Review Chapter
12
1. {x Ix -3}
2. a. 3x + 7 > 19
3x > 12
x>4
{x Ix > 4}
.
I
4
)
b. 6x-44-2x
8x - 4 4
8x8
x1
{x Ix I}
...
c. -5x+212
-5x 10
x -2
{x Ix -2}
'.
I
-2
1
-3
d. -x-6>
-x
2
2
2x-6> 0
2x>6
x>3
{x Ix > 3}
.
I
3
)
e. -9x>x-(2+4x)
-9x>x-2-4x
-9x> -3x-2
-6x> -2
1
x<3
{XIX<}
I.
I
-1
3
2 7
f. 2x----x
3 3
2 7
3x - - 3 3
3x3
x1
{x Ix I}
.
3.
........-2
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Chapter Review
4. a. x + 3 = 10 and x -1
x=7 andx=3
=2
{x 13 ~ x ~ 7}
b.
9. The graph ofYI = abs(X + 1) intersects the graph
ofY2 = 3 at X ="4 and X = 2.
a. The absolute value function is above three
when X <'4 or 2 <X.
.
I
t
-4
2
2x < - 6 or x + 5 = 7
x < -3 or x
179
=2
{xlx<'30r2~x}
c. .!..x~ -3 and 6x< -30
4
x~ -12 and x < -5
{xl'12::::x<-5}
b. The absolute value function is below three
when X is between '4 and 2.
,
It)
-4
2
10. Yl
= "4X1\2
+ 25X, Y2 = 10
Windowsmayvary:Xmin= '9.4, Xmax= 9.4,
Ymin= '10, Ymax= 40.
d. -3x+l>4or5x>20
-3x>30rx>4
x< -1 orx>4
{x Ix < 'lor x> 4}
e. -(8-x)<-3and
-4x=2x+12
-8+x< -3 and -6x=12
x<5andx=-2
{xl "2::::x<5}
r.7<2x-land2x-I=13
8< 2x and 2x=14
4 < x and x = 7
{x 4 < x::::7}
I
?
-4(0)2 +25*0>10
5. x ~ 100° F; at least 100°F means all temperatures
greater than or equal to 100°F.
6. Let x represent the temperature in degrees
Fahrenheit; {x Ix < 32 or 75 < x}.
7. a.
18+ 20+15 +x
4
Values of X between.43 and 5.82, excluding the
endpoints, are where the parabola YI is above the
line Y2; {x 1.43 < x < 5.82}.
Check
"4~ + 25x > 10
53+x
4
b. 19~53+x
4
c. 76~53+x
23~x
The player must score at least 23 points in the
fourth game to average at least 19 points per
game.
.
23
d.
8. a. 50 ~ 125- 5x and 125- 5x ~ 100
b. -75 = -5x and -5x
15 = x and x = 5
= -25
This is the same as 5 :::: x and x::::15. The
cover charge must be at least $5 and no more
than $15.
c. {x I 5 ~ x ~ IS}
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?
-4(3)2 +25*3>10
?
"!
-4(6)2 +25*6>10
?
0>10 False
-36+75>10
-144+150>10
0> 10False
39> 10 True
6> 10 False
As expected, only x = 3 gives a true solution
because it falls within the solution interval.
11. a. 3; the graph of the basic quadratic function is
at or above 9 when x :::: -3 or x 23.
Yl =XI\2, Y2=9
Windows may vary: Xmin = '4.7, Xmax= 4.7,
Ymin = '15, Ymax = 15.
180
Chapter 12: Inequalities
b. 1; the graph of the basic quadratic function
does not go below O.
Y1 =X"2, Y2="1
Windows may vary: Xmin = "4.7,Xmax = 4.7,
Ymin = -2, Ymax = 2.
b. y} - 8x > 0; we can algebraically solve the
quadratic x2 - 8x = 0 to fmd the exact
endpoints of the solution intervals.
x(x-8)=0
x=O orx-8=0
x=Oorx=8
The exact endpoints of the intervals are 0, 8.
Use a graph to view the intervals.
Y1 =X"2-8X, Y2=0
Windows may vary: Xmin = "18.8,
Xmax = 18.8, Ymin = "20, Ymax = 20.
c. 2; the graph of the basic quadratic function is
below 9 when x is between -3 and 3. See the
graph for 11a.
