Problem 101 - College Algebra with Modeling and Visualization

Question

Solve the nonlinear system of equations $$(a) symbolically\quad and\quad (b)\quad graphically.$$ $$ \begin{array}{r} x y=12 \\ x-y=4 \end{array} $$

Step-by-Step Solution

Answer

Solutions: \((6, 2)\) and \((-2, -6)\).

Step 1: Identify the Equations

The system of equations we need to solve is composed of: 1. Equation 1: \( xy = 12 \)2. Equation 2: \( x - y = 4 \)

Step 2: Express y in terms of x using Equation 2

Rearrange Equation 2 to express \( y \) in terms of \( x \):\[ y = x - 4 \]

Step 3: Substitute y in Equation 1

Replace \( y \) in Equation 1 with the expression found from Equation 2:\[ x(x - 4) = 12 \]

Step 4: Expand and Solve the Quadratic Equation

Now, expand the equation and solve for \( x \):\[ x^2 - 4x = 12 \]\[ x^2 - 4x - 12 = 0 \]Using the quadratic formula \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, \text{ and } c = -12 \), solve for \( x \):\[ x = \frac{4 \, \pm \, \sqrt{16 + 48}}{2} \]\[ x = \frac{4 \, \pm \, \, 8}{2} \]Thus, \( x_1 = 6 \) and \( x_2 = -2 \).

Step 5: Find y Values for each x

Substitute the \( x \) values back into the equation \( y = x - 4 \) to find \( y \):- For \( x = 6 \):\[ y = 6 - 4 = 2 \]- For \( x = -2 \):\[ y = -2 - 4 = -6 \]

Step 6: Solution to the System

The solutions to the system of equations are:- \( (x, y) = (6, 2) \)- \( (x, y) = (-2, -6) \)

Step 7: Graphical Solution

Graph the system of equations. The first equation, \( xy = 12 \), is a hyperbola, and the second equation, \( x - y = 4 \), is a line. Plotting these on a graph will show intersections at the points (6, 2) and (-2, -6), verifying the solutions found symbolically.

Key Concepts

quadratic equation

A quadratic equation is a polynomial equation of the second degree. This means its highest exponent of the unknown variable, usually denoted as \( x \), is 2. A typical form of a quadratic equation looks like \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations are fundamental in algebra because they often appear in various mathematical contexts. In the context of solving nonlinear systems, a quadratic equation arises when substituting one variable in terms of another.In our problem, after substituting \( y = x - 4 \) into \( xy = 12 \), we obtain the quadratic equation \( x^2 - 4x - 12 = 0 \). Solving such quadratic equations can often be done using the quadratic formula:

• The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• In this formula, \( b^2 - 4ac \) is called the discriminant.
• The discriminant can tell us the nature of the roots (real and distinct, real and repeated, or complex).

Applying this formula to our problem enabled us to find the roots \( x_1 = 6 \) and \( x_2 = -2 \). These roots help us move forward to find the corresponding \( y \) values.

symbolic solution

A symbolic solution involves finding an exact answer using algebraic manipulations, without relying on numerical approximations. It allows you to see clearly how different parts of an equation relate to each other.In our exercise, we started by identifying the system of equations:

• \( xy = 12 \)
• \( x - y = 4 \)

The next step was to isolate one variable. From the second equation, we expressed \( y \) in terms of \( x \), getting \( y = x - 4 \). By substituting \( y \) in the first equation, we transformed it into a quadratic equation, which we then solved using the quadratic formula.The solutions \( (x, y) = (6, 2) \) and \( (x, y) = (-2, -6) \) are found by substituting back the \( x \) values into the expression for \( y \). Solving symbolically is powerful because it clearly shows how the two solutions satisfy both original equations.

graphical solution

A graphical solution represents equations visually, usually on a coordinate plane, which helps to identify where solutions (intersections) exist between equations.For the system of nonlinear equations given:

• \( xy = 12 \)
• \( x - y = 4 \)

The first equation, \( xy = 12 \), describes a hyperbola. The second equation, \( x - y = 4 \), is a straight line. When graphing these, each point of intersection corresponds to a solution of the system.Plotting these on the graph, they intersect at the points \( (6, 2) \) and \( (-2, -6) \). These intersections verify the solutions found symbolically.Graphical solutions are especially helpful in understanding the nature of solutions, showing:

• The relationship between variables.
• The number of solutions.
• The type of graphs produced by each equation.

Graphing offers a visual verification of the symbolic solutions and can also highlight if any extraneous solutions exist.