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 = 6y – 6
Solve by elimination.
2.
Holt Algebra 2
3x + 4y = 57
5x – 4y = – 1
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=
Holt Algebra 2
1
(x – 3)2
2
by graphing.
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
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
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
elimination method.
Holt Algebra 2
by using the