12. a. 2x2- 5 < 13; we can algebraically solve the
quadratic 2x2 - 5
= 13 to
find the exact
endpoints of the solution interval.
2X2 =18
x2 =9
x=:t3
The exact endpoints of the interval are -3 and
3. Use a graph to view the interval.
Y1 = 2X"2 - 5, Y2 = 13
Windows may vary: Xmin = -4.7, Xmax = 4.7,
Ymin="20, Ymax = 20.
Values of X between "3 and 3 are where the
parabola YI is below the line Y2;
{x I-3 < x < 3}.
Values of X less than 0 or greater than 8 are
where the parabola YI is above the line Y2; {x
Ix<Oor8<x}.
c. x2 + x ~ 12; we can algebraically solve the
quadratic x2 + x = 12 to fmd the exact
endpoints of the solution interval.
x2 +x-12=0
x2-3x+4x-12=0
(x2 -3x)+(4x-12)=0
x(x-3)+4(x-3)=0
(x-3)(x+4) =0
x-3=Oorx+4=0
x=30rx=-4
The exact endpoints are -4 and 3. Use a graph
to view the interval.
Y1 =X"2+X, Y2= 12
Windows may vary: Xmin = "9.4,Xmax= 9.4,
Ymin = "20,Ymax = 20.
Values of X between "4 and 3 are where the
parabola Y1 is at or below the line Y2;
{x 1"4~ x ~ 3}.
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c:
,
Chapter Review
d. Although this could be solved algebraically to
find the exact endpoints of the solution
intervals, we solve the inequality with a graph.
Yl = abs(X - 3), Y2 = 7
Windows may vary: Xmin = '18.8,
Xmax = 18.8, Ymin = 'IS, Ymax = 15.
181
f. E::; ~ ; we can algebraically solve the square
2
root
E =~2 to fmd the exact endpoints of the
solution interval.
E=~2
(Et
Values of X less than or equal "4 or greater
than or equal to 10 are where the absolute
value function Yl is at or above the line Y2;
{xlx::;'4orlO::;x}.
=(%J
25
x=-=6.25
4
The exact endpoint of the interval is 6.25. Use
a graph to view the interval.
Yl ="l/x, Y2=5/2
Windows may vary: Xmin = '9.4, Xmax = 9.4,
Ymin = '5, Ymax = 5.
e. x3 < 64; we can algebraically solve the cubic
x3 = 64 to findthe exactendpointsof the
solution interval.
x3 =64
V=~
x={/4*4*4
x=4
The endpointof the solutionintervalis 4. Use
a graphto viewthe interval.
Yl = X"3, Y2 = 64
Windowsmayvary:Xmin= '9.4, Xmax= 9.4,
Ymin= '100,Ymax= 100.
Il'Ihr5.;:c:~i(Jh
M=fU:S
IY=:::.S
Values of X less than or equal to 6.25 are
where the square root function Yl is at or
below the line Y2; {x I0::; x::; 25
4 }.
13. '7.3p2 + 320p > 3000
Yl = '7.3X"2 + 320X, Y2 = 3000
Windows may vary: Xmin = 0, Xmax = 47,
Ymin = 0, Ymax = 5000.
Il'Ihr5.;:c:~i(Jh
M=&t i
IY=6&t
Valuesof X lessthan 4 arewherethe cubicYl
is belowthe lineY2; {x Ix < 4}.
The solutions are the values of X where the
parabola Yl is above the line Y2.
Any price between $13.59 and $30.25 generates
revenue above $3000.
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.-.I
182
Chapter
12: Inequalities
14. Y1 = abs(X - 5), Y2 = 2
Windows may vary: Xmin = 0, Xmax= 9.4,
Ymin = 0, Ymax= 3.
I
/
17. Only b and d are solutions.
y::::;3
y>x+5
Test a. p, 3)
?
3 > '2 + 5
3::::;3
3>3
True
False
Test b. C5, 1)
?
1::::;3
Values of X less than 3 or greater than 7 are
where the absolute value function Y1 is above the
line Y2. Remove grapefruits less than 3 inches in
diameter or greater than 7 inches in diameter.
15. 40000::::;i and i ::::;50625; we can solve the
following quadratics to fmd the exact endpoints
of the solution interval.
