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Chapter 6
Conic Sections
Section 6.4
Nonlinear Systems of Equations
Nonlinear Systems of Equations
• The graphs of the equations in a
nonlinear system of equations can
have no point of intersection or one
or more points of intersection.
• The coordinates of each point of
intersection represent a solution of
the system of equations.
• When no point of intersection
exists, the system of equations has
no real-number solution.
• We can solve nonlinear systems of
equations by using the substitution
or elimination method.
Example
• Solve the following system of equations:
x  y 9
2x  y  3
2
2
Example continued
• We use the substitution method.
First, we solve equation (2) for y.
Example continued
• Next, we substitute
y = 2x  3 in equation (1) and solve for x:
Example continued
• Now, we substitute these numbers for x in
equation (2) and solve for y.
• x=0
x = 12 / 5
Example continued
Check: (0, 3)
x2  y 2  9
2x  y  3
0 3 9
99
2(0)  (3)  3
33
2
3
 12 9 
 , 
Check:
 5 5
x2  y 2  9
2x  y  3
 
12 2
5


9 2
5
9
99
2( 125 )  ( 95 )  3
33
• Visualizing the
Solution
Example
• Solve the following system of equations:
xy = 4
3x + 2y = 10
Example continued
Solve xy = 4 for y.
Substitute into
3x + 2y = 10.
Example continued
• Use the quadratic formula to solve:
3 x  10 x  8  0
2
Example continued
• Substitute values of x to find y.
3x + 2y = 10
x = 4/3
The solutions are
x = 2
• Visualizing the
Solution
Example
• Solve the system of equations:
5 x  2 y  13
2
2
3 x  4 y  39
2
2
Example continued
• Solve by elimination.
Example continued
• Substituting x = 1 in equation (2) gives us:
x=1
x = -1
• The possible solutions are
Example continued
All four pairs check,
so they are the
solutions.
• Visualizing the Solution
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