Section 13.5 Systems of Nonlinear Equations in Two Variables
3.
2
2
44
xy
yx x
+=
=−+
Substitute
2
4 4 for
xx y
−+ in the first equation and
solve for x.
2
442
xx x
+−+=
Substitute 1 and 2 for x to find y.
2or 1
xx
==
The solution set is
()( )
{
}
1,1 , 2, 0 .
4. Substitute
26 7 for xx y++
in the first equation and
solve for x.
()
2
2
2675
2675
xx x
xx x
+++=−
+++=−
The solution set is
()( )
{
}
6, 7 , 2, 1 .−−−
5.
2
2
410
214
yx x
yx x
=−−
=− − +
Substitute
2
214 for xx y−− + in the first equation
and solve for x.
22
43
xx
==−
Substitute
3−
and 4 for x to find y.
() ()
2
3
34310
91210 11
x
y
=−
=− − − −
=+ − =
The solution set is
()( )
{
}
3,11 , 4, 10 .−−
22 1 for xx y+−
Chapter 13 Conic Sections and Systems of Nonlinear Equations
7.
22
25
1
xy
xy
+=
−=
Solve the second equation for x.
Substitute 1 for yx+to find y.
()
22
22
22
25
125
21 25
xy
yy
yy y
+=
++=
+++=
Substitute –4 and 3 for y to find x.
43
41 31
34
yy
xx
xx
=− =
=− + = +
=− =
The solution set is
()
{
3, 4 ,−−
()
}
4, 3 .
8. Solve the second equation for y.
35
xy
−=
()()
2
2
10 30 20 0
320
210
xx
xx
xx
−+=
−+=
−−=
9.
6
21
xy
xy
=
−=
Solve the first equation for y.
yx
=
Substitute
6for y
x
in the second equation and solve
for x.
6
21
xx
−=
3
2
20or2 30
223
xx
xx
x
−= +=
==−
=−
Substitute 2 and
3
2
−
for x to find y.
3
2or
2
26 36
32
xx
y
y
y
==−
=
−=
=
10. Solve the first equation for y.
12
12
xy
yx
=−
=−
Section 13.5 Systems of Nonlinear Equations in Two Variables
12
2140
24 14 0
xx
xx
−− +=
++=
11.
22
9
23
yx
yx
=−
=−
Solve the second equation for x.
23
23
yx
yx
=−
+=
Substitute
23 for yx+
to find y.
12. Solve the first equation for y.
2
2
4
4
xy
yx
+=
=−
13.
22
3
10
xy
xy
=
+=
Solve the first equation for y.
3
3
xy
yx
=
=
Chapter 13 Conic Sections and Systems of Nonlinear Equations
30 30
33
10 10
11
xx
xx
xx
xx
+= −=
=− =
+= −=
=− =
14. Solve the first equation for y.
4
4
xy
yx
=
=
Substitute
4 for
y
x
to find x.
20 or 20
22
xx
xx
+= −=
=− =
Substitute –2 and 2 for x to find y.
15.
22
1
5
xy
xxyy
+=
+− =−
Solve the first equation for y.
1
1
xy
yx
+=
=− +
41
xx
==−
Substitute –1 and 4 for x to find y.
()
1
4or
11
41
311
2
x
x
y
y
yy
y
=−
=
=− − +
=− +
=− =+
=
()()
2
324360
8120
620
xx
xx
xx
++=
++=
++=
Section 13.5 Systems of Nonlinear Equations in Two Variables
Substitute –6 and –2 for x to find y.
17.
()( )
22
1
1210
xy
xy
+=
−++ =
Solve the first equation for y.
1
1
xy
yx
+=
=− +
Substitute
1 for xy−+
to find x.
Substitute 0 and 4 for x to find y.
0or4
01 41
13
xx
yy
yy
==
=− + =− +
==−
18. Solve the first equation for y.
()
()()
2
2
22
2
2
21 22 4
214 8 44
5654
5610
51 10
xx x
xx xx
xx
xx
xx
+++−+ =
+++ −+=
−+=
−+=
−−=
510or 10
51 1
xx
xx
−= −=
==
The solution set is
()
118
,,1,2.
