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Addition or
Subtraction
Multiplication
or Division
Story
Problems
Synthetic or
Long Division
Simplify
100
100
100
100
100
200
200
200
200
200
300
300
300
300
300
400
400
400
400
400
500
500
500
500
500
Perform the indicated operation and
simplify your answer completely
2t  1 2t  4

5t
t2
LCD : 5t (t  2)
t  2 2t  1 2t  4 5t



t  2 5t
t  2 5t
(2t  1)(t  2) 5t (2t  4)

5t (t  2)
5t (t  2)
2t 2  t  4t  2 10t 2  20t

5t (t  2)
5t (t  2)
8t  17t  2
5t (t  2)
2
Perform the indicated operation and
simplify your answer completely.
3x
5

2
2
x 1 x  2x 1
3x
5

( x  1)( x  1) ( x  1)( x  1)
x 1
3x
5
x 1



x  1 ( x  1)( x  1) ( x  1)( x  1) x  1
3x( x  1)  5( x  1)
( x  1)( x  1)( x  1)
3x 2  3x  5 x  5
( x  1)( x  1)( x  1)
3x  2 x  5
( x  1)( x  1)( x  1)
2
Perform the indicated operation and
simplify your answer completely.
3
4x  7

x  2 x  2
3
4 x  7 1


x  2  x  2 1
3
(4 x  7)

x2
x2
3  4x  7
x2
4 x  10
x2
2(2 x  5)
x2
Solve the following equation. Do not
forget to check for extraneous solutions.
1
4
3x  2

 2
x  6 x 1 x  7x  6
1
4
3x  2


x  6 x  1 ( x  6)( x  1)
1 ( x  6)( x  1)
4 ( x  6)( x  1)
3x  2
( x  6)( x  1)





x6
1
x 1
1
( x  6)( x  1)
1
1( x  1)  4( x  6)  3x  2
x 1 4x  24  3x  2
5x  25  3x  2
2x  23
23
x
2
Solve the following equation. Do not
forget to check for extraneous solutions.
5
x

1
x 5 x 5
5 x 5
x x 5
x 5



 1
x 5 1
x 5 1
1
5  x  ( x  5)
5  x  x5
55
All real numbers except
x5
Perform the indicated operation and
simplify your answers completely.
x  6x  8 x  3

2
x  4x  3 x  2
2
( x  2)( x  4) x  3

( x  1)( x  3) x  2
( x  4)
( x  1)
Perform the indicated operation and
simplify your answers completely.
x  9 x  81

2x  4 x  2
2
x 9
( x  9)( x  9)

2( x  2)
x2
Keep it…Switch it…flip it.
x 9
x2

2( x  2) ( x  9)( x  9)
1
2( x  9)
Perform the indicated operation and
simplify your answer completely.
a  b a  2ab  b

2 2
ab
2a b
2
2
2
2
(a  b)(a  b) (a  b)(a  b)

2 2
ab
2a b
(a  b)(a  b) 2  a  a  b  b

ab
(a  b)(a  b)
2ab(a  b)
( a  b)
Perform the indicated operation and
simplify your answer completely.
4 x  9 y 4 x  6 xy  9 y

3
3
2
2
8x  27 y 4 x  12 xy  9 y
2
2
2
2
(2 x  3 y)(2 x  3 y)
4 x  6 xy  9 y

2
2
(2 x  3 y)(4 x  6 xy  9 y ) (2 x  3 y)(2 x  3 y)
2
1
2x  3y
2
Perform the indicated operation and
simplify your answer completely.
8 y  27
4y 9

3
2
64 y  1 16 y  4 y  1
3
2
(2 y  3)(4 y  6 y  9) (2 y  3)(2 y  3)

2
2
(4 y  1)(16 y  4 y  1) 16 y  4 y  1
2
(2 y  3)(4 y  6 y  9) 16 y  4 y  1

2
(4 y  1)(16 y  4 y  1) (2 y  3)(2 y  3)
2
2
(4 y  6 y  9)
(4 y  1)(2 y  3)
2
The weight (M) of an object on the
moon varies directly as its weight (E)
on Earth. A person who weighs 95 kg
on Earth weighs 15.2 kg on the moon.
How much would a 105 kg person
weigh on the moon?
Moon weight = k • Earth weight
15.2  k  95
.16  k
Moon weight = 0.16 • Earth weight
y  0.16 105
y  16.8kg
The time (t) required to drive a fixed
distance varies inversely as the speed
(r). It takes 5 hours at 80km/h to
drive a fixed distance. How long
would it take to drive the fixed
distance at 60 km/h?
k
time =
speed
k
5
80
400km / h  k
400
time = speed
400
t
60
2
t  6 hours
3
A panda can eat leaves from a
certain bamboo stem in 14 minutes.
Together, two pandas could eat the
leaves from the same stem in 9
minutes. How long would it have
taken the second
panda to eat the
leaves from the
stem by itself?
9
1 1 1
 
