Intermediate Algebra (11th edition)

Since neither equation has a squared term, we can solve either equation for one of

the variables and then substitute the result into the other equation. Solving for

xgives We substitute for xin equation (1).

(1)

Let.

Multiply.

Multiply by y,.

Standard form

Factor.

or Zero-factor property

We substitute these results into to obtain the corresponding values of x.

If then

If then

The solution set of the system is

See the graph in FIGURE 29.

NOW TRY

OBJECTIVE 2 Solve a nonlinear system by elimination.We can often use

the elimination method (Section 4.1)when both equations of a nonlinear system are

second degree.

Solving a Nonlinear System by Elimination

Solve the system.

(1)

(2)

The graph of (1) is a circle, while the graph of (2) is a hyperbola. By analyzing

the possibilities, we conclude that there may be zero, one, two, three, or four points of

intersection. Adding the two equations will eliminate y.

(1)

(2)

Add.

Divide by 3.

x= 1 or x=- 1 Square root property

x^2 = 1

3 x^2 = 3

2 x^2 - y^2 =- 6

x^2 + y^2 = 9

2 x^2 - y^2 =- 6

x^2 +y^2 = 9

EXAMPLE 3

ea

4

3

, 3b, a-

1

2

, - 8 bf.

x=-

1

2

y=- 8 ,.

x=

4

3

y= 3 ,.

x=

4

y

y= 3 y=- 8

1 y- 321 y+ 82 = 0

y^2 + 5 y- 24 = 0

24 - y^2 = 5 y yZ 0

24

y

- y= 5

6 a x=^4 y

4

y

b -y= 5

6 x-y= 5

4

x= y

4

y.

xy= 4

(^43 , 3)

x

y

Hyperbola: xy = 4

Line: 6x – y = 5

0

(–, –8^12 )

FIGURE 29

NOW TRY
EXERCISE 2
Solve the system.

x- 3 y= 1

xy= 2

NOW TRY ANSWER

2. E 1 - 2, - 12 , A3,^23 BF

SECTION 11.4 Nonlinear Systems of Equations 659