) 2. \( x = \sqrt{13}, y = -2\sqrt{3} \) 3. \( x = -\sqrt{13}, y = 2\sqrt{3} \) 4. \( x = -\sqrt{13}, y = -2\sqrt{3} \).

Key Concepts

Elimination Method

The elimination method is a popular strategy to solve a system of equations, making it easier to find the values of the variables involved. In this method, the primary goal is to eliminate one of the variables to simplify the system. This involves combining the equations in a way that cancels out one of the unknowns by addition or subtraction.

• Look for coefficients that can be easily manipulated to cancel a variable.
• You may need to multiply one or both of the equations by a certain number to align coefficients.
• Once one variable is eliminated, solve for the other and back-substitute to find the remaining variable.

In the exercise provided, we have a pair of equations where the term \( y^2 \) can be easily eliminated by adding the two equations. This is possible because the variable \( y^2 \) in the first equation has an opposite sign compared to the second equation, allowing for a straightforward elimination.

System of Equations

A system of equations consists of two or more equations that share two or more unknowns. Solving a system means finding values for the variables that satisfy all equations simultaneously. These systems can be linear or nonlinear.

• Linear systems consist of linear equations, where variables are not raised to any power greater than one.
• Nonlinear systems have at least one equation that is not linear, involving exponents, products, or other nonlinear operations.

In the given problem, we have a nonlinear system of equations, given by:

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

To solve this system involves finding pairs of \((x, y)\) that satisfy both equations at once. The elimination process helps in simplifying such systems by removing one of the variables, allowing us to solve for each variable step by step.

Algebraic Solutions

Finding algebraic solutions means using algebraic techniques to solve equations or systems of equations involving variables. Mathematics offers various techniques, but for systems like in our exercise, the elimination method is often efficient.

• Start by rewriting equations if needed to align terms for elimination.
• After eliminating a variable, solve the resulting single-variable equation by algebraic manipulation.
• Use the solution found to back-substitute into one of the original equations to find the other variable.

In our initial problem, after eliminating \( y \), we simplified to find \( x^2 = 13 \), which gives real roots \( x = \pm \sqrt{13} \). We then substituted \( x^2 \) back to solve for \( y \), finding real roots \( y = \pm 2\sqrt{3} \). This systematic approach of algebraic manipulation and substitution delivers precise solution pairs for the initial nonlinear system.