Problem 25 - Precalculus: Functions and Graphs

Question

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

Step 1: Simplify the Equations

Let's start with the given system of equations:1. \( x^2 - y^2 = 4 \)2. \( x^2 + y^2 = 12 \)We can express \( y^2 \) from each equation. From the first equation, \( y^2 = x^2 - 4 \). And from the second equation, \( y^2 = 12 - x^2 \).

Step 2: Substitute and Equate

Since we have two expressions for \( y^2 \), we can set them equal to each other:\[ x^2 - 4 = 12 - x^2 \]

Step 3: Solve for \( x^2 \)

To solve for \( x^2 \), rearrange the equation:\[ 2x^2 = 16 \]Now, divide both sides by 2:\[ x^2 = 8 \]

Step 4: Find \( x \)

Take the square root of both sides to solve for \( x \):\[ x = \pm \sqrt{8} = \pm 2\sqrt{2} \]

Step 5: Solve for \( y \) Using \( x^2 \)

Substitute \( x^2 = 8 \) back into either original equation. Let's substitute into the second equation:\[ 8 + y^2 = 12 \]Now, solve for \( y^2 \):\[ y^2 = 4 \]Take the square root of both sides:\[ y = \pm 2 \]

Step 6: Identify Solutions

Now we match each solution for \( x \) with each value for \( y \):- If \( x = 2\sqrt{2} \), \( y = 2 \) and \( y = -2 \).- If \( x = -2\sqrt{2} \), \( y = 2 \) and \( y = -2 \).Therefore, the solutions are \( (2\sqrt{2}, 2) \), \( (2\sqrt{2}, -2) \), \( (-2\sqrt{2}, 2) \), and \( (-2\sqrt{2}, -2) \).

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations that are solved together, such that the solution is shared among them. In our exercise, we have:

• The first equation: \( x^2 - y^2 = 4 \)
• The second equation: \( x^2 + y^2 = 12 \)

To solve systems of equations, various methods such as substitution, elimination, or graphing can be used. In this context, the substitution method is chosen.
This involves solving one equation for one variable and substituting this expression into the other equation. The ultimate goal is to find values for \( x \) and \( y \) that satisfy both equations simultaneously.

Quadratic Equations

Quadratic equations are those which can be expressed in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Though, the equations in the exercise don't possess a linear term, they involve quadratic expressions (\( x^2 \) and \( y^2 \)), hence are quadratic in nature.

• Quadratics are solved through various methods like factoring, using the quadratic formula, or taking square roots for simple quadratics.
• Here, our focus is on the expression involving \( y^2 \), which is simplified first, allowing substitution and further manipulation to solve for \( x \) and \( y \).

Understanding quadratic equations' properties is crucial in solving such systems, as the presence of squares impacts the number and nature of potential solutions.

Square Roots

Square roots play an essential role in solving the given problem. When you have an equation like \( x^2 = 8 \), square roots are used to find the actual values for \( x \).

• The square root operation determines what number, when squared, would lead to the given value.
• For example, taking the square root of 8 gives \( \sqrt{8} \), which simplifies to \( 2\sqrt{2} \).

It's important to remember that square root operations typically yield two results: one positive and one negative. Thus, \( x = \pm 2\sqrt{2} \) represents the full solution set for \( x \), accounting for both positive and negative roots.

Solution Sets

A solution set is the collection of all possible solutions that satisfy a system of equations. For our exercise, solving the system reveals multiple solutions.

• From the quadratic nature of our equations and the calculations, there are four valid sets of solutions for \( (x, y) \).
• Each set corresponds to the different combinations of positive and negative values derived from both square roots.

In conclusion, the solution set consists of:

• \( (2\sqrt{2}, 2) \)
• \( (2\sqrt{2}, -2) \)
• \( (-2\sqrt{2}, 2) \)
• \( (-2\sqrt{2}, -2) \)

Understanding solution sets helps verify that all calculated solutions meet the conditions set by both equations in the system.