Problem 54 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}+9 y^{2}=1 \\ x^{2}-9 y^{2}=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The system has no real solutions for \(x\) or \(y\).

Step 1: Add the equations

Let's add the two equations:\[(x^2 + 9y^2) + (x^2 - 9y^2) = 1 + 3\]Simplifying this, we get:\[2x^2 = 4\]Now, divide both sides by 2 to solve for \(x^2\):\[x^2 = 2\]

Step 2: Subtract the equations

Next, subtract the second equation from the first equation:\[(x^2 + 9y^2) - (x^2 - 9y^2) = 1 - 3\]This simplifies to:\[18y^2 = -2\]Now, divide both sides by 18:\[y^2 = -\frac{1}{9}\]Given that \(y^2\) must be non-negative for real values of \(y\), there are no real solutions for \(y\) in this case.

Key Concepts

Real Solutions

When we talk about real solutions in the context of a system of equations, we are referring to solutions that are valid within the realm of real numbers. Real numbers include all the rational and irrational numbers that can be found on the number line. However, they do not include imaginary numbers, which involve the square root of negative numbers.

In the exercise above, we aimed to find real values for the variables \(x\) and \(y\). This means we were looking for solutions that do not involve the imaginary unit \(i\), where \(i = \sqrt{-1}\). When solving systems of equations for real solutions, it's crucial to determine if all parts of the system can yield numbers that fit within the real number set.

As you saw in the solution, \(x^2 = 2\) is valid because it allows \(x\) to be the real numbers \(\pm \sqrt{2}\). However, when we calculated \(y^2 = -\frac{1}{9}\), it implied \(y\) would be an imaginary number, since you cannot take the square root of a negative number within the real number system. Thus, the exercise demonstrates how some systems may not have real solutions for specific variables.

Addition and Subtraction of Equations

Addition and subtraction of equations are fundamental operations in solving systems of equations. These operations help us to simplify the system and isolate variables, making the problem easier to solve.

In the original exercise, we first added the two equations. This operation effectively eliminated the terms involving \(y^2\) because they perfectly cancelled each other out. Here is a recap of the process:

• Original Equations: \(x^2 + 9y^2 = 1\) and \(x^2 - 9y^2 = 3\)
• Adding the equations: \((x^2 + 9y^2) + (x^2 - 9y^2) = 1 + 3\)
• Resulting Equation after Addition: \(2x^2 = 4\)

Next, subtraction was used to further simplify the system. By subtracting one equation from the other, we easily isolated the \(y^2\) term.

• Subtracting the equations: \((x^2 + 9y^2) - (x^2 - 9y^2) = 1 - 3\)
• Resulting Equation after Subtraction: \(18y^2 = -2\)

Both of these operations show the power of combining equations to eliminate variables and break down more complex expressions into simpler parts.

Algebraic Methods

Solving systems of equations often involves a variety of algebraic methods, enabling us to find the values of unknown variables. These methods, including substitution, elimination (or combination), and graphical solutions, often intertwine to tackle different scenarios.

For the given exercise, an algebraic method known as the elimination method was applied via addition and subtraction. This choice was particularly strategic because it directly simplified one of the equations to only involve \(x^2\), and another to show the impossibility of a real \(y^2\) solution:

The elimination method helps in:

• Canceling out coefficients, thus reducing two-variable equations into single variable equations.
• Providing a clearer path to solve complex systems by systematically eliminating one variable at a time.

In this example, once we found \(x^2 = 2\), one part of the problem was solved for real solutions. The negative \(y^2\) outcome confirmed no real solutions for \(y\). This reflects how algebraic methods not only solve for real answers but also verify when solutions simply aren’t possible within the given constraints.