5.5 systems involving nonlinear equations 1 in previous sections, we solved systems of linear...

5.5 Systems Involving Nonlinear Equations

1

In previous sections, we solved systems of linear equations using various techniques such as substitution, elimination.

The solution to a system of two linear equations is:1. A point (x,y) where the two lines intersected,2. No solution if the lines are parallel, or3. Infinitely many solutions if the lines are the same line.

Case 1: Case 2: Case 3:

one solution no solution infinitely many solutions

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In this section we will look for the intersection of two equations where we may have graphs such as parabolas, circles, ellipses, hyperbolas and lines.


2

Let’s look at an example of a parabola and an ellipse. How many points of intersection can exist?

4 points 3 points 1 points2 points 0 points

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One method of solving a system of equations is by graphing. This method is very impractical for finding the intersections, however, it does help to understand the solution obtained algebraically. Fortunately we have other methods for solving a system of equations such as elimination and substitution.

Which method? Substitution or Elimination?

Although elimination may seem easier than substitution, some systems must be solved using the substitution method. If one of the equations has both x an y squared and the other equation does not, use substitution. If both equations have the x and y squared, use elimination.


3

Example 1. Solve the system:

2 2x y 10x y 2

If one of the equations has a variable which is not squared use substitution. The reason we can not use elimination is we do not have like terms in the both equations. Let’s solve for x in the bottom equation.Now substitute -y - 2 for ‘x’ in the first equation. This will give an equation with only one variable. Then solve for y.

x y 2

x y 2

22yy 2 10

2 2x y 10

Next, substitute both -3 and 1 for y in the equation x = -y - 2 to obtain the values for the variable x.

The two solutions are the intersections of line and a circle. We will graph the equations to verify the answers are reasonable.

2y 2 y 2 y 10

2 2y 2y 2y 4 y 10

22y 4y 4 10

22y 4y 6 0

2 y 3 y 1 0

y 3,y 1

(x 3) 2x 1

y 3: y 1:x 3 x (1) 2

The solution set is 1, 3 , 3,1 .

The first equation is a circle with center (0,0) and the radius is10 3.2.The second equation is a line: x y 2 or y x 2.

Graphically, the solutions of (-3,1) and (1,-3) are correct.

We also need to check the solutions by substituting them back into the original equations.

2 21 3 101 3 2

1, 3 ; 3,1 ;

2 23 1 103 1 2


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Your Turn Problem #1

2y x 12x y 1

Answer:

x

y

●

●

(2,3)

(0,-1)

0, 1 , 2,3

Solve the system by substitution or elimination. Verify the solution by graphing both equations.


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Example 2. Solve the system:

2y x 1y x 1

Since both of the equations has a variable which is not squared, use substitution.Substitute x-1 for ‘y’ in the first equation. This will give an equation with only one variable. Then solve for x.

2x 1 x 1

Next, substitute both values of x in one of the original equations to obtain the values for y.

Obtaining a non real number solution means that the parabola and the line do not intersect in the real number plane.

1 i 7 1 i 7 1 i 7 1 i 7, , ,

2 2 2 2

20 x x 2

This quadratic equation is not factorable, use quadratic formula to solve.

1 i 7 1 i 7x , x

2 2

i 72

y1

1

1 i 7x ;

2

1 i 7 2y

2 2

1 i 7y

2

i 72

y1

1

1 i 7x ;

2

1 i 7 2y

2 2

1 i 7y

2

We can see the graphs do not intersect. However, we did obtain imaginary solutions.


6

x

yAnswer:

1 i 3 3 i 3 1 i 3 3 i 3, , ,

2 2 2 2

Note: Please do not feel that imaginary solutions have no meaning. For those involved in fields such as engineering and aeronautics, applications involving imaginary solutions are very important.

2y x 1x y 2




7

Example 3. Solve the system :

2 2

2 2x 4y 4x y 9

Since the variables are all the same, elimination will work well. Multiply bottom row by -1 to eliminate the x when the two rows are added together.

Substitute +1 and -1 for ‘y’ in either equation to find x.

2 2

2 2x 4y 4x y 9-1

2 2

2 2x 4y 4x y 9

25y 5

Now, divide by -5 on both sides and use the square root property to solve for y.

-5 -5

2y 1y 1

22x 1 9y 1;

2x 8x 2 2

22x 1 9y 1;

2x 8x 2 2

The solution set is 2 2,1 , 2 2,1 , 2 2, 1 , 2 2, 1 .

The 1st equation is a hyperbola: 2 2x 4y 4.

The center is (0,0), a =±2, b=±1.

2 2x y

14 1

The 2nd equation is a circle.The center is (0,0), r=3. The graphs do intersect. It is consistent with

the solution set obtained using the elimination method.


8

x

yAnswer

●

●

●

●

2, 6

2, 6 2, 6

2, 6

2, 6 , 2, 6 , 2, 6 , 2, 6

The EndB.R.2-1-07


2 2

2 22y 3x 6x y 8


5.5 systems involving nonlinear equations 1 in previous sections, we solved systems of linear...

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