658 CHAPTER 11 Nonlinear Functions, Conic Sections, and Nonlinear Systems
If a system consists of two second-degree equations, then there may be zero, one,
two, three, or four solutions. FIGURE 27shows a case where a system consisting of a
circle and a parabola has four solutions, all made up of ordered pairs of real numbers.
OBJECTIVE 1 Solve a nonlinear system by substitution.We can usually
solve a nonlinear system by the substitution method (Section 4.1) when one equation
is linear.
Solving a Nonlinear System by SubstitutionSolve the system.
(1)
(2)The graph of (1) is a circle and the graph of (2) is a line, so the graphs could
intersect in zero, one, or two points, as in FIGURE 26on the preceding page. We solve
the linear equation (2) for one of the two variables and then substitute the resulting
expression into the nonlinear equation.
(2)
Solve for y. (3)Substitute for yin equation (1).
(1)
Let
Square
Combine like terms. Subtract 9.
Factor. The GCF is x.
Zero-factor propertyLet in equation (3) to get If
then The solution set of the system is
See the graph in FIGURE 28.
e10, - 32 , a
12
5
,
9
5
bf.
y=^95.
x= 0 y=- 3. x=^125 ,
x=
12
5
x= 0 or 5 x- 12 = 0
x 15 x- 122 = 0
5 x^2 - 12 x= 0
x^2 + 4 x^2 - 12 x+ 9 = 9 2 x-3.
x 2 + 12 x- 322 = 9 y= 2 x-3.
x^2 +y^2 = 9
2 x- 3
y= 2 x- 3
2 x-y= 3
2 x-y= 3
x^2 + y^2 = 9
EXAMPLE 1
xy0This system has four
solutions, since there are
four points of intersection.
FIGURE 27xy0(0, –3)x^2 + y^2 = 9^2 x – y = 3Circle: Line:12
59
(), 5FIGURE 28Set bothfactors
equal to 0.Solving a Nonlinear System by SubstitutionSolve the system.
(1)
(2)The graph of (1) is a line. It can be shown by plotting points that the graph of (2)
is a hyperbola. Visualizing a line and a hyperbola indicates that there may be zero,
one, or two points of intersection.
xy= 4
6 x- y= 5
EXAMPLE 2
NOW TRY
EXERCISE 1
Solve the system.
x-y= 34 x^2 +y^2 = 36NOW TRY ANSWER
- E 1 3, 0 2 , A-^95 , -^245 BF
NOW TRY