Problem 31 - Algebra and Trigonometry

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 \((\sqrt{5}, 2)\), \((\sqrt{5}, -2)\), \((-\sqrt{5}, 2)\), and \((-\sqrt{5}, -2)\)."

Step 1: Understand the System of Equations

The given system is comprised of two equations:1. The first equation is a circle equation: \(x^2 + y^2 = 9\),2. The second equation is a hyperbola equation: \(x^2 - y^2 = 1\).

Step 2: Solve by Adding the Equations

To simplify, add the two equations together:\[(x^2 + y^2) + (x^2 - y^2) = 9 + 1\]This simplifies to:\[2x^2 = 10\].

Step 3: Solve for x

From the equation \(2x^2 = 10\), divide both sides by 2 to get:\[x^2 = 5\].Taking the square root, \(x = \pm \sqrt{5}\).

Step 4: Solve for y by Substituting x

Substitute \(x = \pm \sqrt{5}\) back into the first equation, \(x^2 + y^2 = 9\):- If \(x = \sqrt{5}\), then \[5 + y^2 = 9\] \[y^2 = 4\] \[y = \pm 2\].- If \(x = -\sqrt{5}\), then the calculation is the same due to symmetry: \[y = \pm 2\].

Step 5: List All Solutions

We now compile all solutions:- For \(x = \sqrt{5}\): - \(y = 2\) gives the solution \((\sqrt{5}, 2)\) - \(y = -2\) gives the solution \((\sqrt{5}, -2)\)- For \(x = -\sqrt{5}\): - \(y = 2\) gives the solution \((-\sqrt{5}, 2)\) - \(y = -2\) gives the solution \((-\sqrt{5}, -2)\)

Key Concepts

Circle Equation

A circle is a shape where all points are equidistant from a single point known as the center. It can be described mathematically by an equation in the form \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle. In the given problem, the equation \(x^2 + y^2 = 9\) represents a circle centered at the origin \((0, 0)\) with a radius of 3 (since \(r^2 = 9\), so \(r = \sqrt{9} = 3\)).
This means that any point \((x, y)\) that satisfies this equation will lie exactly on this circular boundary.
Recognizing a circle equation helps understand the geometric implications of the solutions, as each solution pair \((x, y)\) will fall along this defined path.

Hyperbola Equation

A hyperbola is a type of conic section that forms an open curve consisting of two separate branches. The standard form of a hyperbola with a horizontal transverse axis is \(x^2/a^2 - y^2/b^2 = 1\). In this exercise, the equation \(x^2 - y^2 = 1\) describes a hyperbola centered at the origin. This can be rewritten as \(x^2/1^2 - y^2/1^2 = 1\), showing that the hyperbola opens horizontally.
A hyperbola differs from a circle in that points on a hyperbola are not equidistant from a center, but rather follow the difference in distances to two focal points. The solutions of this equation will satisfy this path and give insight into the shape's nature when intersected with another conic form, like a circle.

Solve for x and y

To solve a system of equations, we aim to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. In steps provided, after combining the two original equations, we simplify solving for \(x\) first by isolating terms.

• Add the equations: \((x^2 + y^2) + (x^2 - y^2) = 10\), resulting in the simplified form \(2x^2 = 10\).
• Solving for \(x\) gives \(x^2 = 5\), so \(x = \pm \sqrt{5}\).

Once \(x\) is found, substitute back into one of the original equations to find \(y\), allowing us complete solutions for both variables.
The process of solving each for its respective variable involves strategic substitution and algebraic manipulation, ensuring all potential results are covered, particularly noting symmetry which can simplify double checking results.

Adding and Subtracting Equations

Sometimes in systems of equations, adding or subtracting the equations is a powerful tool for simplification. In this exercise, adding was used effectively to eliminate the \(y^2\) terms which can give clarity to the path of solving.
By combining the equations \((x^2 + y^2) + (x^2 - y^2) = 10\), observe how the \(y^2\) terms cancel each other out, simplifying to \(2x^2 = 10\). This is beneficial when looking for straightforward variables.
Adding and subtracting equations help simplify complex problems by reducing them to manageable parts. It utilizes algebraic properties to streamline solving, especially useful when dealing with symmetrical conic sections like circles and hyperbolas that share characteristics in their equations.