14 t 9
LCD :126t
14
1 126t 1 126t 1 126t

 
 
14 1
t 1
9 1
9t 126  14t
126  5t
25.2min.  t
A car travels 300 km in the same
time that a freight train travels
200 km. The speed of the car is
20 km/hr more than the speed of
the train. Find the speed of the
car and the speed of the train.
car
train
d
300 km
200 km
r
r + 20
r
t
t
t
Car time = train time
Train speed  40km / h
Car speed  60km / h
300
200

r  20
r
300r  200(r  20)
300r  200r  4000
100r  4000
r  40
d
t
r
A paddleboat can move at a speed of 2
km/hr in still water. The boat is paddled
4km downstream in a river in the same
time it takes to go 1 km upstream. What
is the speed of the river?
c = speed of the current
upstream
downstream
d
1 km
4 km
r
2-c
2+c
t
t
t
d
t
r
upstream time = downstream time
1
4

2c 2c
1(2  c)  4(2  c)
2  c  8  4c
5c  6
6
c  km / h
5
Divide using long division.
(2 x  3x  5)  ( x  1)
2
2x  1
2
x  1 2 x  3x  5
2x  2x
x5
2
x 1
4
2x  1

4
x 1
Divide using synthetic division.
( x  4 x  6 x  4 x  1)  ( x  1)
4
3
2
1 1 4 6 4 1
1 3 3 1
1 3 3 1 0
x  3x  3x  1
3
2
Divide using synthetic division.
(4 x  x  7)  ( x  2)
3
2
4 0
1 7
8 16 34
4 8 17 41
41
4 x  8 x  17 
x2
2
Divide using long division.
( y  11y  6)  ( y  3)
3
2
y  8 y  24
2
y  3 y  11y  0 y  6
3
2
y  3y
3
2
8 y  0 y
2
8 y  24 y
24 y  6
24 y  72
2
y  8 y  24
2
66

y 3
66
Divide using long division.
( y  y  54)  ( y  3)
4
2
2
y 2
2
y  3 y  0 y  y  0 y  54
4
2
y
3y
2
4
3
2
2 y  0 y  54
2
6
2y
48
2
y 2
2
48
 2
y 3
Simplify.
4x  9
2
4 x  12 x  9
2
(2 x  3)(2 x  3)
(2 x  3)(2 x  3)
(2 x  3)
(2 x  3)
The complex fractional expression
1 1

x y
x2  y2
xy
can be simplified by multiplying the
numerator and denominator by:
A. ( x  y)( x  y)
B. xy
C. x  y
D. x  y
B
Simplify completely.
3
5
x
1
1
2x
3
3 2x
5
  5  2 x 6  10 x
2x x 1
x



1
1 2x
2
x
1

2
x
1
  1 2 x
2x
2x 1
2(3  5 x)
1 2x
Simplify.
5
4
 2
x2 x
3
3
x
5
4
5 x 2 ( x  2) 4 x 2 ( x  2)
 2 2

 2
x  2 x  x ( x  2)  x  2
1
x
1
2
3
3 x  x( x  2)
x 2 ( x  2)
x
(
x

2)
3

 3
x
x
1
1
5x  4( x  2)

2
3x( x  2)  3x ( x  2)
2
5x  4 x  8
 2
3x  6 x  3x3  6 x 2
2
5x2  4 x  8
 3
3x  3x 2  6 x
5x2  4 x  8

3x( x  2)( x  1)
Simplify.
1
1
1
5 x  5 y  10 x y
1
1 1
6 y  12 x y
1
5 5 10
5 xy 5 xy 10 xy
    
 
x y xy
x 1 y 1 xy 1

6 xy 12 xy
6 12
  

y 1 xy 1
y xy
5 y  5 x  10

6 x  12
5( y  x  2)

6( x  2)