Problem 27 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

There are no real solutions for the given system of equations.

Step 1: Understand the System of Equations

We have two equations: \( x^2 + 3y^2 = 6 \) and \( x^2 - 3y^2 = 10 \). Our goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Use the Elimination Method

Let's subtract the second equation from the first. \[(x^2 + 3y^2) - (x^2 - 3y^2) = 6 - 10\]This simplifies to \( 6y^2 = -4 \).

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

Divide both sides of the equation \( 6y^2 = -4 \) by 6 to solve for \( y^2 \).\[y^2 = -\frac{2}{3}\]

Step 4: Analyze the Result

The equation \( y^2 = -\frac{2}{3} \) suggests that \( y^2 \) is negative, which means there are no real numbers \( y \) that satisfy this condition, as the square of a real number cannot be negative.

Step 5: Conclusion

Since \( y^2 \) requires an imaginary solution, there are no real solutions for the given system of equations. The equations cannot be satisfied with real number values for both \( x \) and \( y \).

Key Concepts

Elimination Method

The elimination method is a powerful tool for solving systems of equations, especially when dealing with nonlinear systems. This method involves combining equations in a way that allows us to eliminate one of the variables. In our original exercise, we used the elimination method to deal with two equations \( x^2 + 3y^2 = 6 \) and \( x^2 - 3y^2 = 10 \).
By subtracting the second equation from the first, we eliminate the \( x^2 \) variable, which leaves us with an equation only in terms of \( y^2 \):

• \( (x^2 + 3y^2) - (x^2 - 3y^2) = 6 - 10 \) simplifies to \( 6y^2 = -4 \).

This simplification is vital as it transforms the nonlinear system into a more manageable form to solve.
It's important to choose the correct method, such as elimination, to simplify the problem and find solutions effectively.

Real Solutions

Understanding real solutions is crucial when solving any system of equations. A real solution refers to the set of solutions where all involved numbers are real numbers. In the original exercise, we aimed at finding real solutions for the two equations.
From the elimination method, we derived \( 6y^2 = -4 \) which simplifies to \( y^2 = -\frac{2}{3} \).
Since \( y^2 \) resulted in a negative number, it immediately hints at a problem for real solutions since:

• The square of a real number is never negative.
• Square roots of negative numbers are not real, thus indicating the absence of real solutions in this scenario.

Given this result, our original system of equations ends without real solutions, leading us towards exploring other types of solutions in mathematics.

Imaginary Numbers

Imaginary numbers arise when dealing with equations where real solutions are not found, especially in situations involving negative square roots. In the given problem, arriving at \( y^2 = -\frac{2}{3} \) signaled the need for imaginary numbers.
Imaginary numbers such as \( i \), where \( i = \sqrt{-1} \), help us express solutions for square roots of negative numbers.
The fact that \( y^2 = -\frac{2}{3} \) suggests using imaginary numbers to represent \( y \), because:

• The solution involves \( \sqrt{-\frac{2}{3}} = i \times \sqrt{\frac{2}{3}} \).
• This acknowledges the need for extending beyond real numbers to satisfy the equation condition.

Through this understanding, imaginary numbers become indispensable in analyzing systems where real solutions are impossible.