Systems of Nonlinear Equations and Their Solutions

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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.
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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.
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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.
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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.
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).
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Example Solving a Nonlinear System by the
Addition Method
Solve the system
4x2 y2 13 x2 y2 10
Equation 1.
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 y2 13
x2 y2 10
4x2 y2 13
-x2 y2 -10
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Solution
Steps 3 and 4 Add equations and solve for the
remaining variable.
Add.
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.
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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).
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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.
Subtract x2 from both sides of Equation 1.
-x2 y 3
x2 y2 9
y y2 12
This is Equation 2.
Add the equations.
Add.
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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.
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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 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).
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Examples

• Solve
• 4. Find the length and width of a rectangle
  whose perimeter is 20 ft. an whose area is 21
  sq.ft.