10-7 Solving Nonlinear Systems
Objective
Solve systems of equations in 2 variables that
contain at least 1 second-degree equation.
Holt Algebra 2
10-7 Solving Nonlinear Systems
Notes - Identify the graph and solve systems
1. Solve
2. Solve
3. Solve
4. Solve
4x2 – 9y2 = 108
x2 + y2 = 40
2x2 + y2 = 54
x2 – 3y2 = 13
4x2 + 4y2 = 52
9x2 – 4y2 = 65
x + y = –1
x2 + y2 = 25
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 1: Solving a Nonlinear System by Substitution
x2 + y2 = 100
Solve
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 4 intersection points.
Step 1 It is simplest to solve for x2 because both equations
have x2 terms.
2
x = 2y + 52
Holt Algebra 2
10-7 Solving Nonlinear Systems
Example 1 Continued
Step 2 Use substitution.
(2y + 52) + y2 = 100
2
y + 2y – 48 = 0
(y + 8) (y – 6) = 0
Substitute this value into the
first equation.
y = –8 or y = 6
Step 3 Complete ordered pairs.
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: Solving a Nonlinear System by Addition
Solve
4x2 + 25y2 = 41
36x2 + 25y2 = 169
by using addition.
The graph of both are ellipses, so there may be as many as
four intersection points
Step 1 Make opposites to eliminate y2.
36x2 + 25y2 = 169
–4x2 – 25y2 = –41
32x2
= 128
x2 = 4, so x = ±2
Step 2 Find the values for y.
Holt Algebra 2
The solutions are
(–2, –1),(–2, 1),
(2, –1), (2, 1)
10-7 Solving Nonlinear Systems
Notes - Identify the graph and solve systems
4x2 – 9y2 = 108
1. Solve
2. Solve
3. Solve
4. Solve
2
2
x + y = 40
2x2 + y2 = 54
x2 – 3y2 = 13
4x2 + 4y2 = 52
2
2
9x – 4y = 65
x + y = –1
x2 + y2 = 25
Holt Algebra 2
(±6, ±2)
(±5, ±2)
(±3, ±2)
(3, –4), (–4, 3)
10-7 Solving Nonlinear Systems
Solving Systems: Extra Info
The following power-point slides contain extra
examples and information.
Reminder: Lesson Objectives
Solve systems of nonlinear systems of equations
(by graphing, substitution, and the addition methods).
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 1: Solving a Nonlinear System by Graphing
Solve
x2 + y2 = 25
4x2 + 9y2 = 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.
The points of
intersection are
(–4, –3), (–4, 3),
(4, –3), (4, 3).
Holt Algebra 2