Problem 13 - Algebra 2

Question

Find the exact solution(s) of each system of equations. $$ \begin{array}{l} \frac{x^{2}}{36}-\frac{y^{2}}{4}=1 \\ x=y \end{array} $$

Step-by-Step Solution

Answer

No real solution exists.

Step 1: Analyze the Equations

We have two equations: \( \frac{x^{2}}{36} - \frac{y^{2}}{4} = 1 \) and \( x = y \). The first equation is the equation of a hyperbola, and the second equation is a line with a slope of 1.

Step 2: Substitute the Second Equation into the First

Substitute \( x = y \) into the hyperbola equation. This gives: \( \frac{x^{2}}{36} - \frac{x^{2}}{4} = 1 \).

Step 3: Solve the Resulting Equation

Combine the terms on the left-hand side: \( \frac{x^{2}}{36} - \frac{x^{2}}{4} = 1 \). This can be written as \( x^{2} \left( \frac{1}{36} - \frac{1}{4} \right) = 1 \).

Step 4: Find a Common Denominator

Find a common denominator for the fractions \( \frac{1}{36} \) and \( \frac{1}{4} \). The common denominator is 36. Rewriting, we have: \( \frac{1}{36} - \frac{9}{36} = -\frac{8}{36} = -\frac{2}{9} \).

Step 5: Simplify and Solve for \( x^2 \)

The equation \( x^{2} \left( -\frac{2}{9} \right) = 1 \) simplifies to \( x^{2} = -\frac{9}{2} \).

Step 6: Assess the Solution

Since \( x^2 = -\frac{9}{2} \), there is no real number solution because the square of a real number cannot be negative.

Key Concepts

Understanding Hyperbolas

A **hyperbola** is a type of conic section that appears as two open, opposing curves. It is characterized by its standard equation form, which can look like \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]In our exercise, the equation \( \frac{x^2}{36} - \frac{y^2}{4} = 1 \) represents a hyperbola. Here, 36 and 4 are constants that determine the curve's shape and orientation. The term \( \frac{x^2}{36} \) indicates how spread out the curve is along the x-axis, while \( \frac{y^2}{4} \) does the same for the y-axis.
Hyperbolas differ from other conic sections, like ellipses, because their equations have a subtraction sign between the terms. This unique feature creates curves that show a way the points diverge as they move away from the center.
In practical problems involving hyperbolas, their intersection with other curves or lines often needs to be evaluated, as in this exercise where the hyperbola intersects the line \( x = y \). This intersection leads us to explore solutions that satisfy both conditions simultaneously.

Exploring the Substitution Method

When dealing with a **system of equations**, the substitution method is a powerful strategy. It simplifies multi-variable problems by replacing one variable in terms of another. This method is particularly useful when one of the equations is straightforward, like \( x = y \), making substitution direct and simple.
Here's a quick guide to using the substitution method:

• Identify a variable from one equation that can be easily isolated. In our case, it's \( x \) from the equation \( x = y \).
• Substitute this variable into the other equation. For us, replace \( y \) with \( x \) in \( \frac{x^2}{36} - \frac{y^2}{4} = 1 \).
• After substitution, a new equation emerges that has only one variable. It simplifies solving.

Once substitution is completed, you'll often encounter a simplified equation. For example, substituting \( x = y \) gives us \( \frac{x^2}{36} - \frac{x^2}{4} = 1 \). Solving this results in finding possible solutions for \( x \). However, as seen in this problem, not all solutions turn out to be real numbers.

Understanding Real Number Solutions

**Real number solutions** are those that can exist within the set of real numbers, essentially covering every number on a continuous line, including both positive and negative numbers, as well as zero. However, certain mathematical conditions prevent a solution from being real. This problem illustrates just that.Here's what happens in our exercise:- From the equation \( x^2 = -\frac{9}{2} \), we need to find a solution for \( x \).- In mathematics, the square of any real number is always non-negative.Consequently, if you derive an equation like \( x^2 = -\frac{9}{2} \), it suggests no real number solution is possible. Squaring a real number cannot yield a negative result. Such outcomes are significant in determining the nature of solutions in any set of equations.
Hence, if you encounter a similar result, it's crucial to recognize the lack of real solutions, indicating that the problem space doesn't intersect within real numbers, or any solution must be complex.