55
19. Solve the system by addition.
22
22
13
xy
+=
Chapter 13 Conic Sections and Systems of Nonlinear Equations
20. Solve the system by addition.
22
22
44
44
xy
xy
−=
+=
21.
22
22
47
3 31
xy
xy
−=−
+=
Multiply the first equation by –3 and add to the
second equation.
22
312 21
xy
−+ =
Substitute –2 and 2 for y to find x.
()
2
2
2
2
3231
3431
y
x
x
=±
+± =
+=
22. Multiply the second equation by –2 and add to the
first equation.
22
22
32 5
xy
−=−
The solution set is
()
{
1, 2
−−
,
()
1, 2
−
,
()
1, 2
−
,
()
}
1, 2 .
23.
22
22
34160
23 50
xy
xy
+−=
−−=
2
2
17 68 0
17 68
4
2
x
x
x
x
−=
=
=
=±
Substitute
2
±
for x to find y.
Section 13.5 Systems of Nonlinear Equations in Two Variables
24. Multiply the second equation by –4 and solve by
addition.
22
22
16 4 72 0
44120
xy
xy
−−=
−+ +=
The solution set is
()
{
5, 2
−−
,
()
5, 2 ,
−
()
5, 2
−
,
()
}
5, 2 .
25.
()
22
22
25
841
xy
xy
+=
−+=
Multiply the first equation by –1 and solve by
addition.
Substitute 3 for x to find y.
3
x
=
26. Multiply the first equation by –1 and solve by
addition.
()
22
2
2
4
3 9
xy
xy
−− =−
+− =
99
232
9
32 16 2 4 2
333
x
x
=
⋅
=± =± =±
The solution set is
422 422
,, , .
33 33
−
2
Apply the zero-product principle.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
28. Multiply the first equation by –1 and solve by
addition.
()()
2
22
2
2
2 8
16
28
280
420
xy
xy
yy
yy
yy
−+=−
+=
+=
+−=
+−=
The solution set is
()
()()
{
}
0, 4 , 2 3, 2 , 2 3, 2 .−−
1
1
y
y
=
=±
Substitute
1±
for y to find x.
()
()
2
2
2
2
2
2
1
34116
34116
3416
312
y
x
x
x
x
=±
+± =
+=
+=
=
40or 30
43
xx
xx
−= +=
==−
Substitute 4 and –3 for x to find y.
3
4or
x
x
=−
=
Section 13.5 Systems of Nonlinear Equations in Two Variables
31.
22
218
4
xy
xy
+=
=
Solve the second equation for y.
4
4 xy y x
=→=
Substitute
4 for y
x in the second equation and solve
Apply the zero-product principle.
2
2
10 10
80
11
8
8
22
xx
x
xx
x
x
x
+= −=
−=
=− =
=
=±
=±
Substitute
22±and 1± for x to find y.
32. Solve the second equation for y.
4
4
xy
yx
=
=
Substitute
4for y
x in the first equation and solve
for x.
2
4
40 40
44
20 20
22
xx
xx
xx
xx
+= −=
=− =
+= −=
=− =
Substitute –4, 4, –2, and 2 for x to find y.
4 4
xx
=− =
Chapter 13 Conic Sections and Systems of Nonlinear Equations
33.
22
420
26
xy
xy
+=
+=
Solve the second equation for x.
26
62
xy
xy
+=
=−
Substitute
6 2 for yx− to find y.
Substitute 1 and 2 for y to find x.
21
or
yy
==
34. Solve the second equation for y.
43
43
xy
xy
−=
−=
Substitute
4 3 for xy−to find x.
()
2
2
32431
xx
−−=
()
19 or 1
29 41 3
19 43
43
29 1
76 87 11
29 29 29
x
x
y
y
y
y
y
=
=
=−
=−
=−
=
=−=−
Substitute –1 and 0 for x to find y.
0or 1
xx
==−
36. Eliminate y by adding.
3
2
32
0
20
20
xy
xy
xx
+=
−=
+=
Section 13.5 Systems of Nonlinear Equations in Two Variables
37.
()
2
2
2
24
20
xy
xy
+− =
−=
Substitute
2
2 for yx
in the first equation and solve
for y.
()
2
224
yy
+− =
Substitute 0 and 2 for y to find x.
