Lecture 11 systems of nonlinear equations

Systems of Nonlinear Equations and Their
Solutions
A system of two nonlinear equations in two variables contains at least one
equation that cannot be expressed in the form Ax + By = C. Here are two
examples:
x2 = 2y + 10
3x – y = 9
y = x2 + 3
x2 + y2 = 9
A solution to a nonlinear system in two variables is an ordered pair of real
numbers that satisfies all equations in the system. The solution set to the
system is the set of all such ordered pairs.

A system of a linear equation and a
quadratic equation can have one real
solution, two real solutions or no real
solutions.

2 6
2 2
 y  x  x


  
Solve
y x

2 2 1
3
 y  x  x


  
Solve
y x

2 2 25
4 3
 x  y


 
Solve
y x

7 8
5
2 2
x y
3 4
2
2 2
Solve
x y

  


  

2 2
y x
x 2 y
2
y x x y
y x y x
y x x
2 2 2 2
2 2 2 2
 
2 2 2
2
x
 
2
2
7 8
5
7 8 5
7 5 8
7 5 8
8
7 5
y
x


 
 
 
  
 
x x
 
 
 


2
 
 
 
 
if x
2
2
2
2
2
1,
8 1
4
7 5 1
2
1
8 1
4
7 5 1
2
y
y
if x
y
y

 

 
 

 
 
 
3 4
 
 
2
2
2
2

  
2 2
  
2 2
2
 
2

2 2
  
2 2
2
2
2
8
7 5
3 7 5
4 2
8
3 28 20
2
8 8
24 28 20
2
8
4 20 16
20 16 4
4 4
1
1
x
x
x x
x x
x
x
x x
x x
x
x
x
 


 
Thus the SS
1,2,1,2.

Example: Solving a Nonlinear System by the
Substitution Method
Solve by the substitution method:
The graph is a line.
The graph is a circle.
x – y = 3
(x – 2)2 + (y + 3)2 = 4
Solution Graphically, we are finding the intersection of a line and a circle
whose center is at (2, -3) and whose radius measures 2.
Step 1 Solve one of the equations for one variable in terms of the other.
We will solve for x in the linear equation - that is, the first equation. (We could
also solve for y.)
x – y = 3 This is the first equation in the given system.
x = y + 3 Add y to both sides.

Solution
Step 2 Substitute the expression from step 1 into the other equation.We
substitute y + 3 for x in the second equation.
x = y + 3 ( x – 2)2 + (y + 3)2 = 4
This gives an equation in one variable, namely
(y + 3 – 2)2 + (y + 3)2 = 4.
The variable x has been eliminated.
Step 3 Solve the resulting equation containing one variable.
(y + 3 – 2)2 + (y + 3)2 = 4 This is the equation containing one variable.
(y + 1)2 + (y + 3 )2 = 4 Combine numerical terms in the first parentheses.
y2 + 2y + 1 + y2 + 6y + 9 = 4 Square each binomial.
2y2 + 8y + 10 = 4 Combine like terms on the left.
2y2 + 8y + 6 = 0 Subtract 4 from both sides and set the quadratic
equation equal to 0.

Solution
y2 + 4y + 3 = 0 Simplify by dividing both sides by 2.
(y + 3)(y + 1) = 0 Factor.
y + 3 = 0 or y + 1 = 0 Set each factor equal to 0.
y = -3 or y = -1 Solve for y.
7
6
5
4
3
2
1
x – y = 3
(2, -1)
-5 -4 -3 -2 -1 1 2 3 4 5 6 7
-1
-2
-3
-4
-5
-6
-7
(x – 2)2 + (y + 3)2 = 4
(0, -3)
Step 4 Back-substitute the obtained values into the
equation from step 1. Now that we have the y-coordin-ates
of the solutions, we back-substitute -3 for y and -1 for
y in the equation x = y + 3.
If y = -3: x = -3 + 3 = 0, so (0, -3) is a solution.
If y = -1: x = -1 + 3 = 2, so (2, -1) is a solution.
Step 5 Check the proposed solution in both of the
system's given equations. Take a moment to show that
each ordered pair satisfies both equations. The solution
set of the given system is {(0, -3), (2, -1)}.

