Transcript 10-7

10-7
10-7Solving
SolvingNonlinear
NonlinearSystems
Systems
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
10-7 Solving Nonlinear Systems
Warm Up
Solve by substitution.
1.
3x + 4y = 15
x = 3; y = 1.5
x = 6y – 6
Solve by elimination.
2.
Holt Algebra 2
3x + 4y = 57
5x – 4y = – 1
x = 7; y = 9
10-7 Solving Nonlinear Systems
Objective
Solve systems of equations in two
variables that contain at least one
second-degree equation.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Vocabulary
nonlinear system of equations
Holt Algebra 2
10-7 Solving Nonlinear Systems
A nonlinear system of equations is a system in
which at least one of the equations is not linear.
You have been studying one class of nonlinear
equations, the conic sections.
The solution set of a system of equations is the set
of points that make all of the equations in the
system true, or where the graphs intersect. For
systems of nonlinear equations, you must be aware
of the number of possible solutions.
Holt Algebra 2
10-7 Solving Nonlinear Systems
You can use your graphing calculator to find solutions
to systems of nonlinear equations and to check
algebraic solutions.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 1: Solving a Nonlinear System by Graphing
Solve
x2 + y2 = 25
2
2
4x + 9y = 145
by graphing.
The graph of the first equation is a circle, and the
graph of the second equation is an ellipse, so there
may be as many as four points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 1 Continued
Step 1 Solve each equation for y.
Solve the first equation for y.
Solve the second equation for y.
Step 2 Graph the system on your calculator, and
use the intersect feature to find the solution set.
The points of
intersection are
(–4, –3), (–4, 3),
(4, –3), (4, 3).
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 1
Solve
3x + y = 4.5
y=
1
(x – 3)2
2
by graphing.
The graph of the first equation is a straight line,
and the graph of the second equation is a
parabola, so there may be as many as two
points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 1 Continued
Step 1 Solve each equation for y.
y = –3x + 4.5
y=
1
2
Solve the first equation for y.
(x – 3)2
Step 2 Graph the system on your calculator, and
use the intersect feature to find the solution set.
The point of intersection is
(0, 4.5).
Holt Algebra 2
10-7 Solving Nonlinear Systems
The substitution method for solving linear
systems can also be used to solve nonlinear
systems algebraically.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 2: Solving a Nonlinear System by
Substitution
Solve
x2 + y2 = 100
1
y=
2
2
x – 26
by substitution.
The graph of the first equation is a circle, and the
graph of the second equation is a parabola, so
there may be as many as four points of
intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 2 Continued
Step 1 It is simplest to solve for x2 because both
equations have x2 terms.
2 in the second
2
Solve
for
x
x = 2y + 52
equation.
Step 2 Use substitution.
(2y + 52) + y2 = 100
y2 + 2y – 48 = 0
(y + 8) (y – 6) = 0
y = –8 or y = 6
Holt Algebra 2
Substitute this value into the
first equation.
Simplify, and set equal to 0.
Factor.
10-7 Solving Nonlinear Systems
Example 2 Continued
Step 3 Substitute –8 and 6 into x2 = 2y + 52 to
find values of x.
x2 = 2(–8) + 52 = 36
x = ±6
(6, –8) and (–6, –8) are solutions.
x2 = 2(6) + 52 = 64
x = ±8
(8, 6) and (–8, 6) are solutions.
The solution set of the system is {(6, –8) (–6, –8),
(8, 6), (–8, 6)}.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 2 Continued
Check Use a graphing calculator. The graph supports
that there are four points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2a
Solve the system of equations by using the
substitution method.
x + y = –1
2
2
x + y = 25
The graph of the first equation is a line, and the
graph of the second equation is a circle, so there
may be as many as two points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2a Continued
Step 1 Solve for x.
x = –y – 1
Solve for x in the first equation.
Step 2 Use substitution.
2
2
(–y – 1) + y = 25
y2 + 2y + 1 + y2 – 25 = 0
Substitute this value into the
second equation.
Simplify and set equal to 0.
2y2 +2y – 24 = 0
2[(y2 + y – 12)] = 0
2(y + 4)(y – 3) = 0
y = –4 or y = 3
Holt Algebra 2
Factor.
10-7 Solving Nonlinear Systems
Check It Out! Example 2a Continued
Step 3 Substitute –4 and 3 into x + y = –1 to
find values of x.
x + (–4) = –1
x=3
(3, –4) is a solution.
x + (3) = –1
x = –4
(–4, 3) is a solution.
The solution set of the system is {(3, –4), (–4, 3)}.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2a Continued
Check Use a graphing calculator. The graph supports
that there are two points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2b
Solve the system of equations by using the
substitution method.
x2 + y2 = 25
y – 5 = –x2
The graph of the first equation is a circle, and the
graph of the second equation is a parabola, so there
may be as many as four points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2b Continued
2
Step 1 It is simplest to solve for x because both
equations have x2 terms.
2 in the second
2
Solve
for
x
x = –y + 5
equation.
Step 2 Use substitution.
(–y + 5) + y2 = 25
y2 – y – 20 = 0
(y + 4) (y – 5) = 0
y = –4 or y = 5
Holt Algebra 2
Substitute this value into the
first equation.
Simplify, and set equal to 0.
Factor.
10-7 Solving Nonlinear Systems
Check It Out! Example 2b Continued
Step 3 Substitute –4 and 5 into x2 + y2 = 25 to
find values of x.
x2 + (–4)2 = 25
x = ±3
(3, –4) and (–3, –4) are solutions.
x2 +(5)2 = 25
x=0
(0, 5) is a solution.
