Problem 12 - Algebra 2

Question

Find the exact solution(s) of each system of equations. $$ \begin{array}{l}{\frac{x^{2}}{30}+\frac{y^{2}}{6}=1} \\ {x=y}\end{array} $$

Step-by-Step Solution

Answer

The solutions are \( (\sqrt{5}, \sqrt{5}) \) and \( (-\sqrt{5}, -\sqrt{5}) \).

Step 1: Understand the Equations

We have two equations to solve: \( \frac{x^2}{30} + \frac{y^2}{6} = 1 \) and \( x = y \). The first equation represents an ellipse, and the second equation is a straight line where the x-coordinate equals the y-coordinate.

Step 2: Substitute Equation II into Equation I

Since \( x = y \), substitute \( y \) in the first equation with \( x \). This gives: \( \frac{x^2}{30} + \frac{x^2}{6} = 1 \).

Step 3: Combine Like Terms

Combine the terms with \( x^2 \): \( \frac{x^2}{30} + \frac{x^2}{6} = \frac{x^2}{30} + \frac{5x^2}{30} = \frac{6x^2}{30} = \frac{x^2}{5} \). So, the equation simplifies to \( \frac{x^2}{5} = 1 \).

Step 4: Solve for \(x\)

To find \( x \), multiply both sides by 5 to isolate \( x^2 \): \( x^2 = 5 \). Taking the square root of both sides gives us \( x = \pm \sqrt{5} \).

Step 5: Determine \(y\) from \(x\)

Using the equation \( x = y \), we have \( y = \pm \sqrt{5} \). Thus, the solutions correspond to the points where both \( x \) and \( y \) are equal and satisfy this condition.

Key Concepts

Ellipse

An ellipse is a type of curve on a plane, similar to a circle but stretched along one or more axes. The general equation of an ellipse centered at the origin is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively. In simpler terms, these are half the lengths of the widest and narrowest parts of the ellipse.
In the given exercise, the ellipse is defined by the equation \( \frac{x^2}{30} + \frac{y^2}{6} = 1 \). This equation tells us that:

• The semi-major axis is related to \( x \) due to the larger denominator 30, which means it stretches more in the x-direction.
• The semi-minor axis is associated with \( y \), indicated by the smaller denominator 6.

Understanding ellipses helps visualize the shape and the spread of the values that satisfy its equation.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of another and then substituting this expression into another equation.
In our exercise, we have the system of equations:

• \( \frac{x^2}{30} + \frac{y^2}{6} = 1 \)
• \( x = y \)

The second equation \( x = y \) is used for substitution, which simplifies problem-solving. By substituting \( y \) with \( x \) in the first equation, we transform it into a quadratic equation in terms of \( x \), making it easier to solve. This is beneficial because it reduces the complexity of solving the system by focusing on one variable at a time.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra. A quadratic equation takes the form \( ax^2 + bx + c = 0 \). In this particular case, after substituting and simplifying, the equation became \( \frac{x^2}{5} = 1 \).
To solve this equation:

• Multiply both sides by 5 to isolate \( x^2 \): \( x^2 = 5 \).
• Take the square root of both sides to find \( x \), which results in \( x = \pm \sqrt{5} \).

Quadratic equations often have two solutions, one positive and the one negative, due to the nature of squaring. Understanding how to manipulate and solve these equations allows us to find intersections, vertices, roots, and other important points in various contexts.

Linear Equations

Linear equations are equations of the first degree, meaning they have variables raised only to the power of one. They represent straight lines when graphed on a coordinate plane.
In our system of equations, the equation \( x = y \) is linear. It describes a line that passes through the origin and has a slope of 1, which means it forms a 45-degree angle with both the x-axis and y-axis.

• In graphical terms, this line identifies all points where the x-coordinate is equal to the y-coordinate.
• When this line intersects an ellipse, as is the case in our exercise, it provides solutions that satisfy both the ellipse and the line's equations.

The simplicity of linear equations makes them useful for substitution, as we observed, because they allow us to express one variable directly in terms of another and make solving complex systems more manageable.