Problem 71 - College Algebra with Modeling and Visualization

Question

Solve the system of equations. $$ \begin{aligned} \frac{x^{2}}{2}+\frac{y^{2}}{4} &=1 \\ -x^{2}+2 y &=4 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions for the system of equations are the pairs \((x, y)\) that satisfy both equations.

Step 1: Understand the system of equations

We have two equations, which we need to solve simultaneously. The first equation is an ellipse: \( \frac{x^2}{2} + \frac{y^2}{4} = 1 \). The second equation is a line: \( -x^2 + 2y = 4 \). These need to be solved together to find the values of \( x \) and \( y \) that satisfy both equations.

Step 2: Express y in terms of x from the second equation

From the second equation \( -x^2 + 2y = 4 \), isolate \( y \). Move \( x^2 \) to the other side to get the expression: \( 2y = x^2 + 4 \). Then divide by 2: \( y = \frac{x^2}{2} + 2 \).

Step 3: Substitute y in the first equation

Replace \( y \) in the first equation with \( \frac{x^2}{2} + 2 \). So the first equation becomes: \[ \frac{x^2}{2} + \frac{(\frac{x^2}{2} + 2)^2}{4} = 1 \].

Step 4: Simplify the substituted equation

Expand \( (\frac{x^2}{2} + 2)^2 \) to get \( \frac{x^4}{4} + 2x^2 + 4 \). Substitute this back: \[ \frac{x^2}{2} + \frac{x^4}{16} + \frac{x^2}{2} + 2 = 1 \]. Combine like terms and simplify to form a single polynomial equation.

Step 5: Solve the polynomial equation

Simplify further to get \( \frac{x^4}{16} + x^2 + 2 = 1 \). Bringing all terms to one side, you get \( \frac{x^4}{16} + x^2 + 1 = 0 \). This can be solved by trial or using software to find solutions.

Step 6: Find corresponding y values

For each value of \( x \) found, substitute back into \( y = \frac{x^2}{2} + 2 \) to get the value of \( y \) corresponding to each \( x \).

Step 7: Verify solutions

Plug the pairs \((x, y)\) back into both original equations to ensure they satisfy both conditions. Only valid solutions that satisfy both equations simultaneously are correct answers.

Key Concepts

Understanding an Ellipse

An ellipse is a geometric shape resembling a stretched circle, having two axes of symmetry—the major and minor axes. In mathematical terms, the equation of an ellipse can be 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, respectively. The center of the ellipse is typically at the origin, though it can be translated. In our exercise, the equation is \[ \frac{x^2}{2} + \frac{y^2}{4} = 1 \]. Here, the semi-major axis is along the \( y \)-axis, with length \( 2 \), and the semi-minor axis is along the \( x \)-axis, with length \( \sqrt{2} \). This ellipse is positioned symmetrically on the origin.

• The terms \( \frac{x^2}{2} \) and \( \frac{y^2}{4} \) balance the equation along each axis.
• The ellipse shape means that solutions will either occur at points at which a line intersects this curve, or not at all, depending on the line's position and orientation.

Understanding ellipses is important as they often arise in physics, particularly in planetary orbits and optics.

Polynomial Equations

A polynomial equation is formed when two algebraic expressions with multiple terms are set equal. It involves constants, variables, and exponents but not division by variables. The expression typically looks like \[ a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0 \]. In the exercise, after substituting \( y \) in the ellipse equation, we derived a polynomial equation \[ \frac{x^4}{16} + x^2 + 1 = 0 \].

• Each term in the equation adds a layer of complexity. In our case, it's a degree-four polynomial because the highest power of \( x \) is 4, indicating it can have up to four solutions.
• Solving such an equation often requires simplification followed by methods like trial and error, factoring, or using algebraic software for more complicated cases.

Understanding polynomial equations and their degrees helps in predicting the types of solutions you might encounter—real or complex—and in selecting effective solving methods.

Substitution Method

The substitution method is a widely used algebraic technique for solving systems of equations, ideal when equations can easily be manipulated to express one variable in terms of another. In our exercise, we used this method to solve a system involving an ellipse and a line.

• The process begins by isolating one variable in terms of the others in one of the equations. For example, from the second equation \( -x^2 + 2y = 4 \), isolate \( y \) to obtain \( y = \frac{x^2}{2} + 2 \).
• After substitution into the initial equation, it formed a polynomial equation that we could then solve for \( x \).

The substitution method reduces complex systems to simpler equations, allowing one to focus on one variable at a time. It is especially effective when dealing with non-linear equations like a mix of ellipse equations and linear equations, making it versatile for many algebraic situations.