Problem 31 - College Algebra

Question

For the following exercises, use any method to solve the nonlinear system. $$\begin{aligned} x^{2}+y^{2} &=1 \\ y^{2} &=x^{2} \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right), \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right), \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right), \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 1: Set up Equation 1

Look at the first equation of the system: \[ x^2 + y^2 = 1 \] This equation describes a circle with a radius of 1, centered at the origin (0,0) in the xy-plane.

Step 2: Set up Equation 2

Examine the second equation of the system: \[ y^2 = x^2 \] This equation can be rewritten as \( y = ext{either} \ x \text{ or} \ y = -x \), since squaring both sides will result in the same value.

Step 3: Solve for Intersection with y=x

Using \( y = x \), substitute \( y \) in the first equation: \[ x^2 + x^2 = 1 \] which simplifies to \[ 2x^2 = 1 \] Divide by 2 to get \[ x^2 = \frac{1}{2} \] Take the square root of both sides to find \[ x = \pm \frac{\sqrt{2}}{2} \] Thus, the points of intersection when \( y = x \) are \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \) and \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

Step 4: Solve for Intersection with y=-x

Now using \( y = -x \), substitute \( y \) in the first equation: \[ x^2 + (-x)^2 = 1 \] which again simplifies to \[ 2x^2 = 1 \] This leads to the same result:\[ x = \pm \frac{\sqrt{2}}{2} \] Hence, the points of intersection when \( y = -x \) are \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \) and \( \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 5: List Solutions

Combine the solutions from Steps 3 and 4 to list all intersection points: \( \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \), \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \), \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \), and \( \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Key Concepts

Circle Equations

When you come across an equation like \( x^2 + y^2 = 1 \), you're dealing with a circle equation. This specific equation represents a circle centered at the origin of the coordinate plane with a radius of 1. A circle equation in this form is simple:

• \( x^2 + y^2 = r^2 \)
• Where \( r \) is the radius of the circle.
• The center is at \((0, 0)\) in the Cartesian plane.

Understanding this setup helps when you're tasked with plotting or finding intersection points with other graphs, as it gives you a visual reference—a perfect circle centered on the graph. Each point on this circle is equidistant from the origin.

Intersection Points

Intersection points are the spots where two or more graphs meet on a coordinate plane. These points have the same coordinates for both graphs, meaning they satisfy all equations involved in the system simultaneously.

• To find intersection points, you solve the system of equations either algebraically or graphically.
• In the exercise we're discussing, intersection points are determined by replacing \( y = x \) and \( y = -x \) into the circle equation.
• This led to solutions—specific \( x \) and \( y \) values that satisfy both equations.

Finding intersection points is crucial because it tells us where conditions defined by different equations overlap or coincide.

Substitution Method

The substitution method is a handy technique for solving a system of equations, especially when one or all equations are not linear. Here's how it works:

• Solve one of the equations for one variable in terms of the other.
• Substitute this expression into the other equation(s).
• Solve for the remaining variable.
• Go back to find the other variable using the expression from the first step.

In our example, we see the substitution method applied as we replace \( y \) with \( x \) and \( -x \) from \( y = \pm x \) into the circle equation. This simplifies the process and reduces the original system into simpler equations that are easy to solve.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In this case, we're examining the following system:

• \( x^2 + y^2 = 1 \) (a circle equation)
• \( y^2 = x^2 \) (a linear relationship in disguise)

The goal here is to find solutions—intersections—that simultaneously satisfy both equations. Approaching a system analytically allows us to understand and visualize how different mathematical objects, like circles and lines, relate to each other on the graph. These intersections show us practical, real-world solutions where conditions overlap between different constraints.