Problem 53 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}-2 y^{2}=6 \\ x^{2}+2 y^{2}=2 \end{array}\right. $$

Step-by-Step Solution

Answer

The system has no real solutions because \(y^2 = -1\) is not possible with real numbers.

Step 1: Write down the system of equations

The system of equations given is:\[ \begin{cases} x^{2}-2y^{2} = 6 \ x^{2}+2y^{2} = 2 \end{cases} \]

Step 2: Subtract the second equation from the first

We subtract the second equation from the first equation to eliminate \(x^2\):\[ (x^2 - 2y^2) - (x^2 + 2y^2) = 6 - 2 \]Simplifying gives us:\[ -4y^2 = 4 \]

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

Divide both sides of the equation \(-4y^2 = 4\) by \(-4\) to find \(y^2\):\[ y^2 = -1 \]

Step 4: Determine the possibility for \(y^2\)

Since \(y^2 = -1\), there is no real solution because the square of a real number cannot be negative.

Key Concepts

Real Number Solutions

In mathematics, especially when working with systems of equations, one of the key aspects we often look for is the existence of real number solutions. Real numbers are numbers that can be found on the number line, including both positive and negative integers, fractions, and irrational numbers. Essentially, if a number can be thought of without the imaginary unit "i" (which represents the square root of -1), it is considered a real number.

For a system of equations to have "real number solutions," the solutions must satisfy all equations in the system with real numbers. When examining the system given in the exercise, we used mathematical operations to find if such solutions exist.

• First, we determined that subtracting one of the equations from the other helps to simplify the system.
• This simplification revealed an equation where the square of a variable equaled a negative number.
• Since no real number squared results in a negative value, we concluded that there are no real number solutions to the system.

Thus, it’s important to understand the properties of real numbers to identify whether a system can be solved within this realm.

Algebraic Methods

Algebraic methods are powerful strategies used to manipulate equations and solve systems. They involve the use of operations such as addition, subtraction, multiplication, and division to isolate variables and simplify expressions.

In the given exercise, algebraic methods were employed to eliminate one of the variables in the system of equations. By subtracting the entire second equation from the first, we effectively removed the term for \(x^2\), focusing instead on the variable \(y\). This is a common technique known as the elimination method, which is highly effective when working with systems of equations.

• Elimination simplifies the system by reducing it to fewer variables.
• This often involves aligning coefficients before adding or subtracting equations.
• The goal is to isolate one variable at a time for straightforward solving.

By using these algebraic techniques, one can simplify complex systems, making it easier to determine whether solutions exist and, if so, what those solutions are.

Quadratic Equations

Quadratic equations are polynomials of degree two, generally written in the form \(ax^2 + bx + c = 0\). These types of equations frequently appear in systems like the one we worked on in the exercise.

Quadratics are unique because they can have zero, one, or two real solutions, depending on the discriminant. In the case of the exercise, the involvement of quadratic terms helped form a situation where one of the variables squaring led us to determine impossibility in real solutions.

• Each quadratic equation has a certain form based on the coefficients \(a\), \(b\), and \(c\).
• The discriminant \(b^2 - 4ac\) determines the nature of the roots (real or complex).
• When quadratic equations result in negative values for squared variables, it indicates no real solutions.

The power of quadratic equations in solving systems lies in their ability to define relationships between variables clearly, but it's crucial to recognize when these relationships lead to no feasible real results.