Nonlinear Systems of Equations

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10TH EDITION

LIAL

HORNSBY

SCHNEIDER

COLLEGE ALGEBRA

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5.5 Nonlinear Systems of Equations

Solving Nonlinear Systems with Real SolutionsSolving Nonlinear Equations with Nonreal Complex SolutionsApplying Nonlinear Systems

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Solving Nonlinear Systems with Real Solutions

A system of equations in which at least one equation is not linear is called a nonlinear system. The substitution method works well for solving many such systems, particularly when one of the equations is linear, as in the next example.

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Example 1 SOLVING A NONLINEAR SYSTEMS BY SUBSTITUTION

Solve the system.

When one of the equations in a nonlinear system is linear, it is usually best to begin by solving the linear equation for one of the variables.

Solution

2 4x y 2x y

(1)

(2)

2y x Solve equation (2) for y.

Substitute this result for y in equation (1).

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2 4x y (1)

Let y = – 2 – x. 2 2( ) 4xx 2 2 4x x Distributive property

2 2 0x x Standard form

( 2)( 1) 0x x Factor.

2 0 or 1 0x x Zero factor property

2 or 1x x

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Substituting – 2 for x in equation (2) gives y = 0. If x = 1, then y = – 3. The solution set of the given system is {(– 2, 0),(1, – 3)}.

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Visualizing Graphs

Visualizing the types of graphs involved in a nonlinear system helps predict the possible numbers of ordered pairs of real numbers that may be in the solution set of the system.

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Example 2 SOLVING A NONLINEAR SYSTEM BY ELIMINATION

Solve the system.

Solution

2 2 4x y (1)

(2)2 22 8x y

The graph of equation (1) is a circle and, as we will learn in the next chapter, the graph of equation (2) is a hyperbola. These graphs may intersect in 0, 1, 2, 3, or 4 points. We add to eliminate y2.

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2 2 4x y 2 22 8x y

(1)

(2)23 12x Add.

2 4x Divide by 3.

2x Square root property

Remember to find both

square roots.

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Find y by substituting back into equation (1).

If x = 2, then If x = − 2, then2 22 4y

2 0y

0.y

2 2 42( ) y 2 0y

0.y

The solutions are (2, 0) and (– 2, 0) so the solution set is {(2, 0), (– 2, 0)}.

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Note The elimination method works with the system in Example 2 since the system can be thought of as a system of linear equations where the variables are x2 and y2. In other words, the system is linear in x2 and y2. To see this, substitute u for x2 and v for y2. The resulting system is linear in u and v.

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Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS

Solve the system.

Solution

2 23 22x xy y (1)

(2)2 2 6x xy y

2 23 22x xy y (1)2 2 6x xy y Multiply (2) by – 1

4 16xy Add. (3)

4yx

Solve for y(x ≠ 0).

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Let y = in (2).

Now substitute for y in either equation (1) or (2). We use equation (2).

4x

22 64 4x x

x x

4x

22

164 6xx

Multiply and square.

4 2 24 16 6x x x Multiply x2 to clear fractions.

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4 210 16 0x x Subtract 6x2.

This equation is in quadratic form.

2 2( 2)( 8) 0x x Factor.

2 22 or 8x x Zero factor property.

2 or 2 2x x Square root property;

8 4 2 2 2 For each equation, include both square

roots.

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Substitute these x-values into equation (4) to find corresponding values of y.

If , then2x If , then2x

2 242

.y 2 24 .2

y

If , then2 2x

.242 2

y

If , then2 2x

2 24 2.y

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If , then2x If , then2x

2 242

.y 2 24 .2

y

If , then2x

.242 2

y

If , then2x

2 24 2.y

The solution set of the system is

2,2 2 , 2, 2 2 , 2 2, 2 , 2 2, 2 .

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Example 4 SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION

Solve the system.

Solution

2 2 16x y (1)

(2)4x y

Use the substitution method. Solving equation (2) for x gives

4 .x y (3)

Since x 0 for all x, 4 – y 0 and thus y 4. In equation (1), the first term is x2, which is the same as x2.

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Example 4

Therefore,2 2(4 ) 16y y Substitute 4 – y in (1).

2 2(16 8 ) 16y y y Square the binomial.

Remember the middle

term.

22 8 0y y Combine terms.

2 ( 4) 0y y Factor.

or .0 4 y y Zero factor property.

SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION

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Example 4

To solve for the corresponding values of x, use either equation (1) or (2). We use equation (1).

If y = 0, then If y = 4, then

2 2 160x

.4x

2 16x

2 2 164x 2 0x

0.x


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Example 4

The solution set, {(4,0), (– 4, 0), (0, 4)}, includes the points of intersection shown in the graph.


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Example 5SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS IN ITS SOLUTIONS

Solve the system.

Solution

2 2 5x y (1)

(2)2 24 3 11x y

2 23 3 15x y Multiply (1) by – 3 .2 24 3 11x y 2 4x

(2)

Add.

4x Square root property

2x i 4 4 2 i i

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To find the corresponding values of y, substitute into equation (1).

If x = 2i, then If x = – 2i, then

2 22( ) 5y ii2 = – 4 24 5y

2 9y

.3y

2 22( ) 5y i24 5y

2 9y

.3y

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Checking the solutions in the given system shows that the solution set is

2 ,3 , 2 , 3 , 2 ,3 , 2 , 3 . i i i i

Note that these solutions with nonreal complex number components do not appear as intersection points on the graph of the system.

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Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX

A box with an open top has a square base and four sides of equal height. The volume of the box is 75 in.3 and the surface area is 85 in.2. What are the dimensions of the box?

Solution Step 1 Read the problem. We must find the

dimensions (width, length, and height) of the box.

Step 2 Assign variables. Let x represent the length and width of the square base, and let y represent the height.

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Step 3 Write a system of equations. Use the formula for the volume of a box, V = LWH, to write one equation using the given volume, 75 in.3.

2 75x y Volume formula

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2 4 85x xy Sum of areas of base and sides

The surface consists of the base, whose area is x2, and four sides, each having area xy. The total surface area is 85 in.2, so a second equationis

The system to solve is2 75x y

2 4 85x xy

(1)

(2)

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22 74 855x

xx

Let in (2).

Step 4 Solve the system. Solve equation (1) for y to get and substitute into equation (2).

275,yx

275yx

2 300 85xx

Multiply.

3 300 85x x Multiply by x, x ≠ 0.

3 85 300 0x x Subtract 85x.

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Coefficients of a quadratic polynomial factor

We are restricted to positive values for x, and considering the nature of the problem, any solution should be relatively small. By the rational zeros theorem, factors of 300 are the only possible rational solutions. Using synthetic division, we see that 5 is a solution.

5 1 0 85 3005 25 300

1 5 60 0

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Therefore, one value of x is 5 and We must now solve

275 3.5

y

2 5 60 0x x for any other possible positive solutions.

Using the quadratic formula, the positive solution is

25 5 4(1)( 60) 5.6392(1)

x Quadratic formula with a = 1, b = 5, c = – 60

This value of x leads to y ≈ 2.359.

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Step 5 State the answer. There are two possible answers.

First answer: length = width = 5 in.; height = 3 in.

Second answer: length = width ≈ 5.639 in.’height ≈ 2.359 in.

Step 6 Check. The check is left for Exercise 61.

Nonlinear Systems of Equations

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