() ()
22
22
02
or
20 22
04
02
yy
xx
xx
xx
==
==
==
==±
The solution set is
()
{
0, 0 ,
()()
}
2, 2 , 2, 2 .−
38. Solve by addition.
22
46 40
xy xy
−−+−=
20
2
x
x
−=
=
Substitute 2 for x to find y.
(
)
22
242640
yy
−− +−=
39.
()
2
3lim
22
x
yx
xy
→∞
=+
+=−
()
2
2
2
2692
212182
213182
xxx
xx x
xx
+++=−
+++=−
++=−
5
2
5
2
or 4
42 2
22
22
54 4 1
41
1
4
x
x
y
y
y
y
y
y
y
=−
=−
−+ =−
−+ =−
=
−+ =−
=
=
=
The solution set is
()
()
{
}
51
4,1 , , .
24
−−
()
2
2
22
2
2122 5
214 845
51055
xx x
xx xx
xx
−++ − =
−++ −+=
−+=
Chapter 13 Conic Sections and Systems of Nonlinear Equations
Substitute 0 and 10 for x to find y.
02
or
xx
==
41.
22
322
21
xy y
xy
++=
+=−
Solve the second equation for y.
21
21
xy
yx
+=−
=− −
Substitute
2 1 for xy−−
to find x.
5
Substitute –2 and
12
5
for x to find y.
()
12 or 2
5221
x
x
y
=−
=
=− − −
42. Solve the first equation for x.
35
xy
−=−
()
2
10 30 0
30
30
yy
yy
yy
−=
−=
−=
0or 30
3
yy
y
=−=
=
Substitute 0 and 3 for y to find x.
03
or
yy
==
Solve the second equation for y.
24
24
xy
yx
=
=
Section 13.5 Systems of Nonlinear Equations in Two Variables
Substitute 6 and 4 for x to find y.
64
xx
==
44. Let x = one of the numbers.
Let y = the other number.
20
96
xy
xy
+=
=
for x.
()
()()
2
2
96 20
96 20
96 20
20 96 0
12 8 0
xx
xx x
x
xx
xx
xx
+=
+=
+=
−+=
−−=
12 0 or 8 0
12 8
xx
xx
−= −=
==
45. Let x = one of the numbers.
Let y = the other number.
2
x
=±
Substitute
2± for x to find y.
()
22
2
23
x
y
=±
±−=
–1, or –2 and 1.
46. Let x = one of the numbers.
Let y = the other number.
22
22
5
32 19
xy
xy
−=
−=
Solve for
2
x
in the first equation.
22
22
5
5
xy
xy
−=
=+
Substitute
22
5 for yx+
in the second equation.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
47.
2
2
26
20
xxy
xxy
+=
+=
Multiply the first equation by
2−
and add the two
equations.
For
2x=
:
() ()
2
222 0
44 0
44
1
y
y
y
y
+=
+=
=−
=−
The solution set is
()( )
{
}
2,1 , 2, 1−−
.
48.
2
2
430
39
xxy
xxy
+=
+=−
Multiply the first equation by
3−
and add the
equations.
{
}
For
3x=
:
() ()
2
333 9
99 9
918
2
y
y
y
y
+=
+=−
=−
=−
()()()
3220
xxx
+−+=
3x=−
,
2x=
, or
2x=−
Substitute these values for x in the second equation
and solve for y.
For
3x=−
:
() ()
32
333
27 27
0
y=− + −
=− +
=
For
2x=
:
() ()
32
232
812
20
y=+
=+
=
Section 13.5 Systems of Nonlinear Equations in Two Variables
50.
32
945
5
xy
yx x
−+=
=+
Substitute
32
5xx+
for y in the first equation and
solve for x.
Substitute these values for x in the second equation
and solve for y.
For
5x=−
:
() ()
32
555
125 125 0
y=− + −
=− + =
51.
22
22
31
7
52 3
xy
xy
+=
−=−
Multiply the first equation by 2 and add the
equations.
For
1x=−
: For
1x=
:
()
22
2
31
7
1
1
37
y
y
+=
−
+=
()
22
2
31
7
1
1
37
y
y
+=
+=
1, , 1, , 1, , 1,
2222
−− − −
.
52.
22
22
21
11
42 14
xy
+=
−=−
2
2
8
1
1
x
x
x
=
=
=±
Back-substitute these values for x in the first
equation and solve for y.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
For
1x=
:
1
1, 3
.
