Nonlinear Systems
To solve a system of two equations and two unknowns when the equations are not linear, we use the methods of substitution or elimination and hope that the resulting equation becomes a hidden quadratic or other solvable equation.
Example:
Solve
x2 - y = 0
x2 - 2x + y = 6
Solution
We use substitution: solving for y in the first equation we get:
y = x2
Putting it into the second equation, we have:
x2 - 2x + x2 = 6
so that
2x2 -2x - 6 = 0
or
x2 - x - 3 = 0
The quadratic formula gives us:
so
x = 2.3 or x = -1.3
since
y = x2
we get
y = 5.29 or y = 1.69
We arrive at the two points:
(2.3,5.29) and (-1.3,1.69)
Example 2
Solve
x2 + y2 = 29
x2 - y2 = 3
Solution
We use elimination: adding the two equations, we have
2x2 = 32
x2 = 16
x = 4 or x = -4
Putting
x = 4
into equation 2, we have:
16 - y2 = 3
-y2 = -13
y = 
or y = -
Putting
x = -4
into equation 2, we have:
16 - y2 = 3
-y2 = -13
y =
or y
= -
Therefore we obtain the four points:
(4,),
(4,-),
(-4,),
(-4,-)
Exercises
Solve the following:
-
x2 - y2 = 4
2x2 + y2 = 16
-
y - 2x2 = 0
x2 +5x - y = -6
-
x2 - y2 = 21
x2 + xy - y2 = 31
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