Problem 63 - A Graphical Approach to Precalculus with Limits

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 solutions are the points (1, 2) and (2, 1).

Step 1: Understand the Equations

You are given two equations. The first is a circle: \(x^2 + y^2 = 5\). The second is a straight line: \(x + y = 3\). To solve the system graphically, find where the line intersects the circle.

Step 2: Plot the Circle

The equation \(x^2 + y^2 = 5\) represents a circle centered at the origin with a radius of \(\sqrt{5}\). To plot it, sketch a circle that passes through points like \((\sqrt{5}, 0)\), \((0, \sqrt{5})\), \((-\sqrt{5}, 0)\), and \((0, -\sqrt{5})\).

Step 3: Plot the Line

For the line \(x + y = 3\), find two points by selecting values for \(x\) and solving for \(y\). For example, when \(x = 0\), then \(y = 3\) (point \((0,3)\)), and when \(y = 0\), then \(x = 3\) (point \((3,0)\)). Connect these points with a straight line.

Step 4: Find the Intersection Points

Where the line \(x + y = 3\) intersects the circle \(x^2 + y^2 = 5\) are the solutions to this system. By observing the graph, you can see that the line touches the circle at two points. Calculate these points by substituting \(y = 3 - x\) into the circle's equation.

Step 5: Substitute and Solve

Substitute \(y = 3 - x\) into \(x^2 + y^2 = 5\):\[x^2 + (3 - x)^2 = 5\]Expanding gives:\[x^2 + 9 - 6x + x^2 = 5\]Simplifying:\[2x^2 - 6x + 4 = 0\]Divide by 2:\[x^2 - 3x + 2 = 0\]Solve using factorization: \((x - 1)(x - 2) = 0\). So, \(x = 1\) or \(x = 2\).

Step 6: Solve for \(y\)

Substitute \(x = 1\) and \(x = 2\) back into the line equation \(x + y = 3\).1. For \(x = 1\): \[1 + y = 3\] \[y = 2\]2. For \(x = 2\): \[2 + y = 3\] \[y = 1\]Thus, the intersection points are \((1, 2)\) and \((2, 1)\).

Step 7: Verify the Solutions

Check these points in both equations:1. For \((1,2)\): - **Circle**: \(1^2 + 2^2 = 5\) (True) - **Line**: \(1 + 2 = 3\) (True)2. For \((2,1)\): - **Circle**: \(2^2 + 1^2 = 5\) (True) - **Line**: \(2 + 1 = 3\) (True)Both points satisfy both equations.

Key Concepts

Circle and Line Intersection

When working with systems of equations that involve circles and lines, understanding their intersection points graphically is crucial. A circle is defined by its equation, typically in the form of \(x^2 + y^2 = r^2\), where \(r\) is the radius. In our case, the circle \(x^2 + y^2 = 5\) is centered at the origin with a radius of \(\sqrt{5}\).
The line given by \(x + y = 3\) is a straight line. Intersection points occur where the graph of this line touches or crosses the circle.
When plotted on the same set of axes, these points of intersection are visually evident. They represent the set of points \((x, y)\) that satisfy both equations simultaneously.

• The circle can be plotted using its radius and center which is the origin \((0,0)\).
• The line requires at least two points for accurate construction, like \((0,3)\) and \((3,0)\).

Solving Systems Graphically

Solving a system graphically involves plotting each equation on a graph and identifying where they intersect. This method is especially useful when the equations involve different geometric shapes, like a circle and a line.
To graphically solve the system \(x^2 + y^2 = 5\) and \(x + y = 3\), follow these steps:

• First, map out the circle by plotting points that lie on the circle's perimeter, dictated by the radius \(\sqrt{5}\).
• Determine the line by plotting known points that satisfy \(x + y = 3\).
• Find the intersection visually on the graph. Each intersection indicates a solution to the system.

The graphical solution should reveal the circle and line touching at specific coordinates. These coordinates represent the approximate solutions, which can be validated algebraically for precision. This approach provides an intuitive understanding of how the shapes interact on the coordinate plane.

Verifying Solutions

Verifying the solutions found graphically requires substituting them back into the original equations. This process confirms whether these solutions truly satisfy both equations.
For the system given:\(x^2 + y^2 = 5\) and \(x + y = 3\), the proposed solutions are \((1, 2)\) and \((2, 1)\).
Verification involves:

• Substitute each point into the circle's equation and check if the equality holds.
• Do the same for the line's equation.

For example, with the point \((1, 2)\): - **Circle Check**: \(1^2 + 2^2 = 5\) (This is true, confirming that the point is on the circle.)
- **Line Check**: \(1 + 2 = 3\) (Also true, affirming that the point is on the line.)
Repeating this process for \((2, 1)\) will yield similar results of validity. Accurate verification endorses the graphical solution, proving the intersection points are indeed correct.