1. Nonlinear Systems of Equations and Inequalities
Section 10.4
Nonlinear Systems of Equations and Inequalities
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2. Objectives Solve a nonlinear system of equations.
Use nonlinear systems of equations to solve applied problems.
Graph nonlinear systems of inequalities.

3. Nonlinear Systems of Equations
The graphs of the equations in a nonlinear system of equations can have no point of intersection or one or more points of intersection. The coordinates of each point of intersection represent a solution of the system of equations. When no point of intersection exists, the system of equations has no real-number solution. We can solve nonlinear systems of equations by using the substitution or elimination method.

4. Example Solve the following system of equations:
We use the substitution method. First, we solve equation (2) for y.

5. Example (continued) Next, we substitute
Now, we substitute these numbers for x in equation (3) and solve for x. Check each solution.
in equation (1) and solve for x:

6. Example (continued) Check: (4, 3) TRUE TRUE Check: (–4, –3) TRUE TRUE
They check, so (4, 3) and (–4, –3) are the solutions.

7. Example (continued) - Graphical Solution
Graph both equations and use the INTERSECT feature twice to find both solutions.
This solution is (4, 3).

8. Example (continued) - Graphical Solution
Graph both equations and use the INTERSECT feature twice to find both solutions.
This solution is (–4, –3).

9. Example Solve the following system of equations:
We use the substitution method, substituting 3 – x2 for y in equation (1).
Next, use the quadratic formula:

10. Example (continued)
Now, we substitute these values for x in equation (1) and solve for y.

11. Example (continued) The solutions are
There are no real number solutions.

12. Example (continued) - Graphical Solution
We graph both equations in the viewing window.
There are no points of intersection, so no real-number solutions. Algebra must be used to find all solutions.

13. Example Solve the following system of equations:
We use the elimination method. Multiply equation (2) by 5 and add to eliminate the y2 term.

14. Example (continued)
Now, we substitute these values for x in equation (2) and solve for y.
The possible solutions are:
All four pairs check, so they are the solutions.

15. Example (continued) - Graphical Solution
We graph both equations in the viewing window and use the INTERSECT feature four times to find the four solutions.
The calculator gives approximate decimal solutions.
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16. Modeling and Problem Solving - Example
For a student recreation building at Southport Community College, an architect wants to lay out a rectangular piece of land that has a perimeter of 204 m and an area of 2565 m2. Find the dimensions of the piece of land.
1. Familiarize. Make a drawing and label it.
2. Translate. We now have the following:
Perimeter: 2w + 2l = 204
Area: lw = 2565
Area = lw
= 2565 m2 w l

17. Example (continued)
3. Carry out. Solve the system. Solving the second equation for l gives us l = 2565/w. We then substitute into equation (1) and solve for w.

18. Example (continued) If w = 57, then l = 2565/w = 2565/57 = 45.
Since the length is generally considered to be longer than the width, we have the solution l = 57 and w = 45.
4. Check. If l = 57 and w = 45, the perimeter is
2(45) + 2(57) = 204.
The area is 57(45) = The numbers check.
5. State. The length of the piece of land is 57 m and the width is 45 m.

19. Example Graph the solution set of the system
Solution: Graph x2 + y2 ≤ 25 by first graphing the equation of the circle x2 + y2 = 25. We also use a solid line since the inequality symbol is ≤. The test point (0, 0) is a solution, so the region containing (0,0) is shaded in red.
Now graph the dashed line 3x – 4y > 0. We choose the point (0, 2). It is not a solution of 3x – 4y > 0, so we shade the half plane that does not contain (0, 2) in green.

20. Example (continued)
The solution set is the region shaded both red and green.
To find the points of intersection, solve the system of equations:
In example 1, we found that these points are:
(4, 3) and (–4, –3).