Addition Method
Solve the system:
Equation 1.
Equation 2.
4x2 + y2 = 13
x2 + y2 = 10
SolutionWe can use the same steps that we did when we solved linear systems
by the addition method.
Step 1 Write both equations in the form Ax2 + By2 = C. Both equations
are already in this form, so we can skip this step.
Step 2 If necessary, multiply either equation or both equations by
appropriate numbers so that the sum of the x2-coefficients or the sum of
the y2-coefficients is 0.We can eliminate y by multiplying Equation 2 by -1.
No change.
Multiply by -1.
4x2 + y2 = 13
x2 + y2 = 10
4x2 + y2 = 13
-x2 – y2 = -10

Solution
Steps 3 and 4 Add equations and solve for the remaining variable.
Add.
2 2
x y
x y
x
4  
13
- 2 - 2
-10
3 2
3
2



1
 
1
x
x
Step 5 Back-substitute and find the values for the other variables.We
must back-substitute each value of x into either one of the original equations.
Let's use x2 + y2 = 10, Equation 2. If x = 1,
12 + y2 = 10 Replace x with 1 in Equation 2.
y2 = 9 Subtract 1 from both sides.
y = ±3 Apply the square root method.
(1, 3) and (1, -3) are solutions. If x = -1,
(-1)2 + y2 = 10 Replace x with -1 in Equation 2.
y2 = 9 The steps are the same as before.
y = ±3
(-1, 3) and (-1, -3) are solutions.

Step 6 Check. Take a moment to show that each of the four ordered pairs
satisfies Equation 1 and Equation 2. The solution set of the given system is
{(1, 3), (1, -3), (-1, 3), (-1, -3)}.
7
6
5
4
3
2
1
(-1, 3) (1, 3)
-5 -4 -3 -2 -1 1 2 3 4 5 6 7
-1
-2
-3
-4
-5
-6
-7
4x2 + y2 = 13
x2 + y2 = 10
(-1, -3)
(1, -3)
Solution

Addition Method
Solve the system:
y = x2 + 3 Equation 1 (The graph is a parabola.)
x2 + y2 = 9 Equation 2 (The graph is a circle.)
SolutionWe could use substitution because Equation 1 has y expressed in
terms of x, but this would result in a fourth-degree equation. However, we
can rewrite Equation 1 by subtracting x2 from both sides and adding the
equations to eliminate the x2-terms.
Add.
-x2 + y = 3
x2 + y2 = 9
y + y2 = 12
Subtract x2 from both sides of Equation 1.
This is Equation 2.
Add the equations.

Solution
We now solve this quadratic equation.
y + y2 = 12
y2 + y – 12 = 0 Subtract 12 from both circles and get the quadratic
equation equal to 0.
(y + 4)(y – 3) = 0 Factor.
y + 4 = 0 or y – 3 = 0 Set each factor equal to 0.
y = -4 or y = 3 Solve for y.
To complete the solution, we must back-substitute each value of y into either
one of the original equations. We will use y = x2 + 3, Equation 1. First, we
substitute -4 for y.
-4 = x2 + 3
-7 = x2 Subtract 3 from both sides.

Because the square of a real number cannot be negative, the equation x2 = -7
does not have real-number solutions. Thus, we move on to our other value for
y, 3, and substitute this value into Equation 1.
7
6
5
4
3
2
1
-5 -4 -3 -2 -1 1 2 3 4 5 6 7
-1
-2
-3
-4
-5
-6
-7
y = x2 + 3
x2 + y2 = 9
(0, 3)
Solution
y = x2 + 3 This is Equation 1.
3 = x2 + 3 Back-substitute 3 for y.
0 = x2 Subtract 3 from both sides.
0 = x Solve for x.
We showed that if y = 3, then x = 0. Thus, (0, 3) is
the solution. Take a moment to show that (0, 3)
satisfies Equation 1 and Equation 2. The solution
set of the given system is {(0, 3)}.

Examples
Solve:
x y y x
x y x y
  2  0   
5
 
       
1. 3.
2 2 2 2
   
2
1 1 5 25
2 2
   
x y
x y
3 2 35
2 2
2.
4  3  48


4. Find the length and width of a rectangle whose
perimeter is 20 ft. an whose area is 21 sq.ft.

References
• http://www.regentsprep.org/Regents/math/algtrig/ATE5/QuadLinearSys.htm
• http://www.mathwarehouse.com/system-of-equations/how-to-solve-linear-quadratic-
system.php

Lecture 11 systems of nonlinear equations

Lecture 11 systems of nonlinear equations

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Lecture 11 systems of nonlinear equations