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
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:
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
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:
y = 34 x
y = 34 25
y = 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:
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
y = 2
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
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)}.