Problem 27 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l} x^{2}+y^{2}=9 \\ x^{2}-y^{2}=1 \end{array}\right.$$

Step-by-Step Solution

Answer

Solutions are \((\sqrt{5}, 2), (\sqrt{5}, -2), (-\sqrt{5}, 2), (-\sqrt{5}, -2)\).

Step 1: Examine the System of Equations

The system consists of two equations: \( x^2 + y^2 = 9 \) and \( x^2 - y^2 = 1 \). Notice that these equations can be viewed as equations of conic sections: the first one is a circle, and the second one represents a hyperbola.

Step 2: Add the Equations

By adding the two equations, you can eliminate \( y^2 \) to find an expression for \( x^2 \). Thus, \( (x^2 + y^2) + (x^2 - y^2) = 9 + 1 \), resulting in \( 2x^2 = 10 \). This simplifies to \( x^2 = 5 \).

Step 3: Subtract the Equations

By subtracting the second equation from the first, eliminate \( x^2 \), resulting in an expression for \( y^2 \). Thus, \( (x^2 + y^2) - (x^2 - y^2) = 9 - 1 \), which simplifies to \( 2y^2 = 8 \). Therefore, \( y^2 = 4 \).

Step 4: Solve for x and y

With \( x^2 = 5 \), we find \( x = \pm \sqrt{5} \). With \( y^2 = 4 \), we find \( y = \pm 2 \). Thus, the possible combinations of \( (x, y) \) are \( (\sqrt{5}, 2) \), \( (\sqrt{5}, -2) \), \( (-\sqrt{5}, 2) \), and \( (-\sqrt{5}, -2) \).

Step 5: Verify the Solutions

Double-check each solution against the original system of equations. All combinations satisfy both \( x^2 + y^2 = 9 \) and \( x^2 - y^2 = 1 \), confirming \((\sqrt{5}, 2)\), \((\sqrt{5}, -2)\), \((-\sqrt{5}, 2)\), and \((-\sqrt{5}, -2)\) as valid solutions.

Key Concepts

Understanding Conic Sections

Conic sections are an important concept when dealing with certain types of algebraic equations. They arise from the intersection of a plane and a cone, resulting in shapes like circles, ellipses, parabolas, and hyperbolas.

• Circle: A circle in an equation often appears like this: \( x^2 + y^2 = r^2 \). Here \( r^2 \) represents the radius squared. For example, the equation \( x^2 + y^2 = 9 \) describes a circle with a radius of 3.
• Hyperbola: A hyperbola, on the other hand, is represented as \( x^2 - y^2 = c \). The value of \( c \) determines the distance between its vertices. Such equations depict a curve with two separate branches that open away from each other, typically along the axis of the non-dominant square.

Understanding these shapes helps to visualize and simplify systems of equations. Recognizing that our system includes both a circle and a hyperbola can guide us in solving and interpreting the solutions geometrically.

Solving Algebraic Equations

When solving systems of algebraic equations like the one given, employing methods that simplify calculations can be beneficial. Adding and subtracting equations are powerful techniques.

• Add the Equations: Combining equations \( x^2 + y^2 \) and \( x^2 - y^2 \) by addition eliminates \( y^2 \), letting us solve directly for \( x^2 \). Here, adding gives \( 2x^2 = 10 \), leading to \( x^2 = 5 \).
• Subtract the Equations: Subtracting one equation from the other eliminates \( x^2 \), simplifying the problem to find \( y^2 \). For example, subtracting to get \( 2y^2 = 8 \) results in \( y^2 = 4 \).

The result of solving \( x^2 = 5 \) is that \( x \) can be both positive or negative \( \sqrt{5} \), and for \( y^2 = 4 \), \( y = \pm 2 \). This approach allows for finding all possible solutions, like those in this exercise: \((\sqrt{5}, 2)\), \((\sqrt{5}, -2)\), \((-\sqrt{5}, 2)\), \((-\sqrt{5}, -2)\).

Verification of Solutions

Solution verification is crucial. It ensures calculated solutions satisfy all original equations in the system. Incorrect solutions can result from errors in algebraic manipulation or calculations. Hence, verifying is a must-do final step.

• Substitute Back: Substitute each pair of solutions \( (x, y) \) back into the original equations \( x^2 + y^2 = 9 \) and \( x^2 - y^2 = 1 \) to check their validity.
• Verify Each Step: Check calculations in earlier steps to ensure accuracy. Re-evaluating intermediary results is helpful to spot potential errors.

For this problem, all calculated pairs \((\sqrt{5}, 2)\), \((\sqrt{5}, -2)\), \((-\sqrt{5}, 2)\), and \((-\sqrt{5}, -2)\) meet the system of equations when substituted back. This step assures us of the correctness and reliability of our solutions.