Problem 11 - Precalculus Mathematics for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{aligned} 3 x^{2}-y^{2} &=11 \\ x^{2}+4 y^{2} &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Solutions are \((2, 1)\), \((2, -1)\), \((-2, 1)\), and \((-2, -1)\).

Step 1: Analyze the system of equations

The given system consists of two equations: \(3x^2 - y^2 = 11\) and \(x^2 + 4y^2 = 8\). We will use the elimination method to solve them by eliminating one variable.

Step 2: Multiply equations to prepare for elimination

To eliminate one variable, first align the equations. Multiply the first equation by 4 to help eliminate \(y^2\). The equations become: \(12x^2 - 4y^2 = 44\) and \(x^2 + 4y^2 = 8\).

Step 3: Add the two equations

Add the modified equations: \(12x^2 - 4y^2 + x^2 + 4y^2 = 44 + 8\). This simplifies to \(13x^2 = 52\).

Step 4: Solve for \(x^2\)

The equation \(13x^2 = 52\) can be simplified to \(x^2 = 4\) by dividing both sides by 13.

Step 5: Find values of \(x\)

Since \(x^2 = 4\), the values for \(x\) are \(x = 2\) and \(x = -2\).

Step 6: Substitute back to solve for \(y\) when \(x = 2\)

Substitute \(x = 2\) into the second original equation: \(4 + 4y^2 = 8\), simplify to get \(4y^2 = 4\), and solve to find \(y^2 = 1\). Thus, \(y = 1\) or \(y = -1\).

Step 7: Substitute back to solve for \(y\) when \(x = -2\)

Substitute \(x = -2\) into the second original equation: \(4 + 4y^2 = 8\), simplify to get \(4y^2 = 4\), and solve to find \(y^2 = 1\). Thus, \(y = 1\) or \(y = -1\).

Step 8: Verify solutions in the first equation

Check \((2, 1)\), \((2, -1)\), \((-2, 1)\), \((-2, -1)\) using \(3x^2 - y^2 = 11\). Each satisfies the equation, confirming they are the actual solutions.

Key Concepts

System of Equations

When dealing with a system of equations, you are trying to find sets of variables that satisfy all equations involved. In this case, we have a system of two quadratic equations, which are equations that involve terms with variables raised to the power of 2. The system is:

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

To solve such systems, we can rely on methods like substitution or elimination. The elimination method is particularly useful here, as it allows us to efficiently remove one variable and simplify the solution process. By following a structured approach, we can identify points where both equations intersect, indicating the solutions that satisfy both equations simultaneously.

Quadratic Equations

Quadratic equations are an essential part of algebra and these equations can have various curve shapes called parabolas when graphed. In our exercise, each equation in the system is quadratic, involving terms like \(x^2\) and \(y^2\).
Quadratic equations generally appear as \(ax^2 + bx + c = 0\), but in systems of equations, they can display in different arrangements, like in:

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

These equations are solved by methods such as factoring, completing the square, or using the quadratic formula. However, in our current context, we leverage the elimination method to handle these quadratics efficiently as part of a system. This involves algebraic manipulations, such as aligning coefficients, to simplify the variables and ultimately find the solutions.

Solving Equations

The process of solving equations involves finding all possible values for the variables that make the equations true. In this exercise, we are dealing with the elimination method as a strategic approach to solve our given system.
Here's how the elimination method works in practice:

• Multiply one or both equations by necessary values to align the coefficients for one of the variables.
• Add or subtract the equations to eliminate one variable. This leaves you with a simpler equation.
• Solve the remaining equation to get a value for one variable.
• Substitute that value back into one of the original equations to find the other variable's value.

For example, we first aligned the coefficient of \(y^2\) across our equations and combined them, enabling us to solve for \(x^2\). Once we had \(x^2 = 4\), we determined \(x = 2\) and \(x = -2\). Then we substituted these back to find corresponding \(y\) values, completing the solution to our system.