Problem 29 - Precalculus: Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{aligned} 2 x^{2}-8 y^{3} &=19 \\ 4 x^{2}+16 y^{3} &=34 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Solutions: \((3, -\frac{1}{2})\) and \((-3, -\frac{1}{2})\).

Step 1: Identify the System of Equations

The given system of equations is:1) \( 2x^2 - 8y^3 = 19 \)2) \( 4x^2 + 16y^3 = 34 \).

Step 2: Simplify the Second Equation

Starting with the second equation \(4x^2 + 16y^3 = 34\), we simplify by dividing every term by 2. This gives us: \[ 2x^2 + 8y^3 = 17 \].

Step 3: Set Up for Elimination

We want to eliminate one of the variables, so let's subtract the first equation from the simplified second equation:\( (2x^2 + 8y^3) - (2x^2 - 8y^3) = 17 - 19 \).

Step 4: Solve for One Variable

Subtracting the equations results in:\[ 16y^3 = -2 \].To find \(y\), divide both sides by 16:\[ y^3 = -\frac{1}{8} \].

Step 5: Solve for y

Taking the cube root of both sides, we find:\[ y = -\frac{1}{2} \].

Step 6: Substitute Back to Find x

Using the value of \(y = -\frac{1}{2}\), substitute it back into the first equation:\( 2x^2 - 8(-\frac{1}{2})^3 = 19 \).Calculate \((-\frac{1}{2})^3 = -\frac{1}{8}\), so:\( 2x^2 + 1 = 19 \).

Step 7: Solve for x

Re-arrange the equation:\[ 2x^2 = 18 \],then divide by 2:\[ x^2 = 9 \].Taking the square root gives:\[ x = 3 \quad \text{or} \quad x = -3 \].

Step 8: Verify the Solutions

Check both solutions \((x, y) = (3, -\frac{1}{2})\) and \((-3, -\frac{1}{2})\) in both original equations:- For \((3, -\frac{1}{2})\): Both equations satisfy.- For \((-3, -\frac{1}{2})\): Both equations satisfy.Thus both solutions are correct.

Key Concepts

Elimination Method

The elimination method is a strategy used for solving systems of equations. It involves eliminating one of the variables, making it easier to solve for the other. In the example provided, the strategy was to simplify the equations and then eliminate the variables step by step. Here's a closer look:

• Firstly, simplify the second equation, which was originally more complex.
• Next, subtract one equation from the other. This helps eliminate one of the variables altogether.
• In this system, subtracting the simplified second equation from the first one eliminated the variable \(x\) and helped solve for \(y\).

By carefully setting up the system to eliminate variables, solving the remaining equation becomes much more straightforward.

Cube Root

Taking the cube root is an operation to find a number that, when multiplied by itself three times, gives the original number. In our problem, we arrived at the equation \(y^3 = -\frac{1}{8}\).

• To solve for \(y\), we need the cube root of both sides of the equation. The cube root of \(-\frac{1}{8}\) is \(-\frac{1}{2}\).
• For negative numbers, the cube root function is quite handy, as it maintains the negative sign. Unlike square roots, cube roots provide one real solution for every real input.
• This operation gives us the variable \(y\) directly, which can then be used in further calculations.

Knowing how to take cube roots simplifies solving algebraic equations, especially when dealing with cubic terms.

Substitution Method

The substitution method is another effective approach for solving systems of equations. It entails solving one of the equations for one variable and then substituting it into the other equation. Here's how it was implemented in this case:

• After finding \(y = -\frac{1}{2}\), this value is substituted back into one of the original equations.
• This substitution allows for simplifying the equation to isolate and solve for \(x\).
• The value of \(y\) simplifies the equation to a basic algebra problem, where the only task left is basic arithmetic to find \(x\).

Using the substitution method can often make complex problems much simpler, as it breaks the problem into smaller, manageable parts.

Solving Algebraic Equations

Solving algebraic equations is a fundamental skill in mathematics, involving finding values of variables that satisfy given equations. This system demonstrated several key steps:

• The preparation of equations, either by elimination or substitution, is a preliminary step that eases the process. Simplifying the algebra involved can often reveal the most direct path to a solution.
• Solving for one variable leads directly to it, while maintaining the integrity of the equation. Here, once \(y\) was determined, finding \(x\) became a straightforward calculation.
• Verification is a crucial final step. It involves checking calculated solutions in the original equations to ensure they satisfy the system.

By following structured steps, solving algebraic equations becomes a consistent process that reaches correct solutions efficiently.