Problem 51 - College Algebra

Question

Find the solutions to the nonlinear equations with two variables. $$ \begin{array}{r} x^{2}-x y-2 y^{2}-6=0 \\ x^{2}+y^{2}=1 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \((x = -1, y = 0)\) or \((x, y)\).

Step 1: Analyze the Given Equations

We have two equations: 1) \( x^2 - xy - 2y^2 - 6 = 0 \) and 2) \( x^2 + y^2 = 1 \). The goal is to solve this system of equations for the variables \( x \) and \( y \).

Step 2: Express One Variable in Terms of Another

From equation (2), \( x^2 + y^2 = 1 \), we can isolate \( x^2 \) to get \( x^2 = 1 - y^2 \). We will use this expression to substitute in equation (1).

Step 3: Substitute to Eliminate One Variable

Substitute \( x^2 = 1 - y^2 \) into equation (1): \( (1 - y^2) - xy - 2y^2 - 6 = 0 \). Simplifying gives \( 1 - y^2 - xy - 2y^2 - 6 = 0 \), or \( - xy - 3y^2 - 5 = 0 \).

Step 4: Solve for One Variable

We now have the equation \( - xy - 3y^2 - 5 = 0 \). We can rearrange this to \( xy = -3y^2 - 5 \). Express \( x \) in terms of \( y \): \( x = -\frac{3y^2 + 5}{y} \).

Step 5: Substitute Back into the Circle Equation

Substitute \( x = -\frac{3y^2 + 5}{y} \) into the equation \( x^2 + y^2 = 1 \). It becomes \( \left(-\frac{3y^2 + 5}{y}\right)^2 + y^2 = 1 \).

Step 6: Solve the Equation for 'y'

Simplify \( \left(-\frac{3y^2 + 5}{y}\right)^2 \) to get \( \frac{(3y^2 + 5)^2}{y^2} \). The equation becomes \( \frac{(3y^2 + 5)^2}{y^2} + y^2 = 1 \). Multiply through by \( y^2 \) to avoid fractions, then solve the resulting polynomial for \( y \).

Step 7: Solve Polynomial Equation

Solve the algebraic equation to find possible values of \( y \), which might involve factoring or using the quadratic formula. Substitute these values back into one of the original equations to find corresponding \( x \) values.

Step 8: Check Solutions

Verify each solution pair \((x, y)\) by substituting back into the original equations to ensure they satisfy both.

Key Concepts

System of Equations

In mathematics, a system of equations is a collection of two or more equations with the same set of variables. In this exercise, we dealt with a system of two nonlinear equations:

• \(x^2 - xy - 2y^2 - 6 = 0\)
• \(x^2 + y^2 = 1\)

The goal is to find values for the variables, \(x\) and \(y\), that simultaneously satisfy both equations.

Nonlinear equations include terms like squares or products of variables, making them less straightforward to solve than linear systems.
Because the equations might graph as curves, the solution might contain multiple points or even no real solutions. For this reason, handling these equations requires patience and a structured approach.
By understanding systems of equations, students master solving problems where multiple constraints need to be met. This is a stepping stone to more complex math topics.

Substitution Method

The substitution method is a common technique to solve systems of equations. This method involves rearranging one equation to express one variable in terms of the others, and then replacing the variable in another equation.

In the exercise, we used the second equation \( x^2 + y^2 = 1 \) to express \( x^2 \) in terms of \( y^2 \):

• \(x^2 = 1 - y^2\)

We then substituted this expression into the first equation to eliminate \(x\):

• \((1 - y^2) - xy - 2y^2 - 6 = 0\)

This reduces the problem to a simpler equation with fewer variables.

The substitution method is particularly useful for nonlinear systems, as it can help break down complex equations into solvable components. By learning how to utilize substitution effectively, students can unlock opportunities to solve diverse mathematical challenges.

Polynomial Solving

Polynomial solving refers to finding the roots or solutions of polynomial equations. In this exercise, after substituting to remove \(x\), we arrived at an equation that needed polynomial solving:

• \(- xy - 3y^2 - 5 = 0 \)

This required expressing \(x\) in terms of \(y\) and further simplifying the result. Converting it into a more manageable form through simplification:

• \(x = -\frac{3y^2 + 5}{y}\)

Subsequently, substituting back into the circle equation, you might face quadratic polynomial forms or higher degrees.

Approaches to solving include factoring, the quadratic formula, or other algebraic methods. Developing skills in this area brings an understanding of how polynomial functions behave, their graph shapes, and the profound role they play in diverse scientific fields. This understanding is foundational in advancing to calculus and other higher-level mathematics courses.

Variables

Variables are symbols used to represent unknown values in mathematical expressions and equations. They are fundamental to algebra, allowing us to formulate and solve equations in a general way rather than relying on specific numbers.
In our problem:

• \( x \) and \( y \) are variables representing potential solutions to the equations.

Variables can take on many values, depending on the constraints given by the equations. Here, they were subject to both a quadratic form and a circular constraint in our system.
Understanding how to manipulate variables through techniques like substitution allows us to explore relationships between them.
When students grasp the concept of variables, they learn how to apply these skills not only in algebra but also in real-world scenarios, such as physics and engineering, where variables are used to represent changing quantities.