Download Systems of Nonlinear Equations and Their Solutions

Solving Systems of Equations
Algebraically
1. When you graph, sometimes you cannot find the
exact point of intersection. We can use algebra
to find the exact point.
2. Also, we do not need to put every equation in
slope-intercept form in order to determine if the
lines are parallel or the same line. Algebraic
methods will give us the same information.
Methods of Solving Systems
Algebraically
We will look at TWO methods to solve
systems algebraically:
1) Substitution
2) Elimination
Method 1: Substitution
Steps:
1. Choose one of the two equations
and isolate one of the variables.
2. Substitute the new expression into
the other equation for the variable.
3. Solve for the remaining variable.
4. Substitute the solution into the
other equation to get the solution
to the second variable.
Method 1: Substitution
Example:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Isolate the ‘x’ in equation ‘b’:
x = - 2y + 2
Method 1: Substitution
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Substitute the new expression,
x = - 2y + 2 for x into equation ‘a’:
3(- 2y + 2) + 4y = - 4
Method 1: Substitution
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Solve the new equation:
3(- 2y + 2) + 4y = - 4
- 6y + 6 + 4y = - 4
- 2y + 6 = - 4
- 2y = - 10
y= 5
Method 1: Substitution
Example, continued:
Equation ‘a’:
Equation ‘b’:
3x + 4y = - 4
x + 2y = 2
Substitute y = 5 into either equation ‘a’ or ‘b’:
x + 2 (5) = 2
x + 10 = 2
x=-8
The solution is (-8, 5).
Method 2: Elimination
Steps:
1. Line up the two equations using
standard form (Ax + By = C).
2. GOAL: The coefficients of the same
variable in both equations should have
the same value but opposite signs.
3. If this doesn’t exist, multiply one or
both of the equations by a number that
will make the same variable coefficients
opposite values.
Method 2: Elimination
Steps, continued:
4. Add the two equations (like terms).
5. The variable with opposite
coefficients should be eliminated.
6. Solve for the remaining variable.
7. Substitute that solution into either
of the two equations to solve for
the other variable.
Method 2: Elimination
Example:
Equation ‘a’:
Equation ‘b’:
2x - 4y = 13
4x - 5y = 8
Multiply equation ‘a’ by –2 to eliminate
the x’s:
Equation ‘a’:
Equation ‘b’:
-2(2x - 4y = 13)
4x - 5y = 8
Method 2: Elimination
Example, continued:
Equation ‘a’:
Equation ‘b’:
-2(2x - 4y = 13) ------> -4x + 8y = -26
4x - 5y = 8
------> 4x - 5y = 8
Add the equations (the x’s are eliminated):
-4x + 8y = -26
4x - 5y = 8
3y = -18
y = -6
Method 2: Elimination
Example, continued:
Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26
Equation ‘b’: 4x - 5y = 8
------> 4x - 5y = 8
Substitute y = -6 into either equation:
4x - 5(-6) = 8
4x + 30 = 8
4x = -22
-22
x= 4
-11
x= 2
-11
2
Solution: ( , -6)
Method 2: Elimination
Example 2:
Equation ‘a’: -9x + 6y = 0
Equation ‘b’: -12x + 8y = 0
Multiply equation ‘a’ by –4 and
equation ‘b’ by 3 to eliminate the x’s:
Equation ‘a’: - 4(-9x + 6y = 0)
Equation ‘b’: 3(-12x + 8y = 0)
Method 2: Elimination
Example 2, continued:
Equation ‘a’:
Equation ‘b’:
- 4(-9x + 6y = 0)
3(-12x + 8y = 0)
36x - 24y = 0
-36x + 24y = 0
0=0
What does this answer mean?
Is it true?
Method 2: Elimination
Example 2, continued:
36x - 24y = 0
-36x + 24y = 0
0=0
When both variables are eliminated,
 if the statement is TRUE (like 0 = 0),
then they are the same lines and there
are infinite solutions.
 if the statement is FALSE (like 0 = 1),
then they are parallel lines and there
is no solution.
Method 2: Elimination
Example 2, continued:
36x - 24y = 0
-36x + 24y = 0
0=0
Since 0 = 0 is TRUE,
there are infinite
solutions.
Solving Systems of Three
Equations Algebraically
1. When we have three equations in a
system, we can use the same two
methods to solve them algebraically as
with two equations.
2. Whether you use substitution or
elimination, you should begin by
numbering the equations!
Solving Systems of Three
Equations
Substitution Method
1. Choose one of the three equations
and isolate one of the variables.
2. Substitute the new expression into
each of the other two equations.
3. These two equations now have the
same two variables. Solve this 2 x 2
system as before.
4. Find the third variable by
substituting the two known values
into any equation.
Solving Systems of Three
Equations
Linear Combination Method
1. Choose two of the equations and eliminate
one variable as before.
2. Now choose one of the equations from step 1
and the other equation you didn’t use and
eliminate the same variable.
3. You should now have two equations (one
from step 1 and one from step 2) that you
can solve by elimination.
4. Find the third variable by substituting the
two known values into any equation.
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.
Example: Solving a Nonlinear System by the
Substitution Method
Solve by the substitution method:
x–y=3
(x – 2)2 + (y + 3)2 = 4
The graph is a line.
The graph is a circle.
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
(y + 3)(y + 1) = 0
y + 3 = 0 or y + 1 = 0
y = -3
or y = -1
Simplify by dividing both sides by 2.
Factor.
Set each factor equal to 0.
Solve for y.
Step 4 Back-substitute the obtained values into the
equation from step 1. Now that we have the y-coordinates 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.
7
6
5
x – y =4 3
3
2
1
-5 -4 -3 -2 -1
-1
(2, -1)
1
2 3 4 5 6
7
-2
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)}.
(0, -3) -3
-4
-5
-6
-7
(x – 2)2 + (y + 3)2 = 4
Example: Solving a Nonlinear System by the
Addition Method
Solve the system:
4x2 + y2 = 13 Equation 1.
x2 + y2 = 10 Equation 2.
Solution We 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.
4x2 +
x2 +
y2
y2
=
=
13
10
No change.
Multiply by -1.
4x2 + y2
-x2 – y2
= 13
= -10
Solution
Steps 3 and 4
Add equations and solve for the remaining variable.
4x2 
Add.
y2 
13
- x 2 - y 2  -10

3x 2
x2
x
3
 1
 1
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.
Solution
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
4x2 + y2 = 13
6
(-1, 3)
5
4
(1, 3)
3
2
x2 + y2 = 10
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5 6
-2
(-1, -3)
-3
-4
-5
-6
-7
(1, -3)
7
Example: Solving a Nonlinear System by the
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.)
Solution We 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
(y + 4)(y – 3) = 0
y + 4 = 0 or y – 3 = 0
y = -4
or
y=3
equation equal to 0.
Factor.
Set each factor equal to 0.
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.
Solution
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.
y = x2 + 3
3 = x2 + 3
0 = x2
0=x
This is Equation 1.
7
6
Back-substitute 3 for y.
Subtract 3 from both sides.
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)}.
y=
x2 +
3
5
4
(0, 3)
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5 6
7
-2
-3
-4
-5
-6
-7
x2 + y2 = 9
Examples
 x  2 y  0
Solve: 1. 
2
2
 x  1   y  1  5
 y  x 2  5
3.  2
2
 x  y  25
3x 2  2 y 2  35
2.  2
2
4
x

3
y
 48

4. Find the length and width of a rectangle whose
perimeter is 20 ft. an whose area is 21 sq.ft.