Problem 75 - A Graphical Approach to College Algebra

Question

Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} &x^{2}+x y+y^{2}=3\\\ &x^{2}-x y+y^{2}=1 \end{aligned}$$

Step-by-Step Solution

Answer

The real solutions are \((1, 1)\) and \((-1, -1)\).

Step 1: Subtract the Equations

Let's subtract the second equation from the first.\[ (x^2 + xy + y^2) - (x^2 - xy + y^2) = 3 - 1 \] Simplifying the left side by combining like terms gives us \[ 2xy = 2 \] Divide both sides by 2 to isolate the term \( xy \), which results in \[ xy = 1 \].

Step 2: Solve for \( y \) in Terms of \( x \)

From \( xy = 1 \), we can solve for \( y \) in terms of \( x \) by dividing by \( x \).\[ y = \frac{1}{x} \]

Step 3: Substitute \( y = \frac{1}{x} \) into First Equation

Substitute \( y = \frac{1}{x} \) into the first original equation \( x^2 + xy + y^2 = 3 \).Replacing \( y \), we have \[ x^2 + x \left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 3 \] This simplifies to \[ x^2 + 1 + \frac{1}{x^2} = 3 \]

Step 4: Multiply Through by \( x^2 \)

Multiply the entire equation by \( x^2 \) to eliminate the fraction.\[ x^4 + x^2 + 1 = 3x^2 \] Simplify by rearranging terms:\[ x^4 - 2x^2 + 1 = 0 \]

Step 5: Recognize the Structure

Notice that \( x^4 - 2x^2 + 1 \) can be seen as a quadratic in terms of \( z = x^2 \).\[ z^2 - 2z + 1 = 0 \]

Step 6: Factor the Quadratic

Factor the quadratic equation found in the previous step:\[ (z - 1)^2 = 0 \] Therefore, \( z = 1 \).

Step 7: Solve for \( x \) using \( z = x^2 = 1 \)

Since \( x^2 = 1 \), solve for \( x \): \[ x = \pm 1 \]

Step 8: Find Corresponding \( y \) Values

Using \( xy = 1 \), find corresponding \( y \) values:If \( x = 1 \): \[ y = \frac{1}{1} = 1 \]If \( x = -1 \): \[ y = \frac{1}{-1} = -1 \]

Step 9: Verify Solutions

Verify that both solutions satisfy the original equations:For \((x, y) = (1, 1)\):- First equation: \( 1^2 + 1 \times 1 + 1^2 = 3 \)- Second equation: \( 1^2 - 1 \times 1 + 1^2 = 1 \)For \((x, y) = (-1, -1)\):- First equation: \((-1)^2 + (-1)(-1) + (-1)^2 = 3 \)- Second equation: \((-1)^2 - (-1)(-1) + (-1)^2 = 1 \) Both solutions satisfy the system.

Key Concepts

Analytical Solution

Analytical solution refers to finding an exact solution of a system of equations, using algebraic techniques. Unlike numerical methods that provide approximate answers, analytical solutions yield precise, explicit results. Subtraction is a common technique used here. It helps to eliminate one of the variables by combining equations. For example, in our system of equations:

• The first equation is: \(x^2 + xy + y^2 = 3\)
• The second equation is: \(x^2 - xy + y^2 = 1\)

By subtracting the second equation from the first, we eliminate the \(x^2\) and \(y^2\) terms. This simplifies the system into an equation with a single variable: \(2xy = 2\). Such manipulations are crucial as they simplify complex systems, making it easier to find solutions by reducing them to more straightforward equations or expressions.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree, typically of the form \(ax^2 + bx + c = 0\). Solving a quadratic equation often involves finding the values of \(x\) that satisfy the equation. Quadratic equations can typically be solved through:

• Factoring
• Using the quadratic formula
• Completing the square

In this exercise, we identified a structure that resembles a quadratic when we expressed our equation in terms of \(z = x^2\). The equation \(z^2 - 2z + 1 = 0\) is a quadratic equation. Factoring this equation, we find \((z - 1)^2 = 0\). Therefore, \(z\) has only one solution: \(z = 1\). This step shows how understanding the quadratic structure helps to solve complex nonlinear equations. Once we establish \(z = x^2\), solving for \(x\) follows naturally as it's straightforward.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying algebraic expressions. This is a key skill when solving systems of equations. It involves operations like addition, subtraction, multiplication, division, and factoring, to isolate and solve for variables. In our problem, starting with simplifying terms using subtraction helped isolate the product \(xy = 1\). This is a simple form but required further manipulation. Substituting \(y = \frac{1}{x}\), another form of manipulation, transforms the system and allows us to eliminate fractions by multiplying through by \(x^2\). This leads to forming a polynomial equation, which is more straightforward to solve once expressed as a quadratic. This solution showcases algebraic manipulation's role in transitioning from one form to another, simplifying a complex system to a readily solvable structure. It's about recognizing structures within equations and strategically applying operations to reveal underlying relationships.