Problem 12 - College Algebra

Question

Solve the system of nonlinear equations using elimination. $$ \begin{array}{l} x^{2}+y^{2}=25 \\ x^{2}-y^{2}=1 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \((\sqrt{13}, 2\sqrt{3})\), \((\sqrt{13}, -2\sqrt{3})\), \((-\sqrt{13}, 2\sqrt{3})\), and \((-\sqrt{13}, -2\sqrt{3})\)."

Step 1: Combine Equations by Elimination

The system of equations given is:1. \( x^{2} + y^{2} = 25 \) 2. \( x^{2} - y^{2} = 1 \)Add the two equations to eliminate \(y^2\). This gives:\( (x^2 + y^2) + (x^2 - y^2) = 25 + 1 \) Simplifying, we get:\( 2x^2 = 26 \).

Step 2: Solve for \(x\)

From \( 2x^2 = 26 \), divide both sides by 2 to get:\( x^2 = 13 \).Taking the square root of both sides, we have:\( x = \pm \sqrt{13} \).

Step 3: Substitute \(x\) in First Equation

Using \( x = \pm \sqrt{13} \), substitute into the first equation \( x^2 + y^2 = 25 \).Substituting \( x^2 = 13 \),\( 13 + y^2 = 25 \).

Step 4: Solve for \(y\)

From \( 13 + y^2 = 25 \), subtract 13 from both sides:\( y^2 = 12 \).Taking the square root of both sides, we find:\( y = \pm \sqrt{12} \), which simplifies to \( y = \pm 2\ ext{sqrt}{3} \).

Step 5: List the Solutions

The solutions for \((x, y)\) are when \(x = \sqrt{13} \) giving \( y = \pm 2\sqrt{3} \), and when \(x = -\sqrt{13} \) giving \( y = \pm 2\sqrt{3} \).Therefore, the pairs \((x, y)\) are:1. \((\sqrt{13}, 2\sqrt{3})\)2. \((\sqrt{13}, -2\sqrt{3})\)3. \((-\sqrt{13}, 2\sqrt{3})\)4. \((-\sqrt{13}, -2\sqrt{3})\).

Key Concepts

Elimination Method

When solving a system of equations, one effective approach is the elimination method. This technique is useful for eliminating one of the variables, thus simplifying the problem to a single equation with one variable. Here's how it works:

• First, align the equations vertically so that like terms are in columns. This makes it easier to see which terms can be eliminated.
• Next, add or subtract the equations to eliminate one of the variables.
• Once the variable is eliminated, you'll find yourself with an equation that has only one variable. Solve this equation as usual.

The main goal of the elimination method is to reduce the number of variables so that solving the system becomes more straightforward. It's particularly useful for nonlinear equations, where managing the terms can be more complex. In our example, by adding the equations, we successfully eliminated the term containing \(y^2\), simplifying the problem to solve for \(x\).

System of Equations

A system of equations is a set of two or more equations with the same variables. In order to find solutions, the variables must satisfy each equation in the system simultaneously.

• These systems can be linear or nonlinear. Linear equations have variables raised only to the first power, while nonlinear equations can have variables that are squared or have higher powers.
• Methods to solve these systems include graphing, substitution, and elimination.
• The solutions can be depicted as points where the graphs of the equations intersect on a coordinate plane.

In our exercise, we tackled a system of nonlinear equations, meaning that the equations included squared terms. The goal was to find pairs of \((x, y)\) that satisfied both equations. By using the elimination method, we efficiently simplified the system, making it easier to isolate and solve for each variable.

Square Roots

Understanding the concept of square roots is crucial when dealing with equations that include squared terms. A square root of a number \(a\) is a number \(b\) that satisfies \(b^2 = a\). Here are some important points about square roots:

• Every positive number has two square roots: a positive and a negative one (e.g., \( \sqrt{9} = 3\) and \(-\sqrt{9} = -3\)).
• The square root symbol \(\sqrt{}\) typically represents only the positive square root, but both positive and negative roots need consideration in equations.
• Taking a square root is the inverse operation of squaring a number, which makes it useful when solving equations like \(x^2 = 13\).

In the solution to our exercise, we handled square roots when isolating \(x\) and \(y\). After simplifying the equations, we found \(x = \pm \sqrt{13}\) and \(y = \pm 2\sqrt{3}\), demonstrating how critical square roots are in finding all potential solutions.