Chapter 13 Review Exercises
Back-substitute 0 and
5
2
for x to find y.
5
0or 2
xx
==
48.
22
2
13
7
xy
xy
+=
−=
Solve for
2
x
in the second equation.
2
7
xy
−=
Back-substitute –3 and 2 for y to find x.
22
3or 2
77
yy
xy xy
=− =
=+ =+
49.
22
22
23 21
34 23
xy
xy
+=
−=
Multiply the first equation by 4 and the second
()
()
22
2
2
2
3
23 3 21
29 3 21
18 3 21
33
x
y
y
y
y
=±
±+ =
+=
+=
=
Solve the first equation for y.
22 26
13
13
xy
xy
yx
+=
+=
=−
Chapter 13 Conic Sections and Systems of Nonlinear Equations
51.
28
6
xy
xy
+=
=
Solve the first equation for y.
Substitute –2x + 8 for y in the second equation.
()
2
286
286
xx
xx
−+=
−+=
Back-substitute 1 and 3 for x to find y.
31
or
xx
==
52. Using the formula for the area, we have
22
2900.xy+=
Since there are 240 feet of fencing
available, we have
() ()
240
240
4 2 240.
xxyyyxyx
xxyyyxyx
xy
+++++−+=
+++++−+=
+=
()
()()
2
2
22
2
2 120 2900
4 480 14400 2900
5 480 11500 0
50 46 0
xx
xx x
xx
xx
+− + =
+− + =
−+ =
−−=
50 0 or 46 0
50 46
xx
xx
−= −=
==
20 28
yy
==
The solutions are
50 feetx=
and 20y= feet or
46 feetx=
and 28y= feet.
()( )
22
3225
xy
−++ =
2.
()()
() ()
()
22
2
22
5349
537
xy
xy
−++=
−+−− =
The center is
()
5, 3−
and the radius is 7 units.
Chapter 13 Test
4.
()
2
237
xy
=− + +
5.
210 23xy y=+ +
The y–coordinate of the vertex is
6.
22
1
49
xy
−=
Because
22
and xy
have opposite signs, the
equation’s graph is a hyperbola.
The equation is in the form
22
22
1
xy
−=
with
7.
22
49 36xy+=
22
22
49 36
36 36 36
1
xy
xy
+=
+=
right. The vertex of the parabola is
()
4, 1 .−−
Replace y with 0 to find the x–intercept.
() ()
22
01 4 1 4 14 3.x= + −= −=−=−
The x–intercept is
3−
. Replace x with 0 to find the
y–intercepts.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
The y–intercepts are –3 and 1.
9.
22
16 16xy+=
Because
22
and xy
have different positive
coefficients, the equation’s graph is an ellipse.
10.
22
22
25 9 225
25 9 225
yx
yx
=+
−=
22
9, and 25ab==
. We know the transverse axis
lies on the y-axis and the vertices are
() ()
0, 3 and 0, 3 .−
Because
22
9 and 25,ab==
()
2
3 6 3 9 18 9.x=− + =− + =
The vertex of the parabola is
()
9,3 .
Replace y with 0 to find the x–intercept
()
2
060000x=− + = + =
The x–intercept is 0. Replace x with 0 to find the
y–intercepts.
()
2
06
06
yy
yy
=− +
=− −
0or 60
yy
−= −=
Chapter 13 Test
12.
()()
22
23
1
16 9
xy−+
+=
Because
22
and xy
have different positive
coefficients, the equation’s graph is an ellipse.
Center Vertices Endpoints of
Minor Axis
()
2, 3−
()
()
24,3
2, 3
−−
=− −
()
()
2, 3 3
2, 6
−−
=−
13.
()( )
22
129xy+++ =
Because
22
and xy
have the same positive
coefficient, the equation’s graph is a circle.
The center of the circle is
()
1, 2−−
and the radius is
3.
14.
