Download 7.4 Systems of Nonlinear Equations in Two Variables.

4/11/2012
7.4 Systems of Nonlinear Equations in Two Variables.
Objective 1: Recognize systems of nonlinear equations in two Objective
1: Recognize systems of nonlinear equations in two
variables.
A system of two nonlinear equations in two variables, also
called a nonlinear system, contains at least one equation
that cannot be expressed in the form Ax+By=C.
A solution of a nonlinear system in two variables is an
ordered pair of real numbers that satisfies both equations
in the system. The solution set of the system is the set of
all such ordered pairs.
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Objective 2: Solve nonlinear systems by substitution.
The substitution method involves converting a nonlinear
system to one equation in one variable by an appropriate
substitution. The steps in the solution process are exactly
the same as those used to solve a linear system by substitution.
Since at least one equation is nonlinear you may get more
than one point of intersection.
Solve by substitution
y-x=-2
y=x 2 − 4
y – x = ‐2
2
2
x – x – 2 = 0
Since the second equation has y = x2 – 4, we can substitute x2 – 4 for the y in the first equation.
2 – x – 2 = 0 for x.
Now, solve x
N
l
2 0f
(x – 2)(x + 1) = 0
x – 2 = 0 or x + 1 = 0
x = 2 or x = ‐1
Use y – x = ‐2 to find the y‐values for x = 2 and x = ‐1.
y – 2 y 2 = ‐2
2 y y – ((‐1)
1) = ‐2
2
y = 0 y + 1 = ‐2
y = ‐3 The solutions are: (2, 0) and (‐1, ‐3)
Sometimes when solving nonlinear systems, you may get extra solutions that do not satisfy both equations in the system. Because of this, you should check all solution pairs in each of the two equations.
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Solve each system by the substitution method
⎧ x − y = −1
⎨
2
⎩ y = x +1
⎧2 x + y = −5
⎨
2
⎩ y = x + 6x + 7
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⎧⎪ y = x 2 + 4 x + 5
⎨
2
⎪⎩ y = x + 2 x − 1
⎧ xy = −12
⎨
⎩ x − 2 y + 14 = 0
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Objective 3: Solve nonlinear systems by addition.
For nonlinear systems, the addition method can be used when
each equation is in the form Ax 2 + By 2 = C. If necessary, we
will multiply either equation or both equations by appropriate
numbers so that the coefficients
coefficients of x 2 or y 2 will have a sum of 0.
0
Again you may get more than one point of intersection.
Solve each system by the addition method.
Solve the system:
x 2 + y 2 = 16
y = x −4
2
x2
+ y2 = 16
‐x2 + y = ‐4
y + y2 = 12
Arrange like terms in columns.
Add the equations.
Now, solve this quadratic equation.
y2 + y – 12 = 0
Next substitute each of the values (y + 4)(y – 3) = 0
for y into either of the original y = ‐4 or y = 3
equations.
y = x2 – 4 y = x
y = x
4
y = x2 ‐ 4
2
‐4 = x ‐4 3 = x2 ‐ 4
x2 = 0 7 = x2
x = 0 x = ‐√7 and x = √7
The solutions are: (0, ‐4), (‐√7, 3), and √7, 3)
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⎧⎪4 x 2 − y 2 = 4
⎨ 2
2
⎪⎩4 x + y = 4
⎧⎪16 x 2 − 4 y 2 − 72 = 0
⎨ 2
2
⎪⎩ x − y − 3 = 0
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⎧⎪ x 2 + y 2 = 5
⎨ 2
2
⎪⎩ x + ( y − 8) = 41
Solve each system by the method of your choice.
⎧⎪ x 2 − y 2 − 4 x + 6 y − 4 = 0
⎨ 2
2
⎪⎩ x + y − 4 x − 6 y + 12 = 0
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