Problem 37 - Precalculus with Limits

Question

Solving a System of Equations Graphically In Exercises \(33-44,\) solve the system graphically. $$\left\\{\begin{array}{r}{x+y=4} \\ {x^{2}+y^{2}-4 x=0}\end{array}\right.$$

Step-by-Step Solution

Answer

There isn't a concise numerical answer to the problem. The solution includes all points where the given line intersects the given circle when both are plotted on the same graph.

Step 1: Graph the linear equation

Start by graphing the linear equation \(x+y=4\). This equation is in slope-intercept form (\(y = mx+c\)), where \(m\) is the slope and \(c\) is the y-intercept. Here, rearranging this equation for \(y\) gives \(y = 4-x\), which has a slope of -1 and a y-intercept of 4.

Step 2: Graph the quadratic equation

Next, graph the quadratic equation \(x^{2}+y^{2}-4x=0\). This equation represents a circle. To see this, rewrite the equation in the simpler form. Completing the square on \(x\) results in \((x-2)^{2} + y^{2} = 2^{2}\), or \((x-2)^{2} + y^{2} = 4\). This is a circle with center at (2,0) and radius 2.

Step 3: Identify the intersection points

Lastly, identify the point or points where the two graphs intersect. These points are the solutions to the system of equations. Based on the properties of these graphs, our solutions are the points at which the straight line intersects the circle.

Key Concepts

Graphing Linear Equations

Graphing linear equations involves plotting straight lines on a coordinate plane. The equation typically takes the form of \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. The slope indicates how steep the line is, and the y-intercept tells us where the line crosses the y-axis. To graph the equation \( x + y = 4 \), it can be rewritten as \( y = 4 - x \). From this, we know:

• Slope \( m = -1 \)
• Y-intercept \( c = 4 \)

Start by plotting the y-intercept on the graph at (0, 4). Then, use the slope to determine the direction and steepness of the line. Since the slope is -1, for every unit you move right on the x-axis, you move down one unit on the y-axis.Finally, draw the line through these points, extending it across the graph to show all potential solutions that satisfy the equation.

Graphing Quadratic Equations

Quadratic equations often form parabola shapes on the graph, but in the case of a system like this, they can also form a circle, as seen here. The equation \( x^2 + y^2 - 4x = 0 \) can be transformed to the canonical circle equation by completing the square. Initially, the equation is manipulated to:

• \( (x^2 - 4x) + y^2 = 0 \)

To complete the square:

• Transform \( x^2 - 4x \) into \( (x - 2)^2 - 4 \).
• Combine to get \( (x-2)^2 + y^2 = 4 \).

This represents a circle centered at (2,0) with a radius of 2. Plot the center and use the radius to mark points all around this center at a constant distance, forming a circle. Ensure it is accurately placed to reflect the true geometry of the equation on the graph.

Intersection Points

Intersection points are crucial when solving systems of equations graphically, as they provide the solutions to the system. These are points where the graphs of the equations meet. In this particular system, you need to find where the line \( x + y = 4 \) intersects the circle from equation \( (x-2)^2 + y^2 = 4 \). To identify these points, overlay the graph of the linear equation (a straight line) onto the graph of the quadratic equation (a circle). Look for points where both equations are satisfied simultaneously. These intersections give possible solutions to the equations:

• Any point(s) where the line intersects the circle.

Check graphically, by calculating coordinates of intersection using both equations, or a combination of both to verify the exact points.

Completing the Square

Completing the square is a technique used to simplify expressions, making them easier to graph and understand. This method is especially useful when dealing with quadratics, including those representing circles or parabolas. The process involves turning a quadratic expression into the sum of a perfect square and a constant. For example, in the equation \( x^2 - 4x + y^2 = 0 \),

• Focus on the \( x \) part: \( x^2 - 4x \).
• To complete the square, add and subtract 4: \( (x - 2)^2 - 4 \).

Now the equation becomes \( (x-2)^2 + y^2 = 4 \). This rearranges the quadratic into a more recognizable format, such as a circle’s equation. Completing the square allows equations to be reorganized so their geometric interpretations, like circles or parabolas, can be more easily understood and graphed.