()
22
22
22
1
44
441
44
4
xy
xy
xy
+=
+=
+=
()()
22
2
2
2
2125
22125
22240
12 0
430
xx x
xx
xx
xx
xx
+−+=
−+=
−−=
−− =
−+=
40 or 30
43
xx
xx
−= +=
==−
Back-substitute –3 and 4 for x to find y.
Chapter 13 Conic Sections and Systems of Nonlinear Equations
16.
22
22
25 2
32 35
xy
xy
−=−
+=
()
2
22
2
9
25 2
29 5 2
x
xy
y
=
−=−
−=−
17.
239
180
xy
xy
+=
=
Solve the first equation for y.
239
39 2
xy
yx
+=
=−
Back-substitute
15
2
and 12 for x to find y.
15 or 12
xx
==
Let y = the width of the rectangle
The system of two equations in two variables is as
follows.
22
25
2214
xy
xy
+=
+=
Solve the second equation for y.
2214
xy
+=
5 feet y
Cumulative Review
Back-substitute 3 and 4 for x to find y.
4or 3
xx
==
Cumulative Review Exercises
(Chapters 1-13)
1.
374or6 1
33 5
15
xx
xx
xx
+> −<
>− − <−
>− >
The solution set is
()
1, .−∞
2.
(
)
2
274
xx
−=
3.
()()
2
530
1
39
530
1
333
xx
xxx
=+
−−
=+
−+−
Chapter 13 Conic Sections and Systems of Nonlinear Equations
4.
2
3850xx++<
The boundary points are
5and 1.
3
−−
Interval Test Value Test Conclusion
5
,3
−∞ −
−2
() ()
2
32 82 50
10, false
−+−+<
<
5
,3
−∞ −
does not belong to the solution set.
5.
21
21 4
381
33
214
25
5
2
x
x
x
x
x
−
−
=
=
−=
=
=
Cumulative Review
7.
22
22
34 39
52 13
xy
xy
+=
−=−
We can back-substitute 1 for
2
x
to find y.
()
2
22
2
1
34 39
31 4 39
x
xy
y
=
+=
+=
8.
()
24
3
24
3
fx x
yx
=− +
=− +
The y–intercept is 4 and the slope is
2
−
. We can
9.
36xy−>
First, find the intercepts to the equation 36xy−=.
Next, use the origin as a test point.
()
30 0 6
00 6
06
−>
−>
>
This is a false statement. This means that the origin
10.
4690xy xy++−+=
Because
22
and xy
have the same positive
coefficient, the equation’s graph is a circle.
()()
22
4 6 9xx yy++− =−
Complete the squares.
22
2
424
b
Chapter 13 Conic Sections and Systems of Nonlinear Equations
11.
22
94 36xy−=
Because
22
and xy
have opposite signs, the
equation’s graph is a hyperbola.
12.
(
)
()
33
2
2 3 12 45 9 3 2 9 12 45 9 3
−−−÷−=−−−÷−
13.
()
()
32
31917434xxx x−++÷−
Rewrite the polynomials in descending order and
divide.
14.
25 2
33
25 2 37
33
44
44 16
xy xy
xy xy xy
⋅
==
()()()
233
xxx
=− + −
18. Since the radicand must be positive, the domain will
exclude all values of x which make the radicand less
than zero.
63 0
36
2
x
x
x
−≥
−≥−
≤
()
()
2
2
2
2
1
1
x
x
−
=
−
Cumulative Review
21.
()
()
32
3521 2
xxx x
−+−÷−
2
−
22.
23 or 23
23 0 23 0
xx
xx
=− =
+= −=
Multiply the factors to obtain the polynomial.
23. Let x = the rate of the slower car
r t d
Fast x + 10 2 2(x + 10)
Slow x 2 2x
24. Let x = the number of miles driven in a day.
39 0.16
R
Cx
=+
175
x
=
The cost is the same when renting from either
company when 175 miles are driven in a day.
()
39 0.16 175 39 28 67
R
C
=+ =+=
Multiply the first equation by –3 and the second
equation by 2 and solve by addition.
9 6 1062
4 6 762
5 300
xy
xy
x
−− =−
+=
−=−