Intermediate Algebra (11th edition)

SECTION 11.4 Nonlinear Systems of Equations 661

Subtract

Factor.

or Zero-factor property

or Solve each equation.

or or or

Substituting these four values into (equation (3)) gives the corresponding val-

ues for y.

then

then

then

then

If we substitute the x-values we found into equation (1)

or (2) instead of into equation (3), we get extraneous

solutions. It is always wise to check all solutions in

both of the given equations.There are four ordered

pairs in the solution set, two with real values and two

with pure imaginary values. The solution set is

The graph of the system, shown in FIGURE 31, shows

only the two real intersection points because the graph

is in the real number plane. In general, if solutions

contain nonreal complex numbers as components,

they do not appear on the graph NOW TRY

NOTE It is not essential to visualize the number of points of intersection of the

graphs in order to solve a nonlinear system. Sometimes we are unfamiliar with the

graphs or, as in Example 4,there are nonreal complex solutions that do not appear as

points of intersection in the real plane. Visualizing the geometry of the graphs is only

an aid to solving these systems.

51 2, 1 2 , 1 - 2, - 12 , 1 i, - 2 i 2 , 1 - i, 2i 26.

If x=-i, y= -^2 i= -^2 i#ii = 2 i.

If x= i, y=^2 i =^2 i # --ii =- 2 i.

If x=-2, y= -^22 =-1.

If x= 2, y=^22 =1.

y=^2 x

x= 2 x=- 2 x= i x=-i

x^2 = 4 x^2 =- 1

x^2 - 4 = 0 x^2 + 1 = 0

1 x^2 - 421 x^2 + 12 = 0

x^4 - 3 x^2 - 4 = 0 3 x^2.

x

y

(–2, –1)

(2, 1)

0

Hyperbola:

x^2 + 2xy – y^2 = 7

Hyperbola:

x^2 – y^2 = 3

FIGURE 31

If the equations in a nonlinear system can be solved for y, then we can graph the

equations of the system with a graphing calculator and use the capabilities of the cal-

culator to identify all intersection points.

For instance, the two equations in Example 3would require graphing four sepa-

rate functions.

and

FIGURE 32indicates the coordinates of one of the points of intersection.

Y 1 = 29 - X^2 , Y 2 =- 29 - X^2 , Y 3 = 2 2X^2 + 6 , Y 4 =- 2 2X^2 + 6

CONNECTIONS

6.2

• 6.2


• 9.4 9.4

FIGURE 32

Multiply by the

complex conjugate

of the denominator.

ii 1 - 2 = 1

NOW TRY
EXERCISE 4
Solve the system.

x^2 - y^2 = 5

x^2 + 3 xy-y^2 = 23

NOW TRY ANSWER

4. 12 i, - 3 i 2 , 1 - 2 i, 3i 26

51 3, 2 2 , 1 - 3, - 22 ,