Problem 50 - College Algebra

Question

Find the solutions to the nonlinear equations with two variables. $$ \begin{array}{r} x^{2}-x y+y^{2}-2=0 \\ x+3 y=4 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \( \left( \frac{5 - 3\sqrt{65}}{20}, \frac{25 + \sqrt{65}}{20} \right) \) and \( \left( \frac{5 + 3\sqrt{65}}{20}, \frac{25 - \sqrt{65}}{20} \right) \).

Step 1: Express one variable

From the second equation, express one variable in terms of the other. We can express \(x\) in terms of \(y\):\[ x = 4 - 3y \]

Step 2: Substitute the expression

Substitute \(x = 4 - 3y\) from Step 1 into the first equation:\[ (4 - 3y)^2 - (4 - 3y)y + y^2 - 2 = 0 \]

Step 3: Expand and simplify the equation

First, expand \((4 - 3y)^2 \):\[ (4 - 3y)^2 = 16 - 24y + 9y^2 \]Substitute back into the equation:\[ 16 - 24y + 9y^2 - (4 - 3y)y + y^2 - 2 = 0 \]Simplify the expression:\[ 9y^2 + y^2 - 24y -4y + 3y + 16 - 2 = 0 \] \[ 10y^2 - 25y + 14 = 0 \]

Step 4: Solve the quadratic equation

Solve the quadratic equation \( 10y^2 - 25y + 14 = 0 \) using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 10, b = -25, \text{ and } c = 14 \):\[ y = \frac{-(-25) \pm \sqrt{(-25)^2 - 4 \cdot 10 \cdot 14}}{2 \cdot 10} \]\[ y = \frac{25 \pm \sqrt{625 - 560}}{20} \]\[ y = \frac{25 \pm \sqrt{65}}{20} \]

Step 5: Calculate possible y-values

Calculate the possible solutions for \(y\):\[ y_1 = \frac{25 + \sqrt{65}}{20} \]\[ y_2 = \frac{25 - \sqrt{65}}{20} \]

Step 6: Find corresponding x-values

Now find the corresponding \(x\) values for each \(y\) value using \(x = 4 - 3y\):- For \(y_1 = \frac{25+\sqrt{65}}{20}\),\[ x = 4 - 3\left(\frac{25 + \sqrt{65}}{20}\right) \]- For \(y_2 = \frac{25-\sqrt{65}}{20}\),\[ x = 4 - 3\left(\frac{25 - \sqrt{65}}{20}\right) \]

Step 7: Simplify x-values

Perform the calculations:- For \(y_1\): \[ x_1 = \frac{80 - 75 - 3\sqrt{65}}{20} = \frac{5 - 3\sqrt{65}}{20} \]- For \(y_2\): \[ x_2 = \frac{80 - 75 + 3\sqrt{65}}{20} = \frac{5 + 3\sqrt{65}}{20} \]

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations, which are in the form of \( ax^2 + bx + c = 0 \). It provides an easy way to find the values of the unknown variable "x" when the equation is set to zero. The quadratic formula itself is written as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula calculates the roots of the equation, showing where the parabola represented by the quadratic equation crosses the x-axis.

• The term \( b^2 - 4ac \) is called the discriminant.
• If the discriminant is positive, there are two distinct real roots.
• If zero, there is exactly one real root (the parabola touches the x-axis).
• If negative, the roots are complex, and they do not intersect the x-axis in the real plane.

In the given problem, the quadratic formula helps determine the values of "y" once substituted back into the related equation.

substitution method

The substitution method is an algebraic technique used to solve systems of equations, especially when dealing with nonlinear equations or equations with several variables. In simple terms, it involves expressing one variable in terms of another and then substituting this expression into another equation.
This allows you to simplify the system into a single equation with just one variable. Here's a breakdown of the method applied to the given system:

• Start by choosing one of the equations and express one variable in terms of the other.
• Substitute this expression into the other equation(s).
• Solve the resulting single-variable equation to find one variable's value.
• Back-substitute this value into one of the original expressions to find the other variable.

The goal of substitution is to reduce complexity and solve for both variables step by step.

system of equations

A system of equations is a set of two or more equations with the same set of variables. The main goal is to find the variable values that satisfy all equations simultaneously. These systems can be linear or nonlinear:

• Linear systems consist entirely of first-order equations.
• Nonlinear systems, like the one in this problem, include at least one equation of second-order or higher, such as a quadratic equation.

To solve them, various techniques are used, including the substitution method or elimination method.
In the given exercise, working with both a quadratic and a linear equation involves manipulating one into a simpler form to solve for variables iteratively, leading to solutions for the point of intersection of these equations.

variables

Variables in algebra are symbols used to represent unknown values, typically denoted by letters such as \( x \) and \( y \). They play an essential role in forming equations and formulas. In our problem, we have two variables forming a system of nonlinear equations.
The variables are:

• \( x \): which we solve for in terms of \( y \) using the substitution method.
• \( y \): for which we solve using the quadratic equation derived from substitution.

Understanding the interaction of these variables within the equations helps find solutions where both conditions set by the equations are satisfied. By substituting and solving, you determine the values satisfying both the quadratic and linear relationships in the system.