Problem 27 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} x^{2}-y^{2}=4 \\ x^{2}+y^{2}=12 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (2\sqrt{2}, 2), (-2\sqrt{2}, 2), (2\sqrt{2}, -2), (-2\sqrt{2}, -2)\).

Step 1: Solving the Second Equation for x²

From the equation \(x^2 + y^2 = 12\), we can isolate \(x^2\) by subtracting \(y^2\) from both sides, resulting in \(x^2 = 12 - y^2\).

Step 2: Substituting into the First Equation

Substitute the expression for \(x^2\) from Step 1 into the first equation \(x^2 - y^2 = 4\). This gives us \((12 - y^2) - y^2 = 4\).

Step 3: Simplifying the Equation

Simplify the equation obtained in Step 2: \(12 - y^2 - y^2 = 4\), which simplifies to \(12 - 2y^2 = 4\).

Step 4: Solving for y²

Rearrange the simplified equation: \(12 - 2y^2 = 4\). Subtract 12 from both sides to obtain \(-2y^2 = -8\). Then, divide both sides by -2 to find \(y^2 = 4\).

Step 5: Solving for y

Take the square root of both sides: \(y = ±2\). This gives us two possible values for \(y\): \(y = 2\) and \(y = -2\).

Step 6: Finding x for y = 2

Use \(y = 2\) in the expression \(x^2 = 12 - y^2\): \(x^2 = 12 - 2^2 = 12 - 4 = 8\). Taking the square root, \(x = ±√8 = ±2\sqrt{2}\).

Step 7: Finding x for y = -2

The case for \(y = -2\) is identical to \(y = 2\) because \(y^2 = (-2)^2 = 4\). Therefore, \(x = ±2\sqrt{2}\) as we found in the previous step.

Step 8: Listing the Solutions

From the previous steps, the solutions to the system are \((x, y) = (2\sqrt{2}, 2), (-2\sqrt{2}, 2), (2\sqrt{2}, -2), (-2\sqrt{2}, -2)\).

Key Concepts

System of Equations

A system of equations is composed of two or more equations that share one or more variables. To find a solution to a system of equations, we're looking for a set of variable values that satisfy all the equations simultaneously.
For example, in our problem, we have two equations:

• \(x^2 - y^2 = 4\)
• \(x^2 + y^2 = 12\)

The goal is to find the values of \(x\) and \(y\) that make both of these true at the same time. Various methods can solve such systems, but here we're focusing on the substitution method for simplicity and to show how each equation can help solve the other.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. They play an important role in algebra as they are a step up in complexity from linear equations.
In our case, both equations contain quadratic terms: \(x^2\) and \(y^2\). The presence of squares often indicates a need to solve a quadratic equation. However, not all problems immediately show the classic quadratic form.
You might need to rearrange and simplify before solving. For example, by recognizing that \(y^2 = 4\) from our steps, we directly worked with a simplified quadratic to find \(y = \pm 2\). This is a quick demonstration of identifying and solving quadratic expressions once they're isolated.

Algebraic Solutions

The process of finding algebraic solutions involves algebraic manipulation to solve equations expressively. Using substitution as in our exercise, we replace one variable's expression with another derived from a separate equation.
This systematic approach allows for eliminating variables and simplifying the process. We isolated \(x^2\) in the second equation by expressing it as \(x^2 = 12 - y^2\).
This expression then substituted back into the first equation, helping isolate \(y^2\). This led us to straightforwardly solve for \(y\) and then for \(x\) using basic algebraic steps. Through algebraic solutions, complex-looking systems become manageable and solvable.