Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{l} \frac{1}{x^{4}}+\frac{6}{y^{4}}=6 \\ \frac{2}{x^{4}}-\frac{2}{y^{4}}=19 \end{array}\right. $$

Short answer

The system has no real solution.

Step-by-step solution

- Substitute variables

Let \( a = \frac{1}{x^4} \) and \( b = \frac{1}{y^4} \). Then the system of equations becomes: \( a + 6b = 6 \) \( 2a - 2b = 19 \).

- Solve the second equation for one variable

Rearrange the second equation to solve for \( a \): \( 2a - 2b = 19 \) Divide by 2: \( a - b = 9.5 \) So, \( a = 9.5 + b \).

- Substitute into the first equation

Substitute \( a = 9.5 + b \) into the first equation: \( (9.5 + b) + 6b = 6 \) Combine like terms: \( 9.5 + 7b = 6 \) Solve for \( b \): \( 7b = -3.5 \) \( b = -0.5 \).

- Find the value of the other variable

Now, substitute \( b = -0.5 \) back into \( a = 9.5 + b \): \( a = 9.5 - 0.5 \) \( a = 9 \).

- Back-substitute to find \( x \) and \( y \)

Recall \( a = \frac{1}{x^4} \) and \( b = \frac{1}{y^4} \), so: \( \frac{1}{x^4} = 9 \) \( x^4 = \frac{1}{9} \) Taking the fourth root: \( x = \frac{1}{\root{4} \two 9} \). Similarly, \( \frac{1}{y^4} = -0.5 \) does not have a real solution, confirming that the initial value leads to contradiction.

Key concepts

Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This approach is particularly useful when dealing with systems of equations. For example, in our given system:

• We first substitute the values: let \( a = \frac{1}{x^4} \) and \( b = \frac{1}{y^4} \).
• This transforms our original equations into simpler ones: \( a + 6b = 6 \) and \( 2a - 2b = 19 \).

By solving these simpler equations step by step, we can find the values of \( a \) and \( b \), and eventually back-calculate the original variables \( x \) and \( y \). The substitution method helps break down a complex problem into more manageable parts, making it easier to find a solution.

Algebraic Manipulation

Algebraic manipulation is a crucial mathematical skill for solving equations. It involves rearranging equations, combining like terms, and performing operations to isolate variables. In our current exercise, we see algebraic manipulation in a few key steps:

• Rearranging the second equation: \( 2a - 2b = 19 \)
• Dividing by 2 to simplify: \( a - b = 9.5 \)
• Substituting \( a \) from one equation into the other: \( a = 9.5 + b \)
• Combining like terms and solving for \( b \): \( 9.5 + 7b = 6 \) leading to \( b = -0.5 \).

These algebraic steps help transform and simplify the equations, allowing us to find the unknowns. Good algebraic manipulation can make solving even difficult systems of equations straightforward.

Fourth Root

When solving for variables raised to a power, understanding roots is important. Specifically, we use the fourth root to solve for \( x \) and \( y \) in our exercise. Here's how this works:

• We had the expression \( \frac{1}{x^4} = 9 \), which simplifies to \( x^4 = \frac{1}{9} \).
• To solve for \( x \), we take the fourth root on both sides: \( x = \frac{1}{\sqrt[4]{9}} \).

Taking roots helps us reverse the power operation and find the actual values of the variables. Remember, the fourth root of a number \( a \) is the same as \( a^{1/4} \). For example, the fourth root of 16 is 2, as \( 2^4 = 16 \).

Non-Real Solutions

Sometimes, equations can lead to non-real solutions, especially when dealing with roots of negative numbers. In our exercise, we come across this when solving for \( y \):

• We find \( \frac{1}{y^4} = -0.5 \).
• Since no real number raised to the fourth power results in a negative value, this indicates a non-real solution.

Non-real, or complex, solutions involve imaginary numbers, typically denoted as \( i \) which equals \( \sqrt{-1} \). In real-world contexts, these solutions often represent constraints or scenarios outside usual physical reality. When solving equations, encountering such scenarios indicates the need to check the initial problem setup or rethink the solution approach, ensuring it aligns with real or feasible conditions.