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Chapter 5: Problem 43

Solve the system graphically. $$\left\\{\begin{array}{r}x+y=4 \\ x^{2}+y^{2}-4 x=0\end{array}\right.$$

Short Answer

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

Step by step solution

01

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.
02

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.
03

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).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The given linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the maximum value of the objective function and where it occurs. Objective function: \(z=2.5 x+y\) Constraints: \(\begin{aligned} x & \geq 0 \\ y & \geq 0 \\ 3 x+5 y & \leq 15 \\ 5 x+2 y & \leq 10 \end{aligned}\)

Peregrine Falcons The numbers of nesting pairs \(y\) of peregrine falcons in Yellowstone National Park from 2001 to 2005 can be approximated by the linear model \(y=3.4 t+13, \quad 1 \leq t \leq 5\) where \(t\) represents the year, with \(t=1\) corresponding to 2001\. (Sounce: Yellowstone Bird Report 2005) (a) The total number of nesting pairs during this five-year period can be approximated by finding the area of the trapezoid represented by the following system. \(\left\\{\begin{array}{l}y \leq 3.4 t+13 \\ y \geq 0 \\ t \geq 0.5 \\\ t \leq 5.5\end{array}\right.\) Graph this region using a graphing utility. (b) Use the formula for the area of a trapezoid to approximate the total number of nesting pairs.

Graph the solution set of the system of inequalities. $$\left\\{\begin{array}{r}2 x^{2}+y>4 \\ x<0 \\ y<2\end{array}\right.$$

MAKE A DECISION: DIET SUPPLEMENT A dietitian designs a special diet supplement using two different foods. Each ounce of food \(\mathrm{X}\) contains 20 units of calcium, 10 units of iron, and 15 units of vitamin \(\mathrm{B}\). Each ounce of food \(\mathrm{Y}\) contains 15 units of calcium, 20 units of iron, and 20 units of vitamin \(\mathrm{B}\). The minimum daily requirements for the diet are 400 units of calcium, 250 units of iron, and 220 units of vitamin B. (a) Find a system of inequalities describing the different amounts of food \(\mathrm{X}\) and food \(\mathrm{Y}\) that the dietitian can use in the diet. (b) Sketch the graph of the system. (c) A nutritionist normally gives a patient 18 ounces of food \(\mathrm{X}\) and \(3.5\) ounces of food \(\mathrm{Y}\) per day. Supplies of food \(\mathrm{X}\) are running low. What other combinations of foods \(\mathrm{X}\) and \(\mathrm{Y}\) can be given to the patient to meet the minimum daily requirements?

Optimal Profit A company makes two models of a patio furniture set. The times for assembling, finishing, and packaging model \(\mathrm{A}\) are 3 hours, \(2.5\) hours, and \(0.6\) hour, respectively. The times for model \(\mathrm{B}\) are \(2.75\) hours, 1 hour, and \(1.25\) hours. The total times available for assembling, finishing, and packaging are 3000 hours, 2400 hours, and 1200 hours, respectively. The profit per unit for model \(\mathrm{A}\) is $$\$ 100$$ and the profit per unit for model \(\mathrm{B}\) is $$\$ 85 .$$ What is the optimal production level for each model? What is the optimal profit?
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