Problem 73 - College Algebra with Modeling and Visualization

Question

Solve the system of equations. $$ \begin{aligned} &\frac{x^{2}}{2}+\frac{y^{2}}{4}=1\\\ &\frac{x^{2}}{4}+\frac{y^{2}}{2}=1 \end{aligned} $$

Step-by-Step Solution

Answer

Solutions are (x, y) = (±2√3/3, ±2√3/3) and (±2√3/3, ∓2√3/3).

Step 1: Identify Equations

We have two equations to solve simultaneously: 1) \( \frac{x^{2}}{2} + \frac{y^{2}}{4} = 1 \) 2) \( \frac{x^{2}}{4} + \frac{y^{2}}{2} = 1 \).

Step 2: Multiply Equations to Eliminate Fractions

To eliminate fractions, first multiply Equation (1) by 4 to get \(2x^2 + y^2 = 4\) and Equation (2) by 4 to get \(x^2 + 2y^2 = 4\).

Step 3: Subtract Equations

Subtract Equation (2) from Equation (1) to eliminate one of the variables: \((2x^2 + y^2) - (x^2 + 2y^2) = 4 - 4\). This simplifies to \(x^2 - y^2 = 0\).

Step 4: Solve for One Variable

From \(x^2 - y^2 = 0\), we get \(x^2 = y^2\). Thus, \(x = y\) or \(x = -y\).

Step 5: Substitute to Find Solutions

Substitute \(x = y\) into one of the original equations, say Equation (1): \(\frac{y^2}{2} + \frac{y^2}{4} = 1\). Simplifying gives \(\frac{3y^2}{4} = 1\) or \(y^2 = \frac{4}{3}\). Therefore, \(y = \pm \frac{2\sqrt{3}}{3}\). Hence, \(x = \pm \frac{2\sqrt{3}}{3}\).

Step 6: Check the Second Case

Now, substitute \(x = -y\): Use Equation (1): \(\frac{(-y)^2}{2} + \frac{y^2}{4} = 1\), resulting in the same equation: \(\frac{3y^2}{4} = 1\), yielding \(y^2 = \frac{4}{3}\), so \(x = \pm \frac{2\sqrt{3}}{3}\) and \(y = \mp \frac{2\sqrt{3}}{3}\).

Step 7: Conclude Solutions

Thus, the solutions are \(x = \frac{2\sqrt{3}}{3}, y = \frac{2\sqrt{3}}{3}\) or \(x = -\frac{2\sqrt{3}}{3}, y = -\frac{2\sqrt{3}}{3}\) for \(x = y\) and \(x = \frac{2\sqrt{3}}{3}, y = -\frac{2\sqrt{3}}{3}\) or \(x = -\frac{2\sqrt{3}}{3}, y = \frac{2\sqrt{3}}{3}\) for \(x = -y\).

Key Concepts

Substitution Method

When faced with solving a system of equations, the substitution method is one of the most straightforward techniques you can employ. The idea is to solve one of the equations for a single variable and then substitute that expression into the other equation. By doing this, you reduce the problem to a single equation with one variable, which is much simpler to solve.

Let's consider our exercise where we ended up with two reduced equations after performed multiplications on the original ones to clear fractions:

• Equation after simplification: \(2x^2 + y^2 = 4\)
• Equation after multiplication: \(x^2 + 2y^2 = 4\)

After subtracting these equations, we simplified our system to \(x^2 = y^2\). This is a perfect setup for substitution, where1 you can express \(x\) in terms of \(y\) (\(x = y\) or \(x = -y\)) and substitute this into one of the simplified equations. This reduces the complexity of our original system of two variables, turning it into a straightforward calculation with a single variable.

Solution of Equations

Solving systems of equations essentially means finding the values for the variables that satisfy all equations in the system simultaneously. In our exercise, we aimed to find both \(x\) and \(y\) that satisfy:

• \(\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1\)
• \(\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1\)

Our goal was not just to fiddle with numbers, but to carefully manipulate the equations in such a manner that both equations agree with the constants they equate to (which was 1 in both cases).

Ultimately, by employing algebraic techniques like multiplying to eliminate fractions and substituting one variable in terms of another, we unraveled the possible solutions of \(x = \pm \frac{2\sqrt{3}}{3}\) and \(y = \pm \frac{2\sqrt{3}}{3}\), verifying these solutions to ensure they satisfy the initial conditions of both equations. That's the essence of solving these equations.

Algebraic Manipulation

Algebraic manipulation is the art of working with equations to transform them into a more solvable state, often a necessary step in approaching a problem like our system of equations. The primary tools involve simplifying expressions, eliminating fractions, factoring, and distributing terms, each of which can bring even complex equations within reach.

In our example, we began by clearing fractions, so we scaled both equations:

• Multiply Equation 1: \(\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1\) by 4, resulting in: \(2x^2 + y^2 = 4\)
• Multiply Equation 2: \(\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1\) by 4, resulting in: \(x^2 + 2y^2 = 4\)

By doing this, not only were we able to work without fractions, but also to easily see relationships between terms by subtracting the equations.

This ultimately gave us an equation \(x^2 - y^2 = 0\), revealing that \(x\) is either equal to \(y\) or the negative of \(y\). Converting this recognition into clear pathways through substitution led to simple expressions we resolved accurately, illustrating algebraic manipulation's power in isolating unknowns.