Solving System of Nonlinear Equations


Solving Systems of Nonlinear Equations
A system of equations where at least one equation is not linear is called a nonlinear system. There are several ways to solve systems of nonlinear equations:
  • Substitution
  • Elimination
  • Using a Combination of methods
  • Using absolute value
By Substitution

Solve the following nonlinear equations: 
x2 + y = 6
x − y = 14

To solve this system we first need to isolate one of the variables. In this example, we can use the second equation to solve for y, 
 
x − y = 14
y = x − 14

Now we can substitute this value of y in the second equation: 
 
x2 + y = 6
x2 + (x − 14) = 6
x2 + x − 14 − 6 = 0
x2 + x − 20 = 0

By factoring we find two possible values of x:
 
(x + 5)(x − 4) = 0
x = −5 or x = 4

From the first equation, we can use the value of x to find a value for y:
 
For x = −5; y = x − 14 = −19
For x = 4; y = x − 14 = −10 

As a result, the solution set for the nonlinear system is {(4, −10), (−5, −19)}
By Elimination

Solve the following nonlinear equations:
x2 + y2 = 34
x2 − 2y2 = 7

To solve this system we multiply the first equation by 2
(x2 + y2 = 34) × 2
2x2 + 2y2 = 68

We then add that product to the second equation
  2x2 + 2y2 = 68
   + (x2 − 2y2  =  7)   
           3x2 = 75 

Now we can solve for x 
3x2 = 75
x2 = 25
x = ±5

We can substitute this value of x into the first equation to find all possible values for y. Since we are substituting into a square, x = 5 and x = −5 will give us the same value:
y2 = 34 − x2
y2 = 34 − 25
y2 = 9
y = ±3

The solution set for the nonlinear system is {(5, 3), (5, −3), (−5, 3), (−5, −3)}
Using a Combination of Methods

Solve the following nonlinear equations:
x2 − 2xy + y2 = 3
x2 + xy + y2 = 12

First we use the elimination method to find a value for y. We can multiply the first equation by −1:

−x2 + 2xy − y2 = −3

Then we can add both equations
−x2 + 2xy − y2 = −3
x2 + xy + y2 = 12
3xy = 12
y = 4/x

Now substitute this value of y in the first equation:
x2 + x(4/x) + (4/x)2 = 12
x2 + 4 + 16/x2 = 12

Multiply both sides by x2:
(x2)2 + 4x2 + 16 = 12x2
(x2)2 − 8x2 + 16 = 0

Using the quadratic equation, we can solve for x:
(x2 − 4)(x2 − 4) = 0
x2 = 4
x = ±2

We can then use that result to find the value for y:
y = 4/x
y = ±2

The solution set for the nonlinear system is {(2, 2), (−2, −2)}.
Using absolute value

Solve the following nonlinear equations:
x2 + y2 = 25
|x| + y = 5

First we use the substitution method and solve for the absolute value of x:
 
|x| + y = 5
|x| = 5 − y

We know from the definition of absolute value that |x| ≥ 0. So for all x, 5 − y > 0, and that can be rewritten as 5 > y. In the first equation |x|2 is the same as x2; therefore, we can use substitution:
 
x2 + y2 = 25
(5 − y)2 + y2 = 25
(25 − 10y + y2) + y2 = 25
2y2 − 10y = 0
2y(y − 5) = 0

We can now solve for y:
y = 0 or y = 5

We can use both of these values of y in the first equation:
 
For y = 0
x2 + (0)2 = 25
x2 = 25
x = ±5

For y = 5
x2 + (5)2 = 25
x2 + 25 = 25 
x = 0

The solution set for the nonlinear system is {(5, 0), (0, 5), (−5, 0)}.



Practice

1.  Solve the following system:
x2 + y2 = 8
   y − x = 4


2.  Solve the following system:
2x + 3y + xy = 16
          xy − 5 = 0




















Answers

1.  (−2, 2)

2.  {(, 2), (3, )}