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^{2} + 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^{2} + y = 6
x^{2} + (x − 14) = 6
x^{2} + x − 14 − 6 = 0
x^{2} + 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^{2} + y^{2} = 34
x^{2} − 2y^{2} = 7
To solve this system we multiply the first equation by 2
(x^{2} + y^{2} = 34) × 2
2x^{2} + 2y^{2} = 68
We then add that product to the second equation
2x^{2} + 2y^{2} = 68
+ (x^{2} − 2y^{2} = 7)
3x^{2} = 75
Now we can solve for x
3x^{2} = 75
x^{2} = 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^{2} = 34 − x^{2}
y^{2} = 34 − 25
y^{2} = 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^{2} − 2xy + y^{2} = 3
x^{2} + xy + y^{2} = 12
First we use the elimination method to find a value for y. We can multiply the first equation by −1:
−x^{2} + 2xy − y^{2} = −3
Then we can add both equations
−x^{2} + 2xy − y^{2} = −3
x^{2} + xy + y^{2} = 12
3xy = 12
y = ^{4}/_{x}
Now substitute this value of y in the first equation:
x^{2} + x(^{4}/_{x}) + (^{4}/_{x})^{2} = 12
x^{2} + 4 + 16/x^{2} = 12
Multiply both sides by x^{2}:
(x^{2})^{2} + 4x^{2} + 16 = 12x^{2}
(x^{2})^{2} − 8x^{2} + 16 = 0
Using the quadratic equation, we can solve for x:
(x^{2} − 4)(x^{2} − 4) = 0
x^{2} = 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^{2} + y^{2} = 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 x^{2}; therefore, we can use substitution:,
x^{2} + y^{2} = 25
(5 − y)^{2} + y^{2} = 25
(25 − 10y + y^{2}) + y^{2} = 25
2y^{2} − 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^{2} + (0)^{2} = 25
x^{2} = 25
x = ±5
x = ±5
For y = 5
x^{2} + (5)^{2} = 25
x^{2} + 25 = 25
x = 0
The solution set for the nonlinear system is {(5, 0), (0, 5), (−5, 0)}.