Problem 29 - Algebra and Trigonometry

Question

\(19-32\) . 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

The solutions are \((3, -\frac{1}{2})\) and \((-3, -\frac{1}{2})\).

Step 1: Simplify the second equation

Start by simplifying the second equation. The equation is \(4x^2 + 16y^3 = 34\). We can factor out a 2: \(2(2x^2 + 8y^3) = 34\) which simplifies to \(2x^2 + 8y^3 = 17\) by dividing both sides by 2.

Step 2: Set up a system of equations

Now we have a new system of equations: \(2x^2 - 8y^3 = 19\) and \(2x^2 + 8y^3 = 17\). Let's rewrite them for clarity: \(a - b = 19\) and \(a + b = 17\), where \(a = 2x^2\) and \(b = 8y^3\).

Step 3: Solve for \(a\) and \(b\)

Add the two equations to eliminate \(b\): \((a - b) + (a + b) = 19 + 17\). This simplifies to \(2a = 36\), so \(a = 18\). Substitute \(a = 18\) back into one of the equations, \(a + b = 17\): \(18 + b = 17\), which simplifies to \(b = -1\).

Step 4: Solve for \(x\) and \(y\)

Now substitute back \(a = 2x^2 = 18\) to solve for \(x\): \(x^2 = 9\), giving \(x = 3\) or \(x = -3\). For \(b = 8y^3 = -1\), solve for \(y\): \(y^3 = -\frac{1}{8}\), giving \(y = -\frac{1}{2}\).

Step 5: Find all solutions

Since \(x\) can be both 3 and -3, while \(y\) is the only value of -\(\frac{1}{2}\), the solutions are \((3, -\frac{1}{2})\) and \((-3, -\frac{1}{2})\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically expressed in the standard form:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are coefficients, and \(x\) represents the unknown variable. In the context of our given problem, we formed quadratic expressions when we isolated \(x\) as part of solving the system of equations. Quadratic equations can have two potential solutions, which can be found using methods such as factoring, completing the square, or applying the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In the exercise, after setting \(2x^2 = a\), we eventually solved for \(x^2 = 9\), leading to the solutions \(x = 3\) and \(x = -3\). These solutions reflect the two values where the quadratic expression equals zero.

Cubic Equations

Cubic equations are polynomial equations of degree three, characterized by the standard form:

• \(ax^3 + bx^2 + cx + d = 0\)

In the exercise at hand, we confronted the cubic nature while solving for \(y\) in the expression \(8y^3 = b\). Unlike quadratic equations that typically offer two solutions, cubic equations may present up to three real solutions, depending on the nature of their discriminants. Cubic equations can particularly be solved using methods like factorization, synthetic division, or more advanced techniques like Cardano's formula.
For this problem, solving \(y^3 = -\frac{1}{8}\) yielded a single real solution of \(y = -\frac{1}{2}\) through the decomposition of the cube root. Understanding cubic equations is crucial when dealing with polynomial expressions in systems of equations.

Solution of Equations

When working with systems of equations, the aim is to find a set of values for the variables that satisfy all equations simultaneously. In a system involving more than one equation, substitution or elimination methods can be effectively employed. The solution process usually involves:

• Simplifying each equation where possible.
• Setting up a relationship or pattern between the equations.
• Solving for one variable at a time.

In this example, the solution was achieved by simplifying both equations to form a system where similar terms canceled each other out. By doing so, it became possible to isolate and solve \(a = 2x^2\) and \(b = 8y^3\) independently. This approach gradually broke down a seemingly complex system into simpler steps, ensuring clarity while finding all possible solutions for the variables \(x\) and \(y\) in our original equations.