53. Answers will vary. One example:
Circle:
22
9xy+=
Ellipse:
22
1
949
xy
+=
54. Answers will vary. One example follows:
Line: 2xy−=
Parabola:
2
xy=
22
40or4 0
04
yy
yy
=+=
==−
Substitute 0 and 4 for y to find x.
22
22
04
or
04 44
40
yy
xx
xx
==−
=− −=−
==
1
y
=±
Since the ship is located in the first quadrant, both x
and y are positive. As a result, disregard –1. Back-
substitute 1
for y to find x.
22
211
x
−=
Section 13.5 Systems of Nonlinear Equations in Two Variables
57. Let x = the length of the rectangle.
Let y = the width of the rectangle.
Perimeter: 2 2 36
Area: 77
xy
xy
+=
=
Solve the second equation for y.
()
()( )
2
2
2
154
236
2 154 36
2361540
18 77 0
7110
xx x
x
xx
xx
xx
xx
+=
+=
−+=
−+=
−−=
58. Let x = the length of the rectangle.
Let y = the width of the rectangle.
Perimeter: 2 2 40
Area: 96
xy
xy
+=
=
Solve the second equation for y.
()
()()
2
2
2
192
240
2 192 40
2401920
20 96 0
12 8 0
xx x
x
xx
xx
xx
xx
+=
+=
−+=
−+=
−−=
Chapter 13 Conic Sections and Systems of Nonlinear Equations
59. Let x = the length of the screen.
Let y = the width of the screen.
22 2
10
48
xy
xy
+=
=
()()()()
88660
xxxx
+−+−=
Apply the zero product principle.
80 80
88
xx
xx
+= −=
=− =
48 48
86
yy
==
60. Let x = the length of the rug.
Let y = the width of the rug.
22 2
15
108
xy
xy
+=
=
()()()()
12 12 9 9 0
xxxx
+−+−=
12 0 12 0
12 12
90 90
xx
xx
xx
+= −=
=− =
+= −=
912
The length of the rug is 9 feet and the width of the
Section 13.5 Systems of Nonlinear Equations in Two Variables
61.
22
21
42 24
xy
xy
−=
+=
511
xx
==
Substitute 5 and 11 for x to find y.
() ( )
511
or
12 2 5 12 2 11
12 10 12 22
210
xx
yy
yy
yy
==
=− =−
=− =−
==−
62. Let L = the length of the piece of cardboard.
Let W = the width of the piece of cardboard.
()( )()
216
4 4 2 224
LW
LW
=
−−=
Solve for W in the second equation.
216
216
LW
WL
=
=
()
2
864
232 4 112
232 4 864 112
LL L
L
LL L
−− =
−−=
63. a. It appears from the graph that the number of
violent crimes per 100,000 Americans was the
same as the number of imprisonments per
100,000 Americans between 2000 and 2005.
b.
2
15 300
0.6 28 730
xy
yxx
−+=
=−+
2
2
2
15 (0.6 28 730) 300
15 0.6 28 730 300
0.6 43 430 0
y
xxx
xx x
xx
−+ − + =
−+ − + =
−+=
Use the quadratic formula.
()
2
2
4
2
( 43) 43 4(0.6)(430)
bb ac
xa
x
−± −
=
−− ± − −
=
Chapter 13 Conic Sections and Systems of Nonlinear Equations
15 300
15(12) 300
xy
y
−+=
−+=
64. – 69. Answers will vary.
70. makes sense
71. does not make sense; Explanations will vary.
Sample explanation: Since the orbits of earth and
Mars do not intersect, their system of equations will
have no solution.
75. true
76. false; Changes to make the statement true will vary.
A sample change is: It is possible that a system of
two equations in two variables whose graphs
78.
()
22 2
2
22
10
917
ab
ab
+=
++ =
Substitute
22
100 for ba− to find b.
2
22
8
a
=±
Disregard –8 because we can’t have a negative
length measurement, and conclude that a is 8 and b
is 6.
79.
()
log 3
log 4 5
y
y
x
x
=
=
()
()()
53
32
3
40
40
220
yy
yy
yy y
−=
−=
+−=
3
2
8
x
x
=
=
The solution is
()
8, 2 and the solution set is