Problem 43 - College Algebra and Calculus: An Applied Approach

Question

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

The solutions of the system are the points (0,4) and (4,0).

Step 1: Rearrange Equations

First, rewrite both equations so that y is expressed as a function of x. For the first equation \(x + y = 4\), we subtract x from both sides to get \(y = 4 - x\). For the second equation \(x^{2} + y^{2} - 4x = 0\), we rearrange it to get \(y^{2} = 4x - x^{2}\), then we take the square root of both sides. As there are positive and negative roots for any square, we get two functions for y: \(y = \sqrt{4x - x^{2}}\) and \(y = -\sqrt{4x - x^{2}}\). Note that because equation 2 equals zero when x=0 or x=4, square roots will only be real numbers for x within this interval.

Step 2: Plot the Graphs

Next, plot the functions \(y = 4 - x\), \(y = \sqrt{4x - x^{2}}\), and \(y = -\sqrt{4x - x^{2}}\). The linear function will be a descending line crossing the y-axis at (0,4) and the x-axis at (4,0). The sqrt function will represent the upper half of a circle with radius 2 centered at (2,0), and the -sqrt function will represent the lower half.

Step 3: Identify Points of Intersection

Those points where the line crosses the circle represent the solutions of the system. Looking at the graph, it's easy to see that these points are (0,4) and (4,0).

Key Concepts

Graphical Solution

To solve a system of equations graphically means to plot each equation on the same set of axes and look for any common points, known as intersection points. For this particular problem, we have two equations: one linear and one that forms a circle.

• The first step is to rewrite each equation to solve for one variable in terms of the other, typically y in terms of x.
• Once the equations are rearranged, you plot them on a graph.

When graphed, the visual representation allows you to see where these plots intersect. These intersections show solutions to the system where both equations are simultaneously satisfied. This method is particularly helpful for visual learners and when equations are challenging to solve algebraically.

Intersection Points

Intersection points are key in identifying where the solutions lie in a system of equations. In our problem, they are where the graphs of the linear equation and the circle equation meet.

• Each point of intersection satisfies both equations, meaning if you plug these values back into the original equations, they will make both true.
• Physically, on a graph, these points appear where the line intersects the shape (in this case, a circle).

For the given equations, the graph visually reveals intersection at the points (0,4) and (4,0). It's like finding common meeting points between paths, just for equations on a graph.

Circle Equation

In our system, one equation rearranges to form parts of a circle. A general circle's equation follows the format of \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) the radius.

• After rearranging the given equation \( x^2 + y^2 - 4x = 0 \), we get the circle's parts: \( y^2 = 4x - x^2 \).
• Taking the square root gives two equations, capturing both upper and lower halves of a circle.

In our specific problem, this circle has its center at (2,0), with a radius of 2. The equations \(y = \sqrt{4x - x^2}\) and \(y = -\sqrt{4x - x^2}\) represent the upper and lower semicircles respectively.

Linear Equation

A linear equation describes a straight line in a graph and is often in the slope-intercept form: \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

• The given linear equation \(x + y = 4\) simplifies to \(y = 4 - x\).
• Here, the slope \(m\) is -1, indicating a downward slant as you move from left to right.
• The y-intercept \(b\) is 4, where the line cuts the y-axis.

Graphically, this line straight away tells us it will run exactly through the point (0,4), sloping down to (4,0). It's straightforward but crucial as it determines part of our system's solution.