Problem 31 - Precalculus: Mathematical 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

The solutions are \((\sqrt{5}, 2)\), \((\sqrt{5}, -2)\), \((-\sqrt{5}, 2)\), \((-\sqrt{5}, -2)\)."

Step 1: Identify the System of Equations

We are given two equations: 1. \( x^2 + y^2 = 9 \) 2. \( x^2 - y^2 = 1 \). These equations form a system of simultaneous equations, where both must be true at the same time.

Step 2: Add the Equations

Add the two equations together:\[(x^2 + y^2) + (x^2 - y^2) = 9 + 1\]This simplifies to:\[2x^2 = 10\]Divide both sides by 2 to solve for \( x^2 \):\[x^2 = 5\]

Step 3: Substitute Back into the First Equation

Now that we have \( x^2 = 5 \), substitute this into the first equation:\[x^2 + y^2 = 9\]Substitute \( x^2 = 5 \):\[5 + y^2 = 9\]Solve for \( y^2 \):\[y^2 = 4\]

Step 4: Solve for x and y

From \( x^2 = 5 \), we have:\[x = \pm \sqrt{5}\]From \( y^2 = 4 \), we have:\[y = \pm 2\]This gives us four possible solution pairs.

Step 5: Determine All Solution Pairs

Combine the possible values for \( x \) and \( y \):1. \( (x, y) = (\sqrt{5}, 2) \)2. \( (x, y) = (\sqrt{5}, -2) \)3. \( (x, y) = (-\sqrt{5}, 2) \)4. \( (x, y) = (-\sqrt{5}, -2) \)Check these pairs against the original equations to verify they satisfy both equations.

Key Concepts

Simultaneous Equations

Simultaneous equations are a set of equations that are solved together because they share the same variables. In this exercise, we deal with two equations that need to be satisfied simultaneously:

• First Equation: \( x^2 + y^2 = 9 \)
• Second Equation: \( x^2 - y^2 = 1 \)

Simultaneous equations typically involve two or more equations and their solutions must make each of these equations true at the same time. That's why they are called "simultaneous."

To solve simultaneous equations, we look for values of the unknowns (in this case, \( x \) and \( y \)) that satisfy all equations involved. Various methods can be used to solve them, such as substitution, elimination, or graphical methods. In the given solution, the elimination method was used by adding and manipulating the given equations.

Solution Pairs

Solution pairs are combinations of variable values that satisfy all given equations in a system. For the system in our exercise, each 'pair' consists of specific values for \( x \) and \( y \) that make both equations true. Once we find values of \( x \) and \( y \), we form solution pairs:

• For \( x^2 = 5 \), \( x \) could be either \( \sqrt{5} \) or \( -\sqrt{5} \).
• For \( y^2 = 4 \), \( y \) could be either 2 or -2.

By combining these possibilities, we find four potential solution pairs:

• \( (x, y) = (\sqrt{5}, 2) \)
• \( (x, y) = (\sqrt{5}, -2) \)
• \( (x, y) = (-\sqrt{5}, 2) \)
• \( (x, y) = (-\sqrt{5}, -2) \)

It is crucial to verify each pair within the original system of equations to ensure it satisfies both equations, confirming they are indeed solutions within the system.

Algebraic Manipulation

Algebraic manipulation involves using algebraic techniques to simplify and solve equations or systems of equations. In this exercise, algebraic manipulation helps isolate terms and find numerical solutions.Initially, adding the two given equations eliminates the \( y^2 \) term, simplifying the system. By adding the equation \( x^2 + y^2 = 9 \) to \( x^2 - y^2 = 1 \), we get:\[2x^2 = 10\] This step results in a simpler equation that's easier to manage.Further manipulation involves dividing both sides by 2, hence finding \( x^2 = 5 \). From here, substituting \( x^2 = 5 \) back into the original equation \( x^2 + y^2 = 9 \) allows us to solve for \( y^2 \). Solving for these individual terms helps find values for both \( x \) and \( y \), leading us to the solution pairs.

Algebraic manipulation is foundational in handling different mathematical problems, as it allows us to break down complex problems into simple, solvable forms, much like piecing together a puzzle.