Problem 61 - A Graphical Approach to College Algebra

Question

Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}+y^{2}=5\\\&x+y=3\end{aligned}$$

Step-by-Step Solution

Answer

The system has solutions at the points (1, 2) and (2, 1).

Step 1: Identify Equations

We have two equations to work with: the first one is a circle equation, \(x^2 + y^2 = 5\), and the second one is a linear equation, \(x + y = 3\). Our task is to graph these equations.

Step 2: Graph the Circle

The equation \(x^2 + y^2 = 5\) represents a circle centered at the origin \((0,0)\) with radius \(\sqrt{5}\). Draw a circle with a radius of approximately 2.24 units using this center and radius.

Step 3: Graph the Line

For the equation \(x + y = 3\), rearrange it to the slope-intercept form \(y = -x + 3\). Plot the y-intercept at (0,3) on the graph, with a slope of -1 (meaning for every 1 unit you move right, you move 1 unit down). Draw the line.

Step 4: Find Points of Intersection

Observe where the line \(x + y = 3\) intersects the circle \(x^2 + y^2 = 5\). The intersection points are the solutions to the system of equations.

Step 5: Check Solutions Algebraically

Solve the system algebraically to ensure the graphical solution is correct. Substitute \(y = 3 - x\) into \(x^2 + y^2 = 5\), leading to \(x^2 + (3-x)^2 = 5\). Simplify to get \(2x^2 - 6x + 4 = 0\). Solve this quadratic equation using the quadratic formula.

Step 6: Solve Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(2x^2 - 6x + 4 = 0\). Here, \(a = 2\), \(b = -6\), \(c = 4\). Calculate the discriminant: \((-6)^2 - 4(2)(4) = 4\). The solutions are \(x = 1\) and \(x = 2\).

Step 7: Substitute Found x Values to Find y

Substitute back into \(y = 3 - x\). For \(x = 1\), \(y = 3 - 1 = 2\). For \(x = 2\), \(y = 3 - 2 = 1\). Therefore, the points of intersection are \((1, 2)\) and \((2, 1)\).

Key Concepts

Systems of Equations

Systems of equations are a set of equations with multiple variables that you solve simultaneously to find a common solution. In our exercise, we dealt with two equations: a circle equation and a linear equation. Solving them graphically involves plotting them on a graph to find the intersection points. These points are where the equations meet in the coordinate plane. By solving these systems, you find the

• values of variables that satisfy both equations.
• geometric intersection points on the graph.

You can solve these systems using various methods, such as substitution, elimination, or graphically, as done in this exercise.

Circle Equations

A circle equation is usually represented in the form

• \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle.
• Our equation \(x^2 + y^2 = 5\) represents a circle centered at the origin \((0,0)\).

This circle has a radius of \(\sqrt{5}\), which is about 2.24 units. To graph a circle, you identify its center and use the radius to draw the circle accurately. Graphing is easier if you pinpoint symmetrical points from the center, such as

• top, bottom, left, and right at the radius distance.

Understanding circle equations helps in visualizing the graph and ensuring correct positioning of the circle in the coordinate plane.

Linear Equations

Linear equations describe lines in the form

• \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
• In our context, the line equation \(x + y = 3\) can be rewritten as

\(y = -x + 3\). This equation gives us a slope \(m = -1\), indicating that for each step right, you step one unit down. The y-intercept \(b = 3\) shows where the line crosses the y-axis. Graphing the line involves:

• starting at the y-intercept, \((0,3)\),
• then using the slope to plot other points.

Understanding how to graph linear equations is crucial for finding accurate intersections with other curves.

Intersection Points

Intersection points show where two graphs intersect and are the solutions to the system of equations. Graphically, look at the crossing points between the line \(x + y = 3\) and the circle \(x^2 + y^2 = 5\). These intersections represent solutions that satisfy both equations simultaneously. In our solution, we identified two points:

• \((1,2)\) and
• \((2,1)\)

as intersection points. To confirm, solve algebraically by substituting one equation into the other. This involves finding

• \(x\) and \(y\) values using these points.

Understanding intersection points helps verify accuracy in solving systems of equations both graphically and algebraically.