Problem 11 - Algebra and Trigonometry

Question

\(9-14\) . Use the elimination method to find all solutions of the system of equations. $$ \left\\{\begin{array}{c}{3 x^{2}-y^{2}=11} \\ {x^{2}+4 y^{2}=8}\end{array}\right. $$

Step-by-Step Solution

Answer

The solutions are (2, 1), (2, -1), (-2, 1), (-2, -1).

Step 1: Rearrange and Multiply the First Equation

We start with the first equation: \(3x^2 - y^2 = 11\). Let's express it as \(y^2 = 3x^2 - 11\). We will multiply by 4 to match the coefficient of \(y^2\) in the second equation, thus creating \(4y^2 = 12x^2 - 44\).

Step 2: Set Equations of y² Equal

Our second equation is \(x^2 + 4y^2 = 8\), or \(4y^2 = 8 - x^2\). We can now set our two expressions for \(4y^2\) from Step 1 and this step equal to each other: \(12x^2 - 44 = 8 - x^2\).

Step 3: Solve for x²

Simplify the equation: \(12x^2 - 44 = 8 - x^2\). Add \(x^2\) to both sides to get \(13x^2 = 52\). Divide both sides by 13 to isolate \(x^2\): \(x^2 = 4\).

Step 4: Calculate x

Taking the square root of both sides, we find \(x = 2\) or \(x = -2\).

Step 5: Substitute x Back to Find y²

Substitute \(x = 2\) back into the expression \(4y^2 = 8 - x^2\) to find \(4y^2 = 8 - 4 = 4\), so \(y^2 = 1\).Alternatively, substitute \(x = -2\) and use the same method to find \(y^2 = 1\).

Step 6: Calculate y

Taking the square root of both sides, we find \(y = 1\) or \(y = -1\) for both values of \(x\).

Step 7: Find Solutions

Combine the values of \(x\) and \(y\) to get the solution pairs: \((x, y) = (2, 1), (2, -1), (-2, 1), (-2, -1)\).

Key Concepts

System of Equations

When dealing with equations in mathematics, a **system of equations** refers to a collection of two or more equations that share common variables. In our scenario, we have two different equations, each involving the variables \(x\) and \(y\). The equations are:

• \(3x^2 - y^2 = 11\)
• \(x^2 + 4y^2 = 8\)

The main goal here is to find a set of values for \(x\) and \(y\) that satisfy both equations simultaneously. This means finding the common solutions that work in both. Often, solving such systems leverages a method, like elimination or substitution, to systematically find these values.
Understanding the interdependency between these equations is crucial. By observing changes in one equation, we can infer information about the other, and thus work towards the final solution.

Solve for Variables

To **solve for variables** in a system of equations, we need to manipulate the equations to isolate one variable at a time, making them easier to solve. In this exercise, we are working with the variables \(x\) and \(y\).
Initially, we focus on one of the equations to isolate one variable. For example, we rearrange the equation \(3x^2 - y^2 = 11\) to express \(y^2\) as \(y^2 = 3x^2 - 11\). This rearrangement helps in aligning it with the second equation, \(x^2 + 4y^2 = 8\). By doing so, we make the process of solving for both \(x\) and \(y\) more straightforward.
This method involves logical reasoning and algebraic manipulation, and it's a fundamental skill for tackling more complex equations.

Substitution Method

The **substitution method** is a technique used to solve a system of equations by expressing one variable in terms of the other, and substituting that expression into the other equation. It's particularly useful when dealing with non-linear equations, as in our case.
From the exercise, we took \(y^2 = 3x^2 - 11\) and calculated a suitable expression for \(4y^2\), i.e., \(4y^2 = 12x^2 - 44\). In parallel, we also derived \(4y^2 = 8 - x^2\) from the second equation. Setting these two expressions equal allowed us to eliminate \(y^2\) and solve for \(x^2\).
The substitution method simplifies the complexity by focusing on fewer variables at a time, making calculations more manageable and accurate.

Quadratic Equations

**Quadratic equations** are polynomial equations of the second degree, typically in the form \(ax^2 + bx + c = 0\). They can have two solutions, which may be real or complex numbers.
In this exercise, we ended up with quadratic equations when we isolated \(x^2\). Specifically, after setting \(12x^2 - 44 = 8 - x^2\), we simplified it to \(13x^2 = 52\), leading to the solutions \(x = 2\) and \(x = -2\) after taking the square root.
Solving for \(y\) then also involved dealing with squared terms. Using the quadratic approach enables not just finding the required values but understanding the nature of solutions, such as real versus imaginary or shared versus distinct solutions among a set of equations.