8' 4 Systems of Nonlinear Equations in Two Variables

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8. 4 Systems of Nonlinear Equations in Two
Variables
MAC 1140 Mrs. Kessler
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Systems of Nonlinear Equations and Their Solutions
A system of two nonlinear equations in two
variables contains at least one equation that
cannot be expressed in the form Ax By C. Here
are two examples
x2 2y 10 3x y 9
y x2 3 x2 y2 9
A solution to a nonlinear system in two variables
is an ordered pair of real numbers that satisfies
all equations in the system. The solution set to
the system is the set of all such ordered pairs.
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Example 1
Solve by the substitution method
The graph is a line.
x y 3 (x 2)2 (y 3)2 4
The graph is a circle.
Graphically, we are finding the intersection of a
line and a circle whose center is at (2, -3) and
whose radius measures 2.
Step 1 Solve one of the equations for one
variable in terms of the other. We will solve for
x in the first equation. x y 3
x y 3 \
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Example 1 cont. Substitution method
x y 3 (x 2)2 (y 3)2 4
Step 2 Substitute the expression from step 1
into the other equation. We substitute y 3 for
x in the second equation. x y 3 ( x
2)2 (y 3)2 4
This gives an equation in one variable.
(y 3 2)2 (y 3)2 4.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing
one variable. (y 3 2)2 (y 3)2 4 . (y
1)2 (y 3 )2 4 y2 2y 1 y2 6y
9 4 2y2 8y 10 4 2y2 8y 6
0
This is a quadratic.
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Example 1 cont. Substitution method
x y 3 (x 2)2 (y 3)2 4
Solution
y2 4y 3 0 (y 3)(y 1) 0 Factor.
y -3 or y -1
Step 4 Back-substitute the y values into the
original equation to get x. If y -3 x -3
3 0, (0, -3) is a solution. If y -1 x
-1 3 2, (2, -1) is a solution.
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Example 1 cont. Substitution method
x y 3 (x 2)2 (y 3)2 4
Step 5 Check . The solution set of the given
system is (0, -3), (2, -1).
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Example 2
Solve the system
4x2 y2 13 x2 y2 10
Solution Use the addition (elimination)
method.
Step 1 Line up like terms, if not already
done.
Step 2 We can eliminate y by multiplying
Equation 2 by -1.
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Example 2 cont.
4x2 y2 13 x2 y2 10
Steps 3 and 4 Add equations and solve for the
remaining variable.
Add.
Step 5 Back-substitute and find the values
for the other variables. Use x2 y2 10,
Equation 2. If x 1, 12 y2
10 Replace x with 1 in Equation 2. y2 9
y 3 (1, 3) and (1, -3) are
solutions. If x -1, (-1)2 y2 10
Replace x with -1 in Equation 2.
y2 9 y 3 (-1, 3) and
(-1, -3) are also solutions.
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Example 2 cont.
4x2 y2 13 x2 y2 10
Solution
Step 6 Check. The solution set of the given
system is (1, 3), (1, -3), (-1, 3), (-1, -3).
To graph these on a TI 83/84 to check, solve for
y and graph all 4 equations
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Example 3
Solve the system y x2 3 (The
graph is a parabola.) x2 y2 9 (The graph
is a circle.)
Solution We could use substitution because
Equation 1 has y expressed in terms of x, but
this would result in a fourth-degree equation.
However, we can rewrite Equation 1 , line up like
terms and add.
.
Add.
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Example 3 cont.
y x2 3 x2 y2 9
We now solve this quadratic equation. y y2
12 y2 y 12 0 (y 4)(y 3) 0 y
-4 or y 3
Back-substitute each value of y into either one
of the original equations. will use y x2 3,
Equation 1. -4 x2 3 -7 x2
Now try y 3 y x2 3 3 x2 3
. 0 x
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Example 3 cont'd.
y x2 3 x2 y2 9
(0, 3) is the only solution.
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Example 4

• Solve the following system of equations x y
  -12 x - 2y 14 0
• Solution
• x -142y
• (-14 2y)y -12
• -14y 2y2 -12
• 2y2 - 14y 12 0

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Example 4 cont'd.

• Solve the following system of equations x y
  -12 x - 2y 14 0
• Solution
• 2y2 - 14y 120 (y2- 7y 6)0 (y - 6)(y -
  1)0 y 6, y 1

y 1 x -14 2yx -14 2(1) x -12
y 6 x -142y x -14 2(6)x -2
(-2, 6) and (-12,1)
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Short Quiz tomorrow partial fractions.