Problem 28 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} 6 x^{3}-y^{3}=1 \\ 3 x^{3}+4 y^{3}=5 \end{array}\right. $$

Step-by-Step Solution

Answer

Solution: \( x = \sqrt[3]{\frac{1}{3}}, y = 1 \).

Step 1: Solve for one variable in terms of the other

Let's solve the first equation for \( y^3 \):\[6x^3 - y^3 = 1 \]Rearranging gives:\[y^3 = 6x^3 - 1\]

Step 2: Substitute the expression into the second equation

Substitute \( y^3 = 6x^3 - 1 \) into the second equation:\[3x^3 + 4(6x^3 - 1) = 5\]

Step 3: Simplify and solve for \( x^3 \)

Simplify the equation from Step 2:\[3x^3 + 24x^3 - 4 = 5\]Combine like terms:\[27x^3 - 4 = 5\]Add 4 to both sides:\[27x^3 = 9\]Divide by 27:\[x^3 = \frac{9}{27} = \frac{1}{3}\]

Step 4: Find the value of \( y^3 \)

Now substitute \( x^3 = \frac{1}{3} \) back into the expression for \( y^3 \):\[y^3 = 6\left(\frac{1}{3}\right) - 1\]Simplifying, we have:\[y^3 = 2 - 1 = 1\]

Step 5: Take the cube root to find \( x \) and \( y \)

Since \( x^3 = \frac{1}{3} \) and \( y^3 = 1 \), take the cube roots:\[x = \sqrt[3]{\frac{1}{3}}\]\[y = \sqrt[3]{1} = 1\]Thus, the solutions for \( x \) and \( y \) are real cube roots of these values.

Key Concepts

Solving Systems of Equations Using Substitution

The substitution method is a powerful technique to solve systems of equations, including those involving cubic terms. It allows us to express one variable in terms of another and eliminate variables step by step.
To begin, you solve one of the equations for one variable. Then, you substitute this expression into the other equation. This approach can simplify the solving process, especially when dealing with complex equations, such as cubic equations.
Using the given system:

• First, isolate a variable. In our case, we found \(y^3 = 6x^3 - 1\) from the first equation.
• Substitute the isolated variable into the other equation. This means replacing \(y^3\) in the second equation with \(6x^3 - 1\).
• Finally, solve for the remaining variable. Here, solving gave us \(x^3 = \frac{1}{3}\).

By using substitution, we effectively reduced the system to a single-variable equation, making it easier to handle. This method is particularly useful when dealing with nonlinear equations, such as cubes. It cuts down on the complexity by turning two equations into one.

Algebraic Manipulation to Simplify Expressions

Algebraic manipulation involves rearranging and combining terms to bring equations into a more workable form. When working with systems of equations, especially ones that include cubic terms, this is an essential skill.
In the discussed solution, after substitution, we performed algebraic manipulation to simplify the equation:

• First, substituting \(y^3 = 6x^3 - 1\) into the second equation gave us \(3x^3 + 4(6x^3 - 1) = 5\).
• Next, distribute the 4, resulting in \(3x^3 + 24x^3 - 4 = 5\).
• Combine like terms to get \(27x^3 - 4 = 5\).

Such simplifications are crucial as they transform the equation into a form where standard algebraic operations can be applied, such as adding, subtracting, and dividing, to isolate the variable. Without proper manipulation, solving these equations would be arduous.

Understanding and Solving Cubic Equations

Cubic equations are expressions where the highest degree is three, often taking the form \(ax^3 + bx^2 + cx + d = 0\). They can be more challenging than linear or quadratic equations due to their complexity.
In the system of equations we solved, each equation included cubic terms (\(x^3\) and \(y^3\)), making the use of the substitution method helpful to isolate these terms.
Upon simplifying the system, we obtained values for \(x^3\) and \(y^3\). To find the actual values of \(x\) and \(y\), take the cube roots:

• For \(x\), since \(x^3 = \frac{1}{3}\), find \(x = \sqrt[3]{\frac{1}{3}}\).
• Similarly for \(y\), as \(y^3 = 1\), thus \(y = \sqrt[3]{1} = 1\).

Cubic equations can also have one real root or three real roots, depending on their nature. Understanding cube roots and their properties is essential in solving such equations effectively.