Problem 13 - Precalculus: Mathematical 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

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

Step 1: Understand the Equations

We have two equations: \(3x^2 - y^2 = 11\) and \(x^2 + 4y^2 = 8\). Our goal is to eliminate one variable to solve the system.

Step 2: Align Variables for Elimination

To align for elimination, rewrite the equations: Equation 1: \(3x^2 - y^2 = 11\) Equation 2: \(x^2 + 4y^2 = 8\). We aim to eliminate \(x^2\) by finding a common coefficient.

Step 3: Multiply to Create Matching Coefficients for x^2

Multiply Equation 1 by 1 and Equation 2 by 3 to align \(x^2\). This gives us:1. \(3x^2 - y^2 = 11\)2. \(3x^2 + 12y^2 = 24\).

Step 4: Subtract Equations to Eliminate x^2

Subtract Equation 1 from Equation 2:\( (3x^2 + 12y^2) - (3x^2 - y^2) = 24 - 11 \),which simplifies to \(13y^2 = 13\).

Step 5: Solve for y

Divide both sides by 13 to find \(y^2\):\(y^2 = 1\).Thus, \(y = \pm 1\).

Step 6: Substitute y Values to Solve for x

Use the value of \(y\) to find \(x\):1. Substitute \(y = 1\) into \(x^2 + 4y^2 = 8\): \(x^2 + 4(1)^2 = 8\) \(x^2 + 4 = 8\) \(x^2 = 4\) \(x = \pm 2\).2. Substitute \(y = -1\) into the same equation: \(x^2 + 4(-1)^2 = 8\) \(x^2 + 4 = 8\) \(x^2 = 4\) \(x = \pm 2\).

Step 7: Verify Solutions

Check solutions in both original equations:- For \((x, y) = (2, 1)\) and \((x, y) = (-2, 1)\): Both satisfy \(3x^2 - y^2 = 11\) and \(x^2 + 4y^2 = 8\).- For \((x, y) = (2, -1)\) and \((x, y) = (-2, -1)\): Both satisfy \(3x^2 - y^2 = 11\) and \(x^2 + 4y^2 = 8\).Thus, all solutions are valid.

Key Concepts

System of Equations

A **system of equations** involves finding values for variables that satisfy all given equations simultaneously. In our example, we are working with a system comprised of the two equations:

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

The goal is to find the values of both \(x\) and \(y\) that work for both equations at the same time. Approaching these problems requires strategic methods, like substitution or elimination, to reduce and simplify the equations. Knowing which method works best often depends on how the equations are structured. Understanding the concept of a system of equations is essential since it is the foundation for many real-world problems and higher-level mathematics.

Quadratic Equations

**Quadratic equations** have the form \(ax^2 + bx + c = 0\). In our system, we have quadratic terms in both equations, where the squares of \(x\) and \(y\) appear. Quadratic equations define parabolas when graphed, and finding their intersections often involves solving such systems of equations. In this exercise, we have:

• The term \(3x^2 - y^2\)
• The term \(x^2 + 4y^2\)

These terms are crucial because they influence the method we use to find solutions. Particularly, when equations involve squares, it can lead to multiple solutions since squares have both positive and negative roots. By understanding how quadratics work, especially in conjunction with other equation forms, students can better grasp how to find intersections and solve for unknowns.

Variables Elimination

The **elimination method** is a technique used to remove one variable so another can be solved directly. Here, we aimed to eliminate the \(x^2\) term from the system to focus on \(y\). This is accomplished by making the coefficients of \(x^2\) the same in both equations, allowing subtraction to cancel them out.In our exercise, this was achieved by the following steps:

• We first made the coefficient of \(x^2\) equal by multiplying the equations appropriately.
• Next, subtract the aligned equations, leaving an expression involving only \(y\).

After substitution, the resulting equation was easy to solve for \(y^2 = 1\), giving solutions \(y = \pm 1\). Then, substituting these back into one of the original equations provides solutions for \(x\). The beauty of elimination lies in its systematic approach, ultimately simplifying complex systems and removing extra dimensions (variables) one at a time. Mastering variables elimination empowers students to tackle larger and more intricate systems efficiently.