Lesson 7: Factoring Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 4
Solve.
1. The perimeter of a rectangle is 154 mm and the area of the rectangle is
360 2mm , find the length and the width of the rectangle.
Perimeter = 2 2 154l w+ =
Area = 360l w⋅ =
We can solve this system of equations by substitution. First, we
solve the first equation for l in terms of w , and then substitute that into the second equation.
2 2 154
2 154 2
2 154 2
2 2
77
l w
l w
l w
l w
+ =
= −
−=
= −
( )2
360
77 360
77 360
l w
w w
w w
⋅ =
− =
− =
We now have a quadratic equation in one variable, which we can
solve by factoring.
( ) ( )
2
2
2
77 360
0 360 77
77 360 0
72 5 0
w w
w w
w w
w w
− =
= − +
− + =
− − =
72 0 or 5 0
72 or 5
w w
w w
− = − =
= =
If 72w = , then 77 72 5l = − = . If 5w = , then 77 5 72l = − = .
The dimensions of the rectangle are 5 mm and 72 mm, but we
don’t know which is the length and which is the width.
I: This problem can also be solved by solving the second
equation for l and substituting that into the first equation to get
3602 2 154ww
+ =
. This can be solved by the methods in lesson
8.
Lesson 7: Factoring Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 4
2. The length of the hypotenuse of a right triangle is 7 inches less than 4
times the shorter leg. The longer leg is 2 inches more than twice the
shorter leg. What are the dimensions of the triangle?
By the Pythagorean theorem, we know that
( ) ( )
( ) ( )
2 22
2 2 2
2 2
2
2 2 4 7
4 8 4 16 56 49
5 8 4 16 56 49
0 11 64 45
0 11 9 5
s s s
s s s s s
s s s s
s s
s s
+ + = −
+ + + = − +
+ + = − +
= − +
= − −
11 9 0 or 5 0
11 9 or 5
9 or 5
11
s s
s s
s s
− = − =
= =
= =
If 9
11s = , then
9 402 2 2 2
11 11s
+ = + =
, and
9 414 7 4 7
11 11s
− − = − =
. Length cannot be negative.
If 5s = , then ( )2 2 2 5 2 12s + = + = , and ( )4 7 4 5 7 13s − = − = .
The dimensions of the triangle are 5, 12, and 13. This is a
Pythagorean triple so the answer checks.
Answer: The dimensions of the triangle are 5 inches, 12 inches
and 13 inches.
4s-7
s
2s+2
Lesson 7: Factoring Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 4
3. The product of two consecutive odd integers is 1 less than 6 times their
sum. What are the integers?
Let x be the first odd integer. Then 2x + is the next odd integer.
( ) ( )( )( )
( ) ( )
2
2
2
2 6 2 1
2 6 2 2 1
2 12 12 1
10 11 0
11 1 0
x x x x
x x x
x x x
x x
x x
+ = + + −
+ = + −
+ = + −
− − =
− + =
11 0 or 1 0
11 or 1
x x
x x
− = + =
= = −
If 11x = , then the two numbers are 11 and 13.
If 1x = − , then the two numbers are -1 and 1.
Check:
( )
( )
?
?
?
?
11 13 6 11 13 1
143 6 24 1
143 144 1
143 143
⋅ = ⋅ + −
= ⋅ −
= −
= �
( )
( )
?
?
?
?
1 1 6 1 1 1
1 6 0 1
1 0 1
1 1
− ⋅ = ⋅ − + −
− = ⋅ −
− = −
− =− �
Both sets of numbers check, so there are two possible solutions.
Answer: The integers are 11 and 13, or -1 and 1.
Lesson 7: Factoring Problem Solving Assignment solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 4
4. The product of a number and twice the number plus 11 is 121. What are
the numbers?
Let n be the number.
( )
( ) ( )
2
2
2 11 121
2 11 121
2 11 121 0
2 11 11 0
n n
n n
n n
n n
+ =
+ =
+ − =
− + =
2 11 0 or 11 0
2 11 or 11
11 or 11
2
n n
n n
n n
− = + =
= = −
= = −
If 11
2n = , then
112 11 2 11 22
2n
+ = + =
.
If 11n = − , then ( )2 11 2 11 11 11n + = − + = − .
Check:
( )
?
?
1122 121
2
121 121
⋅ =
= �
( ) ( )
?
?
11 11 121
121 121
− ⋅ − =
= �
Both sets of numbers check, so there are two possible solutions.
Answer: The number is 11
2, or -11.
