Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Solve the system graphically. $$\left\\{\begin{array}{r}-x+y=3 \\ x^{2}-6 x-27+y^{2}=0\end{array}\right.$$

Step-by-Step Solution

Answer

The system is solved by plotting line and circle, and finding the intersection points on the graph.

Step 1: Graphing the Line

The first equation can be written as \( y = x + 3 \). By plotting this line on a graph, we create the first part of the graphical representation of the system

Step 2: Graphing the Circle

The second equation can be written as \( x^2 + y^2 = 6x + 27 \). By completing the square for both x and y, the equation can be transformed into the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), where \( h \), \( k \), and \( r \) are the center coordinates and the radius of the circle. Thus, the equation becomes \( (x - 3)^2 + (y - 0)^2 = 27 \), which stands for a circle with center at (3, 0) and radius \sqrt{27}. By plotting this circle, we complete the graphical representation

Step 3: Finding the Intersections

The solutions to the system are the points where the line intersects the circle. This problem is solved by plotting the line and the circle on the same graph, and identifying the points where they intersect

Key Concepts

System of Equations

A **system of equations** is a set of two or more equations that have common variables. Solving a system involves finding the values of these variables that satisfy all the equations simultaneously. There are several methods to solve systems of equations such as substitution, elimination, and graphical representation.

In graphical solutions, each equation in the system is represented as a graph. The solutions to the system are the points where all the graphs intersect. Visualizing equations can sometimes offer better insights, and it is particularly useful when you want to understand how different equations interact with each other geometrically.

• Equations can be linear, quadratic, or of other forms.
• The graphical representation aids in finding their intersection points visually.

In our example, the system includes a linear equation (line) and a quadratic equation (circle). By plotting both, the solutions are the intersection points between the line and the circle on the graph.

Circle Equation

A **circle equation** typically represents all the points that are at a fixed distance from a common point, known as the center. The standard form of a circle's equation is \[(x-h)^2 + (y-k)^2 = r^2\] where

• \(h\) and \(k\) are the x and y coordinates of the circle's center.
• \(r\) is the radius of the circle.

In our given equation \(x^2 - 6x - 27 + y^2 = 0\), it is rearranged to \((x-3)^2 + (y-0)^2 = 27\) after completing the square. This tells us that the circle's center is at (3, 0) and the radius is \(\sqrt{27}\).

This transformation helps us plot the circle accurately on the graph. Identifying the center and radius allows us to understand its size and position, which are crucial when determining where it might intersect with another graph.

Line Equation

The **line equation** is often expressed in the form \(y = mx + b\), where

• \(m\) is the slope, showing how steep the line is.
• \(b\) is the y-intercept, indicating where the line crosses the y-axis.

For our specific equation \(-x + y = 3\), it can be rearranged to \(y = x + 3\).

This equation represents a line with a slope of 1 (meaning the line rises one unit for every unit it runs) and a y-intercept of 3 (the point where the line crosses the y-axis).

• Graphing this line gives us a visual understanding of how it interacts with other figures, like circles, in a system of equations.
• The slope tells us the angle of intersection with other objects while the intercept provides the position on a graph.

Once plotted, it helps us see where it potentially intersects the other equation's graph – in this case, the circle.

Completing the Square

**Completing the square** is a mathematical process used to convert a quadratic equation into its standard form. It involves creating a perfect square trinomial from the original equation. This technique is pivotal when dealing with circle equations that are not in the standard form. It lets you express the equation in terms of the circle's center and radius, making graphical representation feasible.

For our circle equation \(x^2 - 6x - 27 + y^2 = 0\), we separate the \(x\) and \(y\) terms. The steps usually involve:

• Rearranging the equation to isolate \(x\) and \(y\) terms from constants.
• Completing the square for the \(x\) terms by finding the appropriate constant to add and subtract.
• Doing the same for \(y\) terms, if necessary.

This transforms the equation into a standard circle form \((x-h)^2 + (y-k)^2 = r^2\) so it can be easily graphed. Understanding this technique is key to graphing complex equations and determining their geometric properties.