Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 8
Show all of your work in order to receive full credit. Attach graph paper for your graphs.
1. Simplify. All of the exponents in your final answer should be positive.
a) 4
2 8
8
10
wp
p w
4 2
2 8 7
8 4
10 5
wp p
p w w=
b) ( ) ( )3 6 86 12k k m−
( ) ( )3 6 8 9 86 12 72k k m k m− = −
c) ( )002 2x y+
( )002 2 2 1 1
2 1
3
x y+ = ⋅ +
= +
=
d) 1 3
2 2
21
3
m n
n r
−
−
−
1 3 3 2
2 2 2
2
21 21
3 3
7
m n n r
n r mn
nr
m
−
−
− −=
−=
2. Write 4,321,000 in scientific notation.
64,321,000 4.321 10= ×
3. Write 78.206 10−× in decimal form.
78.206 10 0.0000008206−
× =
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 8
4. Evaluate ( ) ( )16 104.23 10 3.14 10−× × , and write answer in scientific
notation.
( ) ( )16 10 6
7
4.23 10 3.14 10 13.2822 10
1.32822 10
−× × = ×
= ×
5. Perform the indicated operation.
a) ( ) ( )2 24 5 6 3 6 7a a a a− + − + −
( ) ( )2 2 2 2
2
4 5 6 3 6 7 4 5 6 3 6 7
11 13
a a a a a a a a
a a
− + − + − = − + − − +
= − +
b) ( )2
3 4r −
( )2 23 4 9 24 16r r r− = − +
c) ( ) ( )2 1 2 1x x+ −
( ) ( ) 22 1 2 1 4 1x x x+ − = −
d) ( ) ( )22 6a a a+ − +
( ) ( )2 3 2 2
3 2
2 6 6 2 2 12
4 12
a a a a a a a a
a a a
+ − + = − + + − +
= + + +
6. Factor.
a) 23 7 6a a+ −
( ) ( )
23 7 6
3 2 3
a a
a a
+ −
− +
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 8
b) 2 3 3 2 23 6 12x y x y x y+ −
( )
2 3 3 2 2
2 2
3 6 12
3 2 4
x y x y x y
x y y xy
+ −
+ −
c) 28 8 35 10ab ac b c− − +
( ) ( )
( ) ( )
28 8 35 10
4 7 2 5 7 2
7 2 4 5
ab ac b c
a b c b c
b c a
− − +
− − −
− −
d) 3 364 125m n+
( ) ( )
3 3
2 2
64 125
4 5 16 20 25
m n
m n m mn n
+
+ − +
7. Solve by factoring.
a) 22 18 40 0x x+ + =
( ) ( )
2
2
2 18 40 0
9 20 0
5 4 0
x x
x x
x x
+ + =
+ + =
+ + =
5 0 or 4 0
5 or 4
x x
x x
+ = + =
= − = −
b) 225 36 0x − =
( ) ( )
225 36 0
5 6 5 6 0
x
x x
− =
+ − =
5 6 0 or 5 6 0
5 6 or 5 6
6 6 or
5 5
x x
x x
x x
+ = − =
= − =
= − =
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 8
8. P varies inversely as q. If 4P = when 10q = , find P when 16q = .
410
40
kP
q
k
k
=
=
=
40
40
16
5
2
Pq
=
=
=
When 16q = , 5
2P = .
9. Perform the indicated operation and simplify.
a) 3 2
3 2
6
4 12
x x
x x x
+
+ −
( )
( )
( )
( ) ( )
( )
23 2
3 2 2
2
2
66
4 12 4 12
6
6 2
6
x xx x
x x x x x x
x x
x x x
x x
++=
+ − + −
+=
+ −
+=
( )6x x + ( )2
2
x
x
x
−
=−
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 8
b) 2 2
3 2 2
3 12 1 3 6
1 5 6 2 3
x x x x
x x x x x
− + + −⋅ ÷
− + + + −
( )( ) ( ) ( ) ( )
( ) ( )
( )
2 2
3 2 2
2 2 2
3 2
2 2
2
3 12 1 3 6
1 5 6 2 3
3 12 1 2 3
1 5 6 3 6
3 4 3 11
3 2 3 21 1
3
x x x x
x x x x x
x x x x x
x x x x
x x xx x
x x xx x x
− + + −⋅ ÷
− + + + −
− + + + −= ⋅ ⋅
− + + −
− + −+ += ⋅ ⋅
+ + −− + +
=2x +( ) 2x −( )
1x −( ) 2 1x x+ +( )
2 1x x+ +⋅
3x +( ) 2x +( )
3x⋅
+( ) 1x −( )3 2x −( )
1=
10. Solve.
a) 4 5
2 8x x=
−
( )
4 5
2 8
4 2 8 5
8 32 5
3 32
32
3
x x
x x
x x
x
x
=−
− =
− =
=
=
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 6 of 8
b) ( )
3 6 105 201
5 5
x
x x x x
−+ = −
+ +
( )
( )
( ) ( )( )
( ) ( ) ( )
( ) ( )
2 2
2
2
3 6 105 201
5 5
LCD: 5
3 6 105 205 1 5
5 5
3 6 1 5 105 20 5
3 6 5 105 20 100
4 20 5
4 19 5 0
4 1 5 0
4 1 0 or 5 0
4 1 or 5
1 or 5
4
x
x x x x
x x
xx x x x
x x x x
x x x x x
x x x x x
x x x
x x
x x
x x
x x
x x
−+ = −
+ +
+
− + + = + − + +
− + ⋅ + = − +
− + + = − −
− = − +
+ − =
− + =
− = + =
= = −
= = −
Check: 5x = −
( )
( )
?3 5 6 105 201
5 5 5 5 5 5
15 6
0
− −+ = −
− + − − + −
− − ?
1 undefined!+ =
Check: 1
4x =
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 7 of 8
?
?
?
?
?
?
13 6
105 2041
1 11 15 5
4 44 4
3 24105 204 4 1
1 20 1 5 1
4 4 16 4 4
211054 1 80
21 1 20
4 16 16
1051 1 80
21
16
0 80 80
0 0
−
+ = −
+ +
−
+ = −
+ +
−
+ = −
+
− + = −
= −
= �
Answer: 1
4x =
11. Working together, Randy and Joshua can put a roof on a house in 6 hours. Working alone, Randy could do the job in half the time that it
would take Joshua. How long would it take each of them to do the job alone?
Let R be the number of hours it would take Randy to do the job. Then it would take Joshua 2R hours to complete the job.
Working alone, in one hour Randy could do 1
R of the job. Also, if
Joshua is working alone, in one hour he could do 1
2R of the job.
Together, they could do 1
6 of the job in one hour.
This yields the equation:
Unit 3: Lessons 6, 7, and 8 Unit 3 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 8 of 8
1 1 1
2 6R R+ =
Solving for R gives:
1 1 1
2 6
1 1 16 6
2 6
6 3
9
R R
R RR R
R
R
+ =
⋅ + = ⋅
+ =
=
Answer: Randy could do the job in 9 hours, Joshua could do the
job in 9 2 18⋅ = hours.
