Problem 72 - College Algebra with Modeling and Visualization

Question

Solve the system of equations. $$ \begin{aligned} &x^{2}+\frac{1}{9} y^{2}=1\\\ &x+y=3 \end{aligned} $$

Step-by-Step Solution

Answer

Solutions: \((0, 3)\) and \(\left( \frac{3}{5}, \frac{12}{5} \right)\)."

Step 1: Understand the Equations

The system of equations given is \( x^2 + \frac{1}{9}y^2 = 1 \) and \( x + y = 3 \). The first equation represents an ellipse, and the second is a straight line.

Step 2: Express y in Terms of x

From the linear equation \( x + y = 3 \), we can express \( y \) in terms of \( x \) as: \[ y = 3 - x \]

Step 3: Substitute y in the Ellipse Equation

Substitute \( y = 3 - x \) into the ellipse equation \( x^2 + \frac{1}{9}y^2 = 1 \). This results in:\[ x^2 + \frac{1}{9}(3-x)^2 = 1 \]

Step 4: Simplify the Substituted Equation

Expand \( (3-x)^2 \) to get \( 9 - 6x + x^2 \). Substitute back to get:\[ x^2 + \frac{1}{9}(9 - 6x + x^2) = 1 \]Simplifying gives:\[ x^2 + 1 - \frac{2}{3}x + \frac{1}{9}x^2 = 1 \]

Step 5: Combine Like Terms

Combine the terms in the simplified equation:\[ \left(1 + \frac{1}{9}\right) x^2 - \frac{2}{3}x + 1 = 1 \]\[ \frac{10}{9} x^2 - \frac{2}{3} x + 1 = 1 \]

Step 6: Simplify Further

Subtract 1 from both sides:\[ \frac{10}{9} x^2 - \frac{2}{3} x = 0 \]

Step 7: Factorize the Quadratic Equation

Factor out \( x \):\[ x \left( \frac{10}{9}x - \frac{2}{3} \right) = 0 \]

Step 8: Solve for x

Set each factor to zero and solve:1. \( x = 0 \)2. \( \frac{10}{9}x - \frac{2}{3} = 0 \) Solving for \( x \), multiply through by 9 to get \[ 10x = 6 \] which reduces to \( x = \frac{3}{5} \).

Step 9: Find Corresponding y Values

For \( x = 0 \), using \( y = 3 - x \), we have \( y = 3 \).For \( x = \frac{3}{5} \), using \( y = 3 - x \), we have:\[ y = 3 - \frac{3}{5} = \frac{15}{5} - \frac{3}{5} = \frac{12}{5} \]

Step 10: Write the Solution Set

The solutions for the system are:\( (0, 3) \) and \( \left( \frac{3}{5}, \frac{12}{5} \right) \).

Key Concepts

Understanding Ellipses in a System of Equations

An ellipse is a curve on a plane surrounding two focal points, where the sum of the distances to the focal points is constant. In mathematical terms, an ellipse that is aligned with the coordinate axes is expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes. This equation is similar to a circle's equation but stretched along one or both axes.
In our exercise, the equation \( x^2 + \frac{1}{9}y^2 = 1 \) represents an ellipse centered at the origin. Here, \( x^2 \) and \( \frac{1}{9}y^2 \) denote that the ellipse is more stretched along the y-axis compared to the x-axis. By identifying the ellipse equation, we could analyze how it interacts with the linear equation \( x + y = 3 \).

• Ellipses in systems of equations illustrate how curves behave algebraically when combined with other functions.
• Intercepts with linear equations typically represent solutions to the system.

The Role of Factorization in Solving Quadratic Equations

Factorization is a crucial step in solving quadratic equations by finding values that make the equation zero. In this exercise, we encountered the quadratic equation \( \frac{10}{9} x^2 - \frac{2}{3} x = 0 \). When equating to zero, our goal was to find values of \( x \) that satisfy the equation.
Factorization involves expressing the polynomial as a product of its factors. For this quadratic equation, we factored out \( x \) from both terms, resulting in \( x \left( \frac{10}{9}x - \frac{2}{3} \right) = 0 \). This tells us two things:

• \( x = 0 \)
• \( \frac{10}{9}x - \frac{2}{3} = 0 \)

Solving these gives the possible values of \( x \). This method is particularly useful because it breaks down the problem into simpler, more manageable parts and allows us to easily find solutions that satisfy the original equation.

Solving Systems with Linear Equations

Linear equations involve the simplest algebraic terms, typically represented as \( ax + by = c \). In the given exercise, the linear equation \( x + y = 3 \) intersected with an elliptical equation to form a system. These systems are sets of two or more equations with a common solution.
One effective method to solve such systems is the substitution method, where you solve one equation for one variable and substitute this into the other equation.

• From the linear equation, we expressed \( y \) in terms of \( x \), \( y = 3 - x \), and plugged it into the ellipse equation.
• This interconnected approach simplifies a multi-variable system into equations with a single variable.

Once the values of \( x \) were found, they were substituted back into the linear equation to find respective values of \( y \). This method breaks down complex relationships in systems, making them approachable and solvable.