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: 
x + 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: 
x + y = 6
x + (x  14) = 6
x + x  14  6 = 0
x + x  20 = 0

By factoring we find two possible values of x:
(x + 5)(x  4) = 0
x5 or x = 4

From the first equation, we can use the value of x to find a value for y
For x5; 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:
x + y = 34
x  2y = 7

To solve this system we multiply the first equation by 2
(x + y = 34)  2
2x + 2y = 68

We then add that product to the second equation
  2x + 2y = 68
   + x  2y  =  7    
           3x = 75 

Now we can solve for x 
3x = 75
x = 25
x5

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:
y = 34  x
y = 34  25
y = 9
y3

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:
x  2xy + y = 3
x + xy + y = 12

First we use the elimination method to find a value for y. We can multiply the first equation by 1:
x + 2xy  y 3

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

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

Multiply both sides by x:
(x) + 4x + 16 = 12x
(x)  8x + 16 = 0

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

We can then use that result to find the value for y:
y = 4/x
y2

The solution set for the nonlinear system is {(2,2), (2,2)}.

Using absolute value

Solve the following nonlinear equations:
x + y = 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| is the same as x; therefore, we can use substitution:, 
x + y = 25
(5  y) + y = 25
(25  10y + y ) + y = 25
2y  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
x + (0) = 25
x = 25
x
5

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

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