Problem 50 - College Algebra with Corequisite Support

Question

For the following exercises, find the solutions to the nonlinear equations with two variables. $$ \begin{array}{l} x^{2}-x y+y^{2}-2=0 \\ x+3 y=4 \end{array} $$

Step-by-Step Solution

Answer

Solutions: \(x = \frac{10 \pm \sqrt{126}}{13}\), with corresponding \(y\) values.

Step 1: Solve Linear Equation for y

Let's solve the linear equation \(x + 3y = 4\) for \(y\). This gives us:\[ y = \frac{4-x}{3} \].

Step 2: Substitute y in the Nonlinear Equation

Substitute \(y = \frac{4-x}{3}\) into the nonlinear equation \(x^2 - xy + y^2 - 2 = 0\). Substitute the expression for \(y\) to get:\[ x^2 - x\left(\frac{4-x}{3}\right) + \left(\frac{4-x}{3}\right)^2 - 2 = 0 \].

Step 3: Simplify the Equation

Simplify the equation by multiplying everything by 9 (the common denominator) to eliminate fractions:\[ 9x^2 - 3x(4-x) + (4-x)^2 - 18 = 0 \].Expand and simplify:\[ 9x^2 - (12x - 3x^2) + (16 - 8x + x^2) - 18 = 0 \].

Step 4: Combine Like Terms

Combine like terms in the expanded equation:\[ 9x^2 - 12x + 3x^2 + 16 - 8x + x^2 - 18 = 0 \]\[ 13x^2 - 20x - 2 = 0 \].

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(13x^2 - 20x - 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 13\), \(b = -20\), and \(c = -2\). Calculate:\[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 13 \cdot (-2)}}{2 \cdot 13} \]\[ x = \frac{20 \pm \sqrt{400 + 104}}{26} \]\[ x = \frac{20 \pm \sqrt{504}}{26} \]\[ x = \frac{20 \pm 2\sqrt{126}}{26} \]\[ x = \frac{10 \pm \sqrt{126}}{13} \].

Step 6: Calculate Corresponding y-values

For each solution of \(x\), find the corresponding \(y\) using \(y = \frac{4-x}{3}\).1. If \(x = \frac{10 + \sqrt{126}}{13}\), then: \[ y = \frac{4 - \frac{10 + \sqrt{126}}{13}}{3} \] Simplify to find \(y\): \[ y = \frac{39 - 10 - \sqrt{126}}{39} \].2. If \(x = \frac{10 - \sqrt{126}}{13}\), then: \[ y = \frac{4 - \frac{10 - \sqrt{126}}{13}}{3} \] Simplify to find \(y\): \[ y = \frac{39 - 10 + \sqrt{126}}{39} \].

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It allows us to calculate the values of \( x \) directly using the coefficients \( a \), \( b \), and \( c \).
The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

To use the quadratic formula, follow these steps:

• First, identify the coefficients in the quadratic equation (\( a \), \( b \), \( c \)).
• Substitute these values into the formula.
• Calculate the discriminant, \( b^2 - 4ac \), inside the square root.
• Use the plus-minus symbol (±) to find the two possible solutions for \( x \).

In our original problem, the quadratic equation was \( 13x^2 - 20x - 2 = 0 \), where \( a = 13 \), \( b = -20 \), and \( c = -2 \). We applied the quadratic formula to find the solutions for \( x \).
This formula is particularly helpful when factorization is difficult or impossible.

substitution method

The substitution method is a commonly used technique to solve systems of equations, especially when dealing with one linear and one nonlinear equation. This method involves solving one of the equations for one variable and then substituting this expression into the other equation.
In our problem, we initially solved the linear equation \( x + 3y = 4 \) for \( y \):

• \( y = \frac{4-x}{3} \)

Next, we substituted this expression for \( y \) into the nonlinear equation \( x^2 - xy + y^2 - 2 = 0 \).
This substitution reduces the problem into a single equation with one variable, \( x \), making it easier to solve.
Using the substitution method is particularly useful when one of the equations is easy to solve for a variable, allowing simplification of the problem significantly.

linear equations

Linear equations represent algebraic expressions that produce straight lines when graphed. They are in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
In this exercise, the linear equation was \( x + 3y = 4 \).
To solve a linear equation for one variable:

• Rearrange the equation to isolate the desired variable.
• Simplify the isolated equation to find the variable.

For example, in this problem, solving for \( y \) gave us \( y = \frac{4-x}{3} \).
Linear equations are simpler compared to nonlinear ones and often serve as building blocks in solving more complex problems, like systems that include nonlinear equations.

algebraic simplification

Algebraic simplification is the process of making an equation or expression easier to work with by combining like terms, removing parentheses, and reducing coefficients. Simplification leads to a form that's easier to interpret and solve.
In our exercise, after substitution, the resulting equation involved dealing with complex terms:

• \( x^2 - x\left(\frac{4-x}{3}\right) + \left(\frac{4-x}{3}\right)^2 - 2 = 0 \)

Next, we multiplied the equation by 9 to clear fractions, resulting in cleaner whole numbers.
By expanding and combining like terms:

• \( 13x^2 - 20x - 2 = 0 \)

Through simplification, we obtained a quadratic equation that was easier to solve using the quadratic formula.
Effective algebraic simplification is crucial in solving mathematical problems as it reduces complexity and errors.