Transcript 5.5 Powerpoint

```Chapter 5.5
Nonlinear Systems of Equations
Solving Nonlinear Systems with Real Solutions
A system of equations in which at least one
equation is not linear is called a nonlinear
system.
The substitution method works well for
solving many such systems, particularly
when one of the equations is linear, as in the
next example.
Solve the system.
x y4
2
x  y  2
x   2  x  4
2
x 2 x  4
2
x  x2  0
2
y  2  x
Solve the system.
x  x2  0
2
x y4
2
x  y  2
y  2  x
x  2x 1  0
x  2  0 or x  1  0
x  2 or x  1
y  2  - 2
y  2  1
y  3
y0
the solution set of the system is
{(-2, 0), (1, -3)}
Here is a graph
of the system.
Caution: If we had solved for x in equation
(2) to begin the algebraic solution in
Example, we would find y = 0 or y = -3.
Substituting y = 0 into equation 1 gives
x2 = 4, so x = 2 or x = -2.
The ordered pair (2, 0) does not satisfy
equation 3 however.
This shows the necessity of checking by
substituting all potential solutions into each
equations of the system.
Visualizing the types of graphs involved in a
nonlinear system helps predict the possible
numbers of ordered paris of real numbers
that may be in the solution set of the system.
(Graphs of some nonlinear equations were
discussed in Chapter 2)
For example, a line and a parabola may have
0, 1, or 2 points of intersection as shown in
the figure.
Solve the system.
x y 4
2
2
2x  y  8
2
3x
 12
2
x
4
x  2 or x  -2
2
2
Solve the system.
x  3xy  y  22
2
2
x  xy  y  6
4 xy
 16
16
y

4x
  
2
x  3x   
  
2
2
4

x
2

  22

x  12x  16  22x
Solve the system.
4
x  3xy  y  22
2
2
x  xy  y  6
2
2
2
2
x  10x  16  0
4
2
( x  2)(x  8)  0
2
2
2
x 2
2
x 8
2
4  4
x  3x      22
 x  x
x   2 x  2 2
16
2
x  12  2  22
x
2
Solve the system.
x  3xy  y  22
2
2
x 2
4
y
x
4
 2 2
y
 2
x  2 2
4
y
x
4
y
 2
2 2
Solve the system.
x  y  16
2
2
x y4
4  y
2
 y  16
2
16  8 y  y  y  16
2
2y  8y  0
2
2
2 y( y  4)  0
2 y  0 or y - 4  0
y  0 or y  4
x 0  4
x 4 4
x 4
x  4
x 0
x0
The solution set is {(-4,0),(4,0),(0,4)}
Solve the system.
x  y 5
- 3x  3 y  15
4x  3 y  11
2
3x
 12
2
x
4
4x  3 y  11
2
x
-4
x  2i
2
2
2
2
 2i
 y 5
2
4 y  5
2
y 9
2
2
2
2
2
2
y 9
2
y  3
Solve the system.
x  y 5
- 3x  3 y  15
4x  3 y  11
2
3x
 12
2
x
4
4x  3 y  11
2
x
-4
x  2i
2
2
2
2
2
The solution set is
{(2i, 3), (2i, -3), (-2i, 3), (-2i, -3)
2
2
2
y  3
A box with an open top has a square base and
four sides of equal height. What are the
dimensions of the box.
Volume  length  width  height
Volume  x y  75
2
Surface Area 
x  x  xy  xy  xy  xy  85
2
2
x y  75
2
x  4xy  75
2
75
y 2
x
 75 
x  4x 2   75
x 
2
 75 
x  4x 2   75
x 
2
 75 
x  4   75
 x
 300
2
x 
  75
 x 
2
x  300  75x
3
x  300  75x
3
x  75x  300  0
3
We are restricted to positive values for x,
considering the nature of the problem, any
solution should be relatively small.
x  75x  300  0
3
By the rational zeros theorem, factors of 300
are the only possible rational solutions. Using
synthetic division, we see that 5 is a solution.
5
1
0
5
-85
25
1
5
-60
300
-300
x  75x  300  0
3
5
1
0
5
-85
25
1
5
-60
300
-300
75
Therefore one value of x is 5 and y  2  3
5
We must now solve
x2 +5x – 60 = 0
for any other possible positive solutions.
x2 +5x – 60 = 0
Using the quadratic formula, the positive
solution is
- 5  5  41 60 - 5  25  240
x

21
2
2
- 5  265  5  16.28 11 .28


 5.64

2
2
2
75
75
y

 2.36
2
31 .8096
5.64
First answer: length = width = 5 in
height = 3 in
First answer: length = width = 5.64 in
height = 2.36 in
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