Chapter 6 Conic Sections by TgH1xH8

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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
               2   2


             2x  y  3
             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)                                 • Visualizing the
x2  y 2  9               2x  y  3             Solution

0 3 9
 2     3
                       2(0)  (3)  3
       99                         33

                12 9 
 Check:         , 
                5 5
  x2  y 2  9       2x  y  3
 
12 2
 5
           
            9 2
            5
                  9      2( 12 )  ( 9 )  3
                              5       5

             99                       33
               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:
              3x  10 x  8  0
                2
                      Example continued
• Substitute values of x to find y.
     3x + 2y = 10
                                      • Visualizing the
                                        Solution
 x = 4/3            x = 2




The solutions are
                    Example
• Solve the system of equations:
              5 x  2 y  13
                2      2


              3x  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,   • Visualizing the Solution
so they are the
solutions.

								
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