Problem 27 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} x^{2}+3 y^{2}=6 \\ x^{2}-3 y^{2}=10 \end{array}\right. $$

Step-by-Step Solution

Answer

The system has no real solution, as \(y^2\) is negative.

Step 1: Subtract the Equations

First, subtract the second equation from the first equation to eliminate the \(x^2\) terms. This gives: \(x^2 + 3y^2 - (x^2 - 3y^2) = 6 - 10\). Simplify to get \(6y^2 = -4\).

Step 2: Solve for y

Divide both sides by 6 to solve for \(y^2\): \(y^2 = -\frac{2}{3}\). Since \(y^2\) is negative, which is not possible for real numbers, \(y\) has no real solutions.

Step 3: Re-evaluate the problem's context

Because \(y^2\) resulted in a negative number, indicating no real solution for \(y\), the nonlinear system doesn't have a solution in the set of real numbers. This means the problem might involve complex or imaginary solutions that are beyond the current scope.

Key Concepts

Complex Solutions

When we talk about complex solutions in mathematics, we are referring to solutions that involve complex numbers. Complex numbers come into play when an equation doesn't have a real number solution. In the given problem, when we solved for \( y^2 \) and found that it equals \(-\frac{2}{3}\), it indicated an impossibility in the realm of real numbers since squares of real numbers are always non-negative.

• Complex solutions involve numbers that have both a real part and an imaginary part.
• These are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.

To explore the complex solution for \(y\), we can use the imaginary unit \(i\), where \(i^2 = -1\). Thus, \(y^2 = -\frac{2}{3}\) implies that \( y = \pm \sqrt{-\frac{2}{3}} = \pm \sqrt{\frac{2}{3}}i \).
Recognizing when to consider complex numbers is crucial in mathematics, especially in systems where real numbers alone are insufficient.

Real Numbers

Real numbers are the familiar numbers we use every day – they include both rational and irrational numbers. In the context of solving equations, real numbers are essential for understanding when certain solutions are possible.

• Real numbers are numbers that can be located on the number line, including fractions and square roots of non-negative numbers.
• Equations like \( y^2 = -\frac{2}{3} \) don't provide real solutions as the square of a real number can't be negative.

In the given problem, after solving and reaching a negative result for \(y^2\), we determined no real solutions existed. This insight is valuable because it helps us identify problems that require exploring complex numbers instead of real ones.

Imaginary Numbers

Imaginary numbers extend our number system by allowing the square root of negative numbers. The core component of imaginary numbers is the imaginary unit \(i\), defined as the square root of \(-1\).

• Any real number multiplied by \(i\) becomes an imaginary number. For example, \(5i \) where \(i^2 = -1\).
• Imaginary numbers are useful for providing solutions in situations where real numbers fall short, like in the given nonlinear system.

In our problem, the realization that \(y^2 = -\frac{2}{3}\) leads us naturally to consider imaginary numbers. Thus, \(y\) can be expressed as \(\pm \sqrt{\frac{2}{3}}i \). Imaginary numbers often appear in engineering and physics, where they are essential for modeling oscillations and waves.

Solving Equations

Solving equations, especially nonlinear ones, is a primary task in mathematics. It involves finding values for unknowns that make the equations true. Nonlinear systems can provide unique challenges due to their complexity.

• The goal in solving is to isolate variables and simplify the equations.
• In the exercise, subtracting the equations helped to eliminate one variable, simplifying the system.

For complex systems like the one in this problem, finding innovative methods to combine and simplify equations is essential. By initially eliminating the \(x^2\) terms and working with \(y\), we streamlined the solution process. Understanding the steps and logic in equation solving builds skills that are widely applicable across mathematics.

Mathematics Education

Mathematics education helps students develop critical thinking and problem-solving skills by engaging with concepts like nonlinear systems and complex numbers. These challenges help students think analytically.

• Learning the theory behind concepts ensures students can apply them to various problems.
• Working through both real and complex solutions prepares students for more advanced studies.

Educators aim to present problems that stimulate curiosity and analytical thinking. By understanding both the need for and the application of complex solutions, students gain a deeper insight into the nature of mathematics. This comprehensive approach in education fosters confidence and adaptability in students as they encounter new mathematical challenges.