# Section 10-8 Systems of second-degree equations and inequalities

```Section 10-8
Systems of second-degree equations and
inequalities
 If
you had three nonconcentric (not having
the same center)circles, how many points of
intersections can you possibly have?
2
3
1
0
4
5
6
 The
number of solutions of a system of
second degree equations equals the
number of times the graphs of the
equations intersect. If the system is
composed of three circles you can have 0,
1,2,3,4,5,or 6 possible solutions. If the
system is composed of a line and a circle
there may be 0, 1 or 2 solutions.
 Line
and a conic: Possible solutions 0, 1, 2
 Two conics: 0, 1, 2, 3, or 4
(example: parabola and ellipse)
 While
you can determine the possible
number of solutions, to find the exact
solution you must use algebra.







2
2
2
Graph the system of equations 5x - 4y =20 and 4x + 16y 2= 64. Use
the graph to find approximate solutions. Then solve it algebraically.
We have an ellipse and a hyperbola. Putting them in standard form,
we have (x-0) 2 /16 + (y-0) 2 /4 = 1 center (0,0) vertices (&plusmn;4,0) and
(0,&plusmn;2) Also (x-0) 2 /4 - (y-0)2 /5 = 1 Hyperbola with center (0,0)
vertices (&plusmn;2,0)
2
-4
-2
2
4
Approximate intersection points (&plusmn;2.5, 1.5)
and (&plusmn;2.5, -1.5

To get the exact points of intersection, we need to solve the system
algebraically.

5x - 4y = 20
2
2
4x + 16y = 64

2
2



2
2
Multiple thru by 4 20x -16y = 80
2
2
4x +16y =64
__________________
2
2
24x = 144
x =6
so x = &plusmn; 6
2
2
2

Now find y. 5(6)-4y = 20
30 -4y =20
-4y = -10
y 2 = 10/4
y=&plusmn; 10 /2 So the points of intersection are

(&plusmn; 6 , 10 /2 )

and (&plusmn; 6 , - 10 /2) There are 4 solutions!



2
Graph the solutions for the system of inequalities. y-1 &gt; 2x and
x2 /9 + y /4 &lt; 1.
2
The first one is a parabola and the second one is an ellipse. Both
boundary lines are dashed.
2





-3
3
-2

Section 10-8
Pp. 682-684
#13-23
odds, 34,36,45
```