Problem 32 - College Algebra Essentials

Question

In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} x^{2}+4 y^{2}=20 \\ x y=4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution for the system of equations is \( (x, y) = (2\sqrt{2}, \sqrt{2}), (-2\sqrt{2}, -\sqrt{2}) \)

Step 1: Express Variable in Terms of Another Variable

From the second equation \(xy = 4\), we can express x or y in terms of the other variable. For example, let's express x in terms of y: \(x = 4/y\)

Step 2: Substitution into the First Equation

Substitute \(x = 4/y\) from the second equation into the first equation to get \( (4/y)^2 + 4y^2 = 20\). Solving this equation gives: \( 16/y^2 + 4y^2 = 20 \) or \( (16 + 4y^4) / y^2 = 20 \) or \( 4y^4 + 16 - 20y^2 = 0 \).

Step 3: Solve Quadratic Equation

The expression \( 4y^4 + 16 - 20y^2 = 0 \) is a quadratic equation in terms of \( y^2 \) which can be solved using Quadratic Formula. Let \( t = y^2\), then the equation becomes \( 4t^2 - 20t + 16 = 0 \). We solve this to get \( t = 2, 2 \). We substitute \( t = y^2\) to find out \( y = \sqrt{2}, -\sqrt{2} \)

Step 4: Solve for x

Substitute \( y = \sqrt{2}, -\sqrt{2} \) into the equation \( x = 4/y \) to finally find out \( x = 2\sqrt{2}, -2\sqrt{2} \)

Key Concepts

Substitution Method

The substitution method is a strategic approach used to solve systems of equations. This method involves substituting one equation into another to eliminate one variable, simplifying the system to a single equation with one variable. This can be particularly useful when dealing with non-linear equations, as in this exercise.

Here's how it generally works:

• Choose one equation and solve for one variable in terms of the other. This step can vary based on which variable is easier to express.
• Substitute the expression from the first step into the other equation. This creates a new equation that typically has fewer variables, making it easier to solve.
• Solve this new equation to find the value of the variable you isolated.

This method is like peeling layers off an onion, allowing us to focus one layer at a time. In our exercise, by expressing \( x \) in terms of \( y \), it simplifies the original system, enabling an easier path to finding the solution.

Quadratic Equation

Quadratic equations are polynomial equations of degree two. They are usually in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. These equations present a parabolic curve when plotted and can have zero, one, or two solutions.

For solving quadratic equations, several methods can be applied, such as:

• Factoring, which involves finding two numbers that multiply to give the constant coefficient and add to give the linear coefficient.
• Completing the square, which transforms the equation into a perfect square trinomial.
• Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). This formula provides the solutions directly and is particularly handy when factoring is complex.

In the given system, we encounter a quadratic in the form of \( 4t^2 - 20t + 16 = 0 \) after substituting \( y^2 = t \), showcasing the versatility of quadratic solutions in various scenarios beyond simple line graphs.

Algebraic Manipulation

Algebraic manipulation is the skillful arrangement and simplification of algebraic expressions and equations to solve them efficiently. It involves a wide range of techniques aimed at making the equations more manageable.

Key strategies in algebraic manipulation include:

• Simplifying expressions by combining like terms and using basic arithmetic operations.
• Rearranging terms to isolate variables, often a crucial step in methods like substitution or elimination.
• Expanding and factoring expressions to find standard forms, especially helpful in quadratic and polynomial equations.

In the current exercise, algebraic manipulation helps us simplify \( (4/y)^2 + 4y^2 = 20 \) to a more workable form, eventually leading to the quadratic \( 4t^2 - 20t + 16 = 0 \). Mastery of these techniques is essential for tackling complex systems of equations effectively and confidently.

Solution of Equations

The process of solving equations implies finding the values of the unknowns that satisfy all given conditions. This involves applying a logical series of steps and techniques that result in obtaining the values of the sought-after variables.

Here’s a general solution process:

• Simplify the equation if necessary, through methods of algebraic manipulation.
• Isolate the variable and rearrange the equation to facilitate solving, as seen with expression substitution.
• Use appropriate methods (such as quadratic formula, factoring, or graphical methods) to solve the equation for the unknowns.

In the provided system, solving involves coordination between substitution, simplification, and using the quadratic formula to finally determine the values of \( x \) and \( y \). This systematic approach highlights the importance of methodical and strategic thinking in tackling even the most challenging systems of equations.