Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to determine whether the system of equations has one solution, two solutions, or no solution. $$\left\\{\begin{aligned} x^{2}+3 x+y &=4 \\ 3 x+y &=-5 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The number of solutions will depend on the number of intersection points of the two graphs when using a graphing utility.

Step 1: Understand the Equations

The system of equations presented are \(x^{2}+3x+y = 4\) and \(3x+y = -5\). These can be rewritten to isolate \(y\) as \(y = 4 - x^{2} - 3x\) and \(y = -5 - 3x\) respectively.

Step 2: Graph of Equations

Using a graphing utility, graph the two equations on the same coordinate plane. The first equation will be a parabola and the second will be a straight line.

Step 3: Identify Intersection Points

The points of intersection between the two graphs represent the solutions for the system of equations. Count the number of intersection points.

Step 4: Report the Solutions

If there is only one intersection point, the system has one solution. If there are two intersection points, the system has two solutions. If there are no intersection points, the system has no solutions.

Key Concepts

Graphing Utility

Understanding how to use a graphing utility is essential when tackling systems of equations, particularly when dealing with quadratic and linear pairs. These tools, which can be software like GeoGebra, Desmos, or a TI graphing calculator, allow us to visualize complex algebraic relationships easily. When you input the two equations into a graphing utility, the software processes the algebra and displays their graphs on a coordinate plane.

For our exercise involving the equations \(x^2 + 3x + y = 4\) and \(3x + y = -5\), a graphing utility would show one equation as a curved parabola and the other as a straight line. Each tick on the graph represents a potential solution depending on where the two equations intersect. Understanding how to navigate the features of such utilities – zooming in and out, adjusting the scale, or tracing along the curve – can significantly aid in identifying the exact points of intersection, and hence, the solutions.

Quadratic and Linear Systems

Systems of equations can involve different types of functions, such as linear and quadratic functions. A linear equation forms a straight line when graphed, described by an equation of the form \(y = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. In contrast, a quadratic equation forms a parabola and is represented by \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a\) is not zero.

In our example, \(x^2 + 3x + y = 4\) rewrites into \(y = -x^2 - 3x + 4\), showing us a parabola, while \(3x + y = -5\) rewrites into \(y = -3x - 5\), which is linear. Combining these visually through a graph reveals how the parabola opens either up or down and where the line crosses this shape. It's this visual juxtaposition that allows us to discover the exact number of solutions, if any, by locating where the curve meets the line.

Intersection Points

The 'intersection points' refer to the exact spots where the graphs of the two equations in a system meet on the coordinate plane. This point (or these points) is significant because it represents the solution set to the system of equations. If one point exists where both graphs intersect, this means there's a single solution that satisfies both equations. In some cases, there might be two intersection points, indicating two possible solutions. And if the graphs do not intersect at all, it signifies that there's no shared solution, or in other words, the system of equations has no solution.

When you use a graphing utility, you can sometimes directly calculate the coordinates of these intersection points. This is particularly useful as it alleviates the need for solving the system algebraically, which can be complex and time-consuming for nonlinear systems. In our exercise example, by identifying either one or two points where the parabola and line intersect, we can determine the precise solutions without engaging in potentially elaborate algebraic manipulations.