Lesson 7.4, page 746 Nonlinear Systems of Equations

Objective: To solve a nonlinear system of equations.

Review – What?

   1.

2.

3.

4.

System – 2 or more equations together Solution of system – any ordered pair that makes all equations true Possible solutions: One point More than one point No solution Infinite solutions

Review - How?

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

1.

Graphing 2.

3.

Substitution Elimination

Review Steps for using SUBSTITUTION

1.

2.

3.

4.

5.

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

Solve this equation to find a value for the variable.

Substitute again to find the value of the other variable.

Check.

Review Solve using Substitution.

2x – y = 6 y = 5x

Review STEPS for ELIMINATION

Review Solve using elimination.

  3x + 5y = 11 2x + 3y = 7

   

What’s New?

A

non-linear system

one or more of the equations has a graph that is not a line.

is one in which 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.

An Example…

x

2  Solve the following system of equations : 

y

2  9 (1)

The graph is a circle .

2 3 (2)

The graph is a line.

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

2

x

3

y x y



2

x



3 3

An Example…  Next, we substitute y = 2x  3 in equation (1) and solve for x:

x

2 

(2

x



3)

2 

9

x

2 

4

x

2 

12

x

5

x

2 

12

x x

(5

x



12)



0



0 9

x



0 or

x



12 5

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

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

y

 2

x

 3

y

 5 

y

 9 5 (0,  3) SOLUTIONS and 12 9 , 5 5

An Example…

Check: (0,  3)

x

2

y

9 2 0 2 9 3 Check: 

x

2

y

    12 9 , 5 5 9    2

   

5  5 2  12 5   5 3  Visualizing the Solution

See Example 1, page 747

Check Point 1: Solve by substitution.

 x 2 = 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 = 4 3x + 2y =  10

Solve xy = 4 for y.

xy

 4

y

 4

x

Substitute into 3

x

+ 2

y

=  10.

3

x

 2

y

  10 3

x

 2( )

x

  10 3

x x

 3

x

 8

x



x

  10 3

x

2 3

x

2  10

x

0 10

x

 Use the quadratic formula (or factor) to solve: 3

x

2  10

x

0

x x x

  

b

2  4

ac

2

a

10 2  2(3) 100  96 6

x

 4  6

x



x

 2 and 6  4 and 3  2 6 6 2 2

Substitute values of

x

to find

y

. 3x + 2y =  10  Visualizing the Solution x =  4/3 x =  2 3   3 4  2

y

  10 3   2

y

  10 2

y

  10 2

y

  10 2

y

  6

y

  3 2

y

  4

y

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

Need to watch another one?

 Solve the system of equations:

5

x

2 

2

y

2  

13 3

x

2 

4

y

2 

39

5

x

2  2

y

2   13 3

x

2  4

y

2  39  Solve by elimination . Multiply equation (1) by 2 and add to eliminate the y 2 term.

10

x

2 

4

y

2  

26 3

x

2 

4

y

2 

39 13

x

2 

13

x

2 

1

 

1

Substituting

x

=  1 in equation (2) gives us: 3

x

2 x = 1  4

y

2  4

y

2 4

y

2

y

2 2

y

   39 3 9 36  9   3 3

x

2  x = -1 4

y

2  39 3(  1 ) 2  4

y

2  39 4

y

2  36

y

2  9

y

  3  The possible solutions are (1, 3), (  1, 3), (  1,  3) and (1,  3).

All four pairs check, so they are the solutions.

 Visualizing the Solution

See Example 3, page 749

Check Point 3: Solve by elimination.

 3x 2 + 2y 2 = 35  4x 2 + 3y 2 = 48

See Example 4, page 750

Check Point 4: Solve by elimination.

 y = x 2 + 5  x 2 + y 2 = 25