40000
= S2
and
S2
I.J 40000 = f1 and
I 200
=s
and s
= 50625
f1 = I.J
=I
True
1 > '5 + 5
1>0
True
Test c. C5, 4)
4::::;3
False
4 > '5 + 5
4>0
True
Test d. C5, 3)
50625
3::::;3
225
Ignore the negative solution because we can't
have negative lot sizes. The sides of the square
lot are at least 200 feet and no more than 225 feet;
{s 1200 s 225}.
16. Only a. and c. are solutions because they lie in the
shaded region of the graph.
True
3 > '5 + 5
3>0
True
Test e. C5, 0)
0::::;3
0> '5+5
0>0
True
False
Only b. and d. lead to a true statement in both
inequalities.
1
18. y > 3--x
2
Shade above the dashed line for y
y~x-1
Shade above the solid line for y
y.
= 3 _.!.
x
2
=x -
1
~
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--
- -
_.-
~-~---~.....
Chapter Review
19. y~3--x
Sum of two numbers is more
than 100
Sum of two numbers is less
than 150
The larger number is at least
3 times the smaller number
22. a. x + y > 100
1
2
x+y< 150
Shade above the solid line for y
=3 - .!.
x
2
y~3x
y<x-1
Shade below the dashed line for y
y
183
= x-I
b. Solving the inequalities for y gives
y>'x+100
Shade above the dashed line for y = 'x + 100
y<'x+150
Shade below the dashed line for y = 'x + 150
y~3x
Shade above the solid line for y = 3x
Windows may vary:
"
y
"
20..Solving the inequalities for y gives
y~x-5
Shade above the solid line for y = x - 5
y~x+1
Shade below the solid line for y = x + 1
x~O
Shade to the right of the solid line for x
'-ao r40
x
=0
y
c. Answers may vary; (25, 100) and ('40, 150)
are solutions because they fall in the shaded
region of the graph, and (75,50) and (50, 0)
are not solutions because they are outside the
shaded region.
21. Solving the inequalities for y gives
- 1
y~ -x+3
4
Shade below the solid line for y
= - .!.x
4
+3
y < '2x + 6
Shade below the dashed line for y = '2x + 6
x~O
Shade to the right of the solid line for x = 0
y~O
Shade above the solid line for y = 0
x
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23. a. Let x represent the width of the enclosure (in
feet) andy represent the length (in feet).
Recall a formula for the perimeter of a
rectangle, perimeter = 2x + 2y.
2x + 2y > 60
The perimeter is more
than 60
2x + 2y ~ 100 The perimeter is no more
than 100
y ~ 2.5x
The length is at most 2Y2
times the width
y~0
Can't have negative lengths
x~0
Can't have negative widths
184
Chapter 12: Inequalities
b. Solving for y gives
y > -x + 30
Shade above the dashed line for y = -x + 30
y ::;Ox+ 50
Shade below the solid line for y = -x+ 50
y::;2.5x
Shade below the solid line for y = 2.5x
y~O
Shade above the solid line for y = 0
x~O
Shade to the right of the solid line for x = 0
Windows may vary:
y
60_x= 0
e. We can't have partial pants and shirts so the
solutions must be whole numbers. The
ordered pairs (0, 10), (1, 8), (2, 7) when
substituted into each inequality lead to true
statements.
Chapter 12 Test
1. 3x+8<32and3x+8~-1
3x < 24 and 3x ~- 9
x < 8 and x ~- 3
{x 1-3~x<8}
------3
2. - 2x + 1~ 7 or x - 5 > 0
-2x~60rx>5
x::; -3 orx>5
{x Ix::; -3 or 5 < x}
3. Let x represent the number of opposing goals in
the fifth game.
Average less than 4 goals
o
c. (6,30) is not a solution because it falls outside
the shaded region of the graph. Substituting (6,
30) into the inequalities leads to a false
statement in the third inequality.
24. a. 30x + 20y ::;200
b. y ~ 2x
c. There can't be a negative number of pants or
shirts.
d. Solving for y gives
y~-1.5x+ 10
Shade below the solid line for y = -1.5x + 10
y~2x
Shade above the solid line for y = 2x
x~O
Shade to the right of the solid line for x = 0
y~O
Shade above the solid line for y = 0
Windows may vary:
y
3+5+4+6+x<4
5
18+x
-<
4
5
18 + x < 20
x<2
Can't score negative or partial goals, so the
opposing team may score 0 or 1 goal.
