Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, use any method to solve the nonlinear system. $$ \begin{array}{l} x^{2}+y^{2}=25 \\ x^{2}-y^{2}=36 \end{array} $$

Step-by-Step Solution

Answer

There are no real solutions to the system; possible complex solutions exist.

Step 1: Identify the given system of equations

The problem provides a system of nonlinear equations: 1. \( x^2 + y^2 = 25 \) 2. \( x^2 - y^2 = 36 \). These equations represent a circle and a hyperbola, respectively.

Step 2: Solve for one variable

To eliminate one variable, add the two equations:\[(x^2 + y^2) + (x^2 - y^2) = 25 + 36.\]This simplifies to:\[2x^2 = 61.\]Solve for \( x^2 \) by dividing both sides by 2:\[ x^2 = \frac{61}{2}. \]

Step 3: Substitute to find the other variable

With \( x^2 = \frac{61}{2} \), use either of the original equations to find \( y^2 \). Choosing the first equation, \( x^2 + y^2 = 25 \), substitute \( x^2 \):\[\frac{61}{2} + y^2 = 25.\]Rearrange to solve for \( y^2 \):\[y^2 = 25 - \frac{61}{2} = \frac{50}{2} - \frac{61}{2} = -\frac{11}{2}.\]

Step 4: Assess mathematical validity

The resulting expression for \( y^2 = -\frac{11}{2} \) is impossible under real numbers because the square of a real number cannot be negative. This indicates no real solutions for the system. Instead, there might be complex solutions.

Key Concepts

Nonlinear Equations

A system of nonlinear equations involves equations that are not straight lines. They could include parabolas, circles, ellipses, and hyperbolas. These curves are what make solving nonlinear systems more intricate compared to linear systems, where solutions are often single points where lines intersect.
The system provided in the exercise includes two nonlinear equations: one representing a circle and the other representing a hyperbola. The complexity lies in their non-linear nature, which results in curves that don't intersect in straightforward ways.
When solving such systems, we use methods like substitution, elimination, or graphing to find where these curves might intersect, if at all. Each curve represents a locus of points satisfying the corresponding equation, and solutions are the intersections of these loci.

Circle Equation

The circle equation given in the exercise is \( x^2 + y^2 = 25 \). This is a standard form for a circle centered at the origin with a radius. The general form for a circle equation centered at the origin is \( x^2 + y^2 = r^2 \). Here, \( r \) is the radius.

• For our circle, since \( r^2 = 25 \), the radius \( r \) is \( 5 \).
• This circle includes all points that are exactly 5 units away from the origin.

This equation is part of our nonlinear system, and it represents a symmetric shape. When combined with other types of equations, like a hyperbola, determining where intersections occur becomes an engaging challenge.

Hyperbola Equation

The hyperbola equation provided is \( x^2 - y^2 = 36 \). In general terms, a hyperbola is represented by equations of a similar form, usually \( x^2/a^2 - y^2/b^2 = 1 \). However, modifications such as different coefficients and constants in different terms can alter its orientation and dimensions.

• In this case, because there is no coefficient \( a^2 \) or \( b^2 \), it simplifies the structure but keeps the essence of the hyperbola's shape.
• The provided equation outlines a hyperbola centered at the origin and its 'curves' mirror each other.

Hyperbolas consist of two separate curves that can open either up and down or left and right, depending on which term is positive. Solving for intersections with a circle in a nonlinear system might not yield real solutions, as seen here.

Complex Solutions

When dealing with solutions to equations, we often seek real-number results. However, nonlinear systems such as those involving circles and hyperbolas might not intersect at points resulting in real solutions. As in this exercise, we encounter \( y^2 = -\frac{11}{2} \). A negative number under a square root implies complex solutions.

• Complex solutions involve imaginary numbers, where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).
• Thus, solving \( y^2 = -\frac{11}{2} \) leads to \( y = i \sqrt{\frac{11}{2}} \), where the solution is expressed in terms of \( i \).

Nonlinear systems with complex solutions indicate that the intersection points are not visible on the real plane, reflecting the rich structure of mathematical solutions beyond integers and real numbers.