§ 10.5

Systems of Nonlinear Equations in Two Variables

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 10.5

Solving Systems of Nonlinear Equations

We looked at solving systems of linear equations in an earlier section. In this section, we will look at solving systems of nonlinear equations. A system of nonlinear equations contains at least one equation that cannot be expressed in the form Ax +By = C.

As with systems of linear equations in two variables, the solution of a nonlinear system (if there is one) corresponds to the intersection point(s) of the graphs of the equations. Unlike linear systems, the graphs can be circles, ellipses, parabolas, hyperbolas, or anything other than two lines. We will solve nonlinear systems using the substitution method and the addition method.

Blitzer, Intermediate Algebra, 5e – Slide #3 Section 10.5

Solving Systems of Nonlinear EquationsEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve by the substitution method:

1) Solve one of the equations for one variable in terms of the other. We begin by isolating one of the variables raised to the first power in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions.

.0142

12

yx

xy

0142 yx This is the second equation in the given system.

142 yx Add 2y - 14 to both sides.

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 10.5

Solving Systems of Nonlinear Equations

2) Substitute the expression from step 1 into the other equation. We substitute 2y – 14 for x in the first equation.

. 12142 yy

This is the given equation containing one variable.

142 yx

CONTINUECONTINUEDD

12yx

This gives us an equation in one variable, namely

The variable x has been eliminated.3) Solve the resulting equation containing one variable.

12142 yy

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 10.5

Solving Systems of Nonlinear Equations

Use the distributive property.

CONTINUECONTINUEDD 12142 2 yy

Add 12 to both sides.012142 2 yy

Factor 2 out of the left side. 0672 2 yy

Divide both sides by 2.0672 yy

Factor. 016 yy

Set each factor equal to zero.

01or 06 yy

Solve for y.6y 1y

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 6 for y and 1 for y into the equation x = 2y - 14.

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DD

5) Check the proposed solutions in both of the system’s given equations. We begin by checking (-2, 6). Replace x with -2 and y with 6.

If y is 6, ,21462 x so (-2, 6) is a solution.If y is 1, ,121412 x so (-12, 1) is a solution.

12xy 0142 yx1262 014622

1212 014122 00

The ordered pair (-2, 6) satisfies both equations. Thus, (-2, 6) is a solution of the system.

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Blitzer, Intermediate Algebra, 5e – Slide #7 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DDNow let’s check (-12, 1). Replace x with -12 and y with 1.

12xy 0142 yx

12112 0141212

1212 014212 00

The ordered pair (-12, 1) satisfies both equations. Thus, (-12, 1) is a solution of the system.

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Thus, the solution set is {(-2, 6), (-12, 1)}.

The graph of the equations in the system and the solutions as intersection points follows.

truetrue

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 10.5

Solving Systems of Nonlinear Equations

-30

-20

-10

0

10

20

30

-15 -10 -5 0 5

CONTINUECONTINUEDD

(-12, 1)(-2, 6)

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 10.5

Solving Systems of Nonlinear EquationsEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve by the substitution method:

We can use the same steps that we did when we solved linear systems by the addition method.

.22

52322

22

yx

yx

1) Write both equations in the form . Both equations are already in this form, so we can skip this step.

CByAx 22

2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the -coefficients or the sum of the -coefficients is 0. We can eliminate by multiplying Equation 2 by -2.

2x2y 2y

Equation 1

Equation 2

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 10.5

Solving Systems of Nonlinear Equations

523 22 yx

3) & 4) Add equations and solve for the remaining variable.

No ChangeCONTINUECONTINUE

DD

22 22 yx Multiply by -2.

523 22 yx

424 22 yx

523 22 yx

424 22 yx

12 x Add equations.12 x Multiply both sides by -1.

1x Use the square root property.

Blitzer, Intermediate Algebra, 5e – Slide #11 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DD 5) Back-substitute and find the values for the other variable. We must back-substitute each value of x into either one of the original equations. Let’s use Equation 2. If x = 1,

212 22 y Replace x with 1 in Equation 2.22 2 y Simplify.42 y Subtract 2 from both sides.

42 y Multiply both sides by -1.2y Apply the square root property.

(1, 2) and (1, -2) are solutions. If x = -1, 212 22 y Replace x with -1 in Equation 2.

22 2 y Simplify.42 y Subtract 2 from both sides.

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DD 42 y Multiply both sides by -1.2y Apply the square root property.

(-1, 2) and (-1, -2) are solutions.

6) Check. Take a moment to show that each of the four ordered pairs satisfies the given equations,

The solutions are (1, 2), (1, -2), (-1, 2), and (-1, -2), and the solution set of the given system is {(1, 2), (1, -2), (-1, 2), (-1, -2)}.

and 523 22 yx.22 22 yx

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 10.5

Solving Systems of Nonlinear EquationsEXAMPLEEXAMPLE

SOLUTIONSOLUTION

A system for tracking ships indicates that a ship lies on a hyperbolic path described by . The process is repeated and the ship is found to lie on a hyperbolic path described by . If it is known that the ship is located in the first quadrant of the coordinate system, determine its exact location.

12 22 yx

12 22 xy

Since both equations represent the path of the ship, we will determine the points they have in common by solving the system of equations

.12 22 yx12 22 xy

1) Use variables to represent unknown quantities. This has already been done so we may skip this step.

Blitzer, Intermediate Algebra, 5e – Slide #14 Section 10.5

Solving Systems of Nonlinear Equations

We will use the addition method. We will multiply Equation 1 by 2 and then add the equations.

.12 22 yx12 22 xy

2) Write a system of equations describing the problem’s conditions. This has already been done so we may skip this step.

CONTINUECONTINUEDD

3) Solve the system and answer the problem’s question. We must solve the system

Equation 1Equation 2

12 22 yx12 22 xy

12 22 yx224 22 xy

No Change

Multiply by 2.

33 2 y

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DD

112 22 x

33 2 y This is the resultant equation.12 y Divide both sides by 3.

1y Apply the square root property.

Since we are only interested in solutions that lie in the first quadrant, we will ignore solutions having y = -1. We now solve for x given y = 1. We will use Equation 2.

Replace y with 1 in Equation 2.

112 2 x Simplify.22 2 x Add 1 to both sides.

12 x Divide both sides by 2.

1x Apply the square root property.

Blitzer, Intermediate Algebra, 5e – Slide #16 Section 10.5

Solving Systems of Nonlinear EquationsCONTINUECONTINUE

DD Since we are only interested in solutions that lie in the first quadrant, we will ignore solutions having x = -1. Therefore, the only viable solution occurs when x = 1 and when y = 1. Therefore the solution is (1, 1) and the solution set is written {(1, 1)}.

A Study Tip: When solving nonlinear systems, extra solutions may be introduced that do not satisfy both equations in the system. Therefore, you should get into the habit of checking all proposed pairs in each of the system’s two equations.