The solution set of the system is {(3, –4), (–3, –4),
(0, 5)}.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 2b Continued
Check Use a graphing calculator. The graph supports
that there are three points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
The elimination method can also be used to solve
systems of nonlinear equations.
Remember!
In Example 3, you can check your work on a
graphing calculator.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 3: Solving a Nonlinear System by
Elimination
Solve
4x2 + 25y2 = 41
36x2 + 25y2 = 169
by using the
elimination method.
The graph of the first equation is an ellipse, and the
graph of the second equation is an ellipse, There
may be as many as four points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 3 Continued
Step 1 Eliminate y2.
36x2 + 25y2 = 169
–4x2 – 25y2 = –41
32x2
= 128
x2 = 4, so x = ±2
Holt Algebra 2
Subtract the first equation
from the second.
Solve for x.
10-7 Solving Nonlinear Systems
Example 3 Continued
Step 2 Find the values for y.
2
4(4) + 25y = 41
2
16 + 25y = 41
Substitute 4 for x2.
Simplify.
25y2 = 25
y = ±1
The solution set of the system is {(–2, –1),
(–2, 1), (2, –1), (2, 1)}.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 3
Solve
25x2 + 9y2 = 225
25x2 – 16y2 = 400
by using the
elimination method.
The graph of the first equation is an ellipse, and the
graph of the second equation is a hyperbola, There
may be as many as four points of intersection.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 3 Continued
Step 1 Eliminate x2.
25x2 – 16y2 = 400
–25x2 – 9y2
= –225
–25y2 = 175
y2 = –7
Subtract the first equation
from the second.
Solve for y.
There is no real solution of the system.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 4: Problem-Solving Application
Suppose that the paths of two boats are
2
2
modeled by 36x + 25y = 900 and y =
0.25x2 – 6. How many possible collision
points are there?
1
Understand the Problem
There is a potential danger of a collision if the
two paths cross. The paths will cross if the
graphs of the equations intersect.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 4 Continued
1
Understand the Problem
List the important information:
• 36x2 + 25y2 = 900 represents the path of the
first boat.
• y = 0.25x2 – 6 represents the path of the
second boat.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 4 Continued
2
Make a Plan
To see if the graphs intersect, solve the system
36x2 + 25y2 = 900
y = 0.25x2 – 6
Holt Algebra 2
.
10-7 Solving Nonlinear Systems
Example 4 Continued
3
Solve
The graph of the first equation is an ellipse, and the
graph of the second equation is a parabola. There
may be as many as four points of intersection.
x2 = 4y + 24
Solve the second equation
2
for x .
36(4y + 24) + 25y2 = 900 Substitute this value into
the first equation.
2
25y + 144y – 36 = 0
Simplify, and set equal to 0.
Use the quadratic formula.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 4 Continued
Substitute y = 0.24 and y = –6 into x2 = 4y + 24
to find the values for x.
x2 = 4(0.24) + 24
x  ±5
x2 = 4(–6) + 24
x=0
There are three real solutions to the system, (0, –6),
(5, 0.24), and (–5, 0.24), so there are 3 possible
collision points.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 4 Continued
4
Look Back
The graph supports
that there are three
points of intersection.
Because the paths
intersect, the boats
are in danger of
colliding if they arrive
at one of the three
points of intersection
at the same time.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 4
What if …? Suppose the paths of the boats can
be modeled by the system
36x2 + 25y2 = 900
1
y+2=–
10
x2
Is there any danger of collision?
Holt Algebra 2
.
10-7 Solving Nonlinear Systems
Check It Out! Example 4 Continued
1
Understand the Problem
There is a potential danger of a collision if the two
paths cross. The paths will cross if the graphs of
the equations intersect.
List the important information:
• 36x2 + 25y2 = 900 represents the path of the
first boat.
1
• y+2=–
x2 represents the path of the
10
second boat.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 4 Continued
2
Make a Plan
To see if the graphs intersect, solve the system
36x2 + 25y2 = 900
y+2=–
Holt Algebra 2
1
x2
10
.
10-7 Solving Nonlinear Systems
Check It Out! Example 4 Continued
3
Solve
The graph of the first equation is an ellipse, and the
graph of the second equation is a parabola. There
may be as many as four points of intersection.
x2 = –10y – 20
Solve the second equation
for x2.
2
36(–10y – 20) + 25y = 900 Substitute this value into
the first equation.
2
25y – 360y – 1620 = 0
Use the quadratic formula.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 4 Continued
Substitute y = 18 and y = –3.6 into x2 = –10y – 20
to find the values for x.
x2 = –10(18) – 20
x2 = –10(–3.6) – 20
x2 = –200 There are no
real values.
x = ±4
There are two real solutions to the system, (4, –3.6) and
(–4, –3.6), so there are 2 possible collision points.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Check It Out! Example 4 Continued
4
Look Back
The graph supports
that there are two
points of intersection.
Because the paths
intersect, the boats are
in danger of colliding if
they arrive at the
intersections (4, –3.6)
or (–4, –3.6) at the
same time.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Lesson Quiz
1. Solve
4x2 – 9y2 = 108
x2 + y2 = 40
2. Solve
2x2 + y2 = 54
by graphing.
(±6, ±2)
by substitution. (±5, ±2)
x2 – 3y2 = 13
3. Solve
4x2 + 4y2 = 52
2
2
9x – 4y = 65
Holt Algebra 2
by elimination. (±3, ±2)