Problem 57 - College Algebra with Modeling and Visualization

Question

If possible, solve the nonlinear system of equations. $$ \begin{array}{r} x y=8 \\ x+y=6 \end{array} $$

Step-by-Step Solution

Answer

The solutions are \((x, y) = (4, 2)\) and \((x, y) = (2, 4)\).

Step 1: Formulate System

The given system of equations is: \( xy = 8 \) and \( x + y = 6 \). We need to find values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Express y in terms of x

From the second equation, we have \( x + y = 6 \). Solve for \( y \): \( y = 6 - x \).

Step 3: Substitute Expression into First Equation

Substitute \( y = 6 - x \) into the first equation \( xy = 8 \), resulting in the equation \( x(6 - x) = 8 \).

Step 4: Simplify Equation

Simplify the equation to form a quadratic equation: \( 6x - x^2 = 8 \). Rearrange it to standard form: \( -x^2 + 6x - 8 = 0 \) or \( x^2 - 6x + 8 = 0 \).

Step 5: Solve Quadratic Equation

Use the quadratic formula to solve \( x^2 - 6x + 8 = 0 \). The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), \( c = 8 \). Substitute these values: \( x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \).

Step 6: Calculate Discriminant and Roots

Calculate the discriminant: \( (-6)^2 - 4 \times 1 \times 8 = 36 - 32 = 4 \). Since the discriminant is positive, there are two real roots: \( x = \frac{6 + 2}{2} = 4 \) and \( x = \frac{6 - 2}{2} = 2 \).

Step 7: Find Corresponding y-values

For \( x = 4 \), substitute back into \( y = 6 - x \): \( y = 6 - 4 = 2 \). For \( x = 2 \), substitute back into \( y = 6 - x \): \( y = 6 - 2 = 4 \).

Step 8: Verify Solutions

Check both pairs \((x, y) = (4, 2)\) and \((x, y) = (2, 4)\) satisfy the original system: \(xy = 8\) is satisfied by both, and \(x+y=6\) is true in both cases.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the degree is 2. This means the equation includes an expression that looks like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The solutions to quadratic equations are found using several methods such as factoring, completing the square, or the quadratic formula.

In the context of solving a nonlinear system of equations, the first equation we encounter can sometimes take the form of a quadratic equation. For instance, when substituting one variable's expression into another equation, it might lead to a term like \( x^2 \), transforming our problem specifically into a quadratic form. Thus, understanding how to manipulate these equations is key to finding their solutions.

• The standard form is \( x^2 + bx + c = 0 \).
• You can solve it using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Recognizing the quadratic structure helps in applying the right method to find solutions.

Substitution Method

The substitution method is a powerful technique to solve systems of equations. It involves expressing one variable in terms of another and then substituting this expression into a different equation. This can be particularly useful in dealing with nonlinear systems that contain quadratic equations.

In our exercise, we take the second equation, \( x + y = 6 \), and express \( y \) in terms of \( x \): \( y = 6 - x \). This expression substitutes \( y \) in the first equation, allowing us to solve it with just one variable. The substitution simplifies the system and often reduces the complexity, making it easier to find an exact solution.

• Start with expressing one variable from one equation.
• Substitute this expression into the other equation.
• Simplify the equation to find the solution for one variable.
• Back substitute to find the other variable.

Solving Algebraic Equations

Algebraic equations are mathematical statements composed of constants, variables, and arithmetic operations. Solving these equations is foundational in mathematics as it involves finding the values of the variables that make the equation hold true.

In solving a system of equations, especially nonlinear ones including quadratic terms, a systematic approach is important. Simplifying equations after substitution is a crucial step. Once simplified, using methods such as the quadratic formula, we determine the roots of the equation that satisfy all parts of the system.

After obtaining potential solutions for one variable, it's important to substitute these back to check for the corresponding values of the other variable, as seen in the step-by-step solution where values of \( x \) were found and then used to find \( y \).

• Simplification is key after making substitutions.
• Use the quadratic formula where applicable, ensuring each step is correctly calculated.
• Always verify solutions within the context of the original equations to ensure validity.