Lesson 7.4, page 746Nonlinear Systems of Equations

Objective: To solve a nonlinear system of equations.

System – 2 or more equations together

Solution of system – any ordered pair that makes all equations true

Possible solutions:1. One point2. More than one point3. No solution4. Infinite solutions

Review – What?

Review - How?

What methods have we used to solve linear systems of equations?

1. Graphing2. Substitution3. Elimination

ReviewSteps for using SUBSTITUTION

1. Solve one equation for one variable. (Hint: Look for an equation already solved for a variable or for a variable with a coefficient of 1 or -1.)

2. Substitute into the other equation.3. Solve this equation to find a value for the

variable.4. Substitute again to find the value of the

other variable.5. Check.

ReviewSolve using Substitution.

2x – y = 6 y = 5x

ReviewSTEPS for ELIMINATION

ReviewSolve using elimination.

3x + 5y = 11 2x + 3y = 7

What’s New?

A non-linear system is one in which one or more of the equations has a graph that is not a line.

With non-linear systems, the solution could be one or more points of intersection or no point of intersection.

We’ll solve non-linear systems using substitution or elimination.

A graph of the system will show the points of intersection.

Solve the following system of equations:

2 2 The graph is a circle9 (1)

2 3 (2)

.

The graph is a line.

x y

x y

An Example…

An Example…

We use the substitution method. First, we solve equation (2) for y.

2 3

2 3

2 3

x y

y x

y x

An Example…

Next, we substitute y = 2x 3 in equation (1) and solve for x: 2 2

2 2

2

(2 3) 9

4 12 9 9

5 12 0

(5 12) 0

120 or

5

x x

x x x

x x

x x

x x

An Example… …

Now, we substitute these numbers for x in equation (2) and solve for y.

x = 0 x = 12 / 5y = 2x 3y = 2(0) 3y = 3

SOLUTIONS

(0, 3) and

125

2 3

2( ) 3

9

5

y x

y

y

12 9,

5 5

An Example…

Check: (0, 3)

Check:

Visualizing the Solution

12 9,

5 5

2 2

2 3

9 2 3

0 3 9 2(0) ( 3) 3

9 9 3 3

x y x y

2 2

2 29 912 125 5 5 5

9 2 3

9 2( ) ( ) 3

9 9 3 3

x y x y

See Example 1, page 747

Check Point 1: Solve by substitution. x2 = y – 1 4x – y = -1

See Example 2, page 748

Check Point 2: Solve by substitution.x + 2y = 0(x – 1)2

+ (y – 1)2 = 5

Another example to watch…

Solve the following system of equations: xy = 43x + 2y = 10

Solve xy = 4 for y. 4

4

xy

yx

Substitute into 3x + 2y = 10.

4

8

8

2

2

3 2 10

3 2( ) 10

3 10

3 10( )

3 8 10

3 10 8 0

x

x

x

x y

x

x

x x x

x x

x x

Use the quadratic formula (or factor)to solve:

2

2

4

2

10 10 4(3 8)

2(3)

10 100 96

6

10 4 10 2

6 610 2 10 2

and 6 6

4 and 2

3

b b acx

a

x

x

x

x

x

23 10 8 0x x

Substitute values of x to find y.

3x + 2y = 10

x = 4/3 x = 2

The solutions are (4/3, 3) and (2, 2).

433 2 10

4 2 10

2 6

3

y

y

y

y

3 2 2 10

6 2 10

2 4

2

y

y

y

y

Visualizing the Solution

Need to watch another one?

Solve the system of equations:

2 2

2 2

5 2 13

3 4 39

x y

x y

Solve by elimination. Multiply equation (1) by 2 and add to eliminate the y2 term.

2 2

2 2

2

2

10 4 26

3 4 39

13 13

1

1

x y

x y

x

x

x

2 2

2 2

5 2 13

3 4 39

x y

x y

Substituting x = 1 in equation (2) gives us:

x = 1 x = -1

The possible solutions are (1, 3), (1, 3), (1, 3) and (1, 3).

2 2

2 2

2

2

3 4 39

3( ) 4 3

3

9

4 36

9

1

x y

y

y

y

y

2 2

2 2

2

2

3

3 4 39

3( ) 4 39

4 36

9

1

x y

y

y

y

y

All four pairs check, so they are the solutions.

Visualizing the Solution

See Example 3, page 749

Check Point 3: Solve by elimination. 3x2 + 2y2 = 35 4x2 + 3y2 = 48

See Example 4, page 750

Check Point 4: Solve by elimination. y = x2 + 5 x2 + y2 = 25