Intermediate Algebra (11th edition)

Each value of xgives corresponding values for ywhen substituted into one of the

original equations. Using equation (1) gives the following.

660 CHAPTER 11 Nonlinear Functions, Conic Sections, and Nonlinear Systems

(1)

Let

or

or

(1)

Let

y= 222 or y=- 222

y^2 = 8

1 - 122 +y^2 = 9 x=-1.

x^2 +y^2 = 9

y= 222 y=- 222

y= 28 y=- 28

y^2 = 8

12 +y^2 = 9 x=1.

x^2 +y^2 = 9

The solution set is

FIGURE 30shows the four points of intersection.

NOW TRY

OBJECTIVE 3 Solve a nonlinear system that requires a combination of

methods.

Solving a Nonlinear System by a Combination of Methods

Solve the system.

(1)

(2)

While we have not graphed equations like (1), its graph is a hyperbola. The graph

of (2) is also a hyperbola. Two hyperbolas may have zero, one, two, three, or four

points of intersection. We use the elimination method here in combination with the

substitution method.

(1)

Multiply (2) by.

2 xy = 4 Add.

- x^2 +y^2 =- 3 - 1

x^2 + 2 xy-y^2 = 7

x^2 - y^2 = 3

x^2 + 2 xy-y^2 = 7

EXAMPLE 4

A-1, 2 22 B, A-1, - 222 BF.

EA1, 2 22 B, A1, - 222 B,

x

y

0

Circle: x^2 + y^2 = 9

(–1, 2√2) (1, 2√2)

(–1, –2√2) (1, –2√2)

Hyperbola: 2x^2 – y^2 = –6

FIGURE 30

The - and -terms

were eliminated.

x^2 y^2

Next, we solve for one of the variables. We choose y.

Divide by 2x. (3)

Now, we substitute into one of the original equations.

The substitution is easier in (2).

Let

Square

x^4 - 4 = 3 x^2 Multiply by x 2 , xZ0.

2

x x.

(^2) -^4

x^2

= 3

x 2 - a y=^2 x.

2

x

b

2

= 3

x^2 - y^2 = 3

y=

2

x

y=

2

x

2 xy= 4

2 xy= 4

NOW TRY
EXERCISE 3
Solve the system.

4 x^2 + 13 y^2 = 100

x^2 + y^2 = 16

NOW TRY ANSWER

3. 
   

A- 223 , 2B, A- 223 , - 2 BF

EA 223 , 2B, A 223 , - 2 B,