4. Let x represent the number of hours the carpenter
plans to work.
1400::;30x + 200 and 30x + 200::; 1700
1200::;30x and 30x::; 1500
40 ::;x and:X::;50
The carpenter will need between 40 and 50 hours
to complete the job.
5. Yl = X"3, Y2 = 4
Windows may vary: Xmin = -4.7, Xmax = 4.7,
Ymin = -8, Ymax = 8
14
II"I~lj:r5r;:c~i(lh
~=1,S:B7't(J11 I't'='t
Values of X greater 1.59 is where the cubic Yl is
above the line Y2; {x Ix > 1.59}.
o
8
=0
10 12~
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.
---
Chapter Test
6. Y1= 5XI\2- 13X,Y2 = 8
Windowsmay vary:Xmin= -4.7,Xmax=4.7,
Ymin= -15,Ymax= 15.
(- .51..8)
9. y>2x-7
x-y:::;4
Solvingfory gives
y>2x-7
Shadeabovea dashedlinefory = 2x - 7
yo::x-4
Shadeabovea solidline fory =x - 4
y
..
..
S
xValues of X between -.51 and 3.11 is where the
parabola Y1 is below the line Y2;
{xl-.51 ~x~3.11}.
7. Y1 = abs(2X + 5), Y2 = 9
Windows may vary: Xmin = -9.4, Xmax = 9.4,
Ymin = -15, Ymax = 15.
x
..
"
I
h
I
=12x-
"
10. xo::O
Shade to the right of a solid line for x = 0
yo::0
Shade above a solid line for y = 0
y:::;5
Shade below a solid line for y = 5
3x+1>y;
this is the same as y < 3x + 1 Shade below a
dashed line for y = 3x + I
y
Values of X less than '7 or greater than 2 is where
the absolute value function Y1 is above the line
Y2; {xlx<"7or2<x}.
8. '4.9r + 2St + 10 0::40
Y1 = -4.9XI\2 + 28X + 10, Y2 = 40
Windows may vary: Xmin = -9.4, Xmax = 9.4,
Ymin = -SO,Ymax = 80.
Values of X between 1.43 and 4.29 is where the
parabola Y1 is above the line Y2. The object is at
least 40 meters high anytime between 1.43
seconds and 4.29 seconds after release.
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i
I
i
-
I
I
1
I
L-3
-.l
, I
-
x
!
1
Xl'"
I
If
185
186
Chapter 12: Inequalities
11. a. x + y ~ 250
The sumof thenumbersis at most250
y~4x
The larger numberis at least4 timesthe
smaller number
x > 0 The smaller number is positive
y > 0 The larger number is positive
b. Solving for y gives
y ~ Ox+ 250
Shade below the solid line for y = Ox + 250
y~4x
Shade above the solid line for y = 4x
x>O
Shade to the right of the dashed line for x = 0
y>O
Shade above the dashed line for y = 0
y,
13. a. x + y < 8
The sum of the driving times is less than 8
hours
40x + 60y ~ 360
The sum of the driving distances is at least 360
miles
x~O
Slowerpersondrivesa minimumof 0 hours
y~O
Faster person drives a minimum of 0 hours
b. Solving for y gives
y < Ox+ 8
Shadebelowthe dashedline for y = .x + 8
y~
-2
-x+6
3
Shade above the solid line for y = - ~x + 6
3
x~O
Shade to the right of a solid line for x = 0
y~O
Shadeabovea solidlinefory = 0
x
c. (10,240) is a solution because substitution
into each inequality leads to a true statement.
12. Let x represent the number of bearings and let y
represent the number of washers.
x + y ~ 5000
Must ship at least 5000 parts
x + y ~ 10,000
Ship at most 10,000 parts
x ~ 0 Can't make negative number of bearings
y ~ 0 Can't make negative number of washers
Solving for y gives
y ~ Ox+ 5000
Shade above the solid line for y = Ox + 5000
y ~ Ox + 10,000
Y
10
4
o
2
4
6
'~I ~=~
8
ill x
c. (4.5,3) is a solution because substitution into
each inequality leads to a true statement.
Shadebelowthe solidlinefory = Ox + 10,000
x~O
Shadeto the right of the solidlineforx = 0
y~O
Shade above the solid line for y = 0
y.
14,000
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