Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 1 of 9
Show all of your work in order to receive full credit. Attach graph paper for your graphs.
1. Give the slope of the line that passes through the points ( )4, 7− and
( )2,4− .
( )
7 4
4 2
11
6
m− −
=− −
−=
The slope is 11
6− .
2. For the linear equation 6 8 10x y− = , find the x- and y- intercepts, and
the slope of the line.
a) x-intercept
To find the x-intercept, we let 0y = and solve for x.
( )
6 8 10
6 8 0 10
6 10
10
6
5
3
x y
x
x
x
x
− =
− =
=
=
=
The x-intercept is 5,0
3
.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 9
b) y-intercept
To find the y-intercept, we let 0x = and solve for y.
( )
6 8 10
6 0 8 10
8 10
10
8
5
4
x y
y
y
y
y
− =
− =
− =
= −
= −
The y-intercept is 5
0,4
−
.
c) Slope
6 8 10
8 6 10
6 10
8 8
3 5
4 4
x y
y x
y x
y x
− =
− = − +
−= +
− −
= −
The slope is 3
4.
3. Given points A ( )4,8 and B ( )9, 4− ,
a) Find the distance between A and B.
( ) ( )( )
( ) ( )
222
2 22
2
2
4 9 8 4
5 12
25 144
169
13
d
d
d
d
d
= − + − −
= − +
= +
=
=
The distance is 13 units.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 9
b) Find the equation of a circle that passes through point B and whose center is at point A.
From part (a), the radius is 13. The center is ( )4,8 . The equation
is:
( ) ( )
( ) ( )
2 2 2
2 2
4 8 13
4 8 169
x y
x y
− + − =
− + − =
4. Write the equation of the line, in slope-intercept form, which has slope 2
and passes through ( )5, 7− .
( )7 2 5
7 10
17
y mx b
b
b
b
= +
− = +
− = +
− =
The equation of the line is 2 17y x= − .
5. Graph the following on separate graph paper:
a) 2
43
y x= −
The slope is 2
3 and the y-intercept is -4.
See next page for graph.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 9
b) 4y =
The slope is 0 and the y-intercept is 4. This is a horizontal line.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 9
c) 6 5 15x y− ≤
6 5 15
5 6 15
63
5
x y
y x
y x
− ≤
− ≤ − +
≥ −
y-intercept is ( )0, 3− ;
slope is 6
5
6. Write the equation of the line, in standard form, that passes through
( )3, 4− and is perpendicular to the line 3 2 18x y+ = .
3 2 18
2 3 18
39
2
x y
y x
y x
+ =
= − +
= − +
The slope of the given line is 3
2− , so the slope of the new line is
2
3.
( )
24 3
3
4 2
6
y mx b
b
b
b
= +
− = +
− = +
− =
The equation of the line is 2
63
y x= − .
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 6 of 9
7. A right triangle has legs of length 320 and 600. Find the length of the hypotenuse.
There are two ways to solve this problem. (1) Using Pythagorean triples: The ratio of the legs is
320 : 600 8 :15= . This is a multiple of the ( )8,15,17
Pythagorean triple. Since 320 8 40= ⋅ and 600 15 40= ⋅ , the
hypotenuse is equal to 17 40⋅ . The hypotenuse is
17 40 680⋅ = .
(2) Using the Pythagorean Theorem:
2 2 2
2 2 2
2
2
320 600
102400 360000
462400
680
a b c
c
c
c
c
+ =
+ =
+ =
=
=
The hypotenuse is 680.
8. Lucy wants to limit the number of calories she eats for lunch to 800. She is having chicken nuggets and ice cream. Each nugget contains 80
calories, and each scoop of ice cream contains 120 calories.
a) Write, in standard form, the inequality that represents the number of nuggets (n) and the number of scoops of ice cream (c) that she can
eat and stick to her diet.
80 120 800n c+ ≤
b) How many nuggets can she eat if she has 4 scoops of ice cream?
( )
80 120 800
80 120 4 800
80 480 800
80 320
4
n c
n
n
n
n
+ ≤
+ ≤
+ ≤
≤
≤
Lucy can eat up to 4 nuggets.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 7 of 9
9. Solve the following systems algebraically.
a) 2 1
4 2 6
y x
x y
= +
− =
Using substitution,
( )4 2 2 1 6
4 4 2 6
2 6
x x
x x
− + =
− − =
− ≠
No solution!
b) 2 3
3 4 13
x y
x y
+ =
− − =
Using elimination,
( )
2 3
3 4 13
4 2 3
3 4 13
8 4 12
3 4 13
5 25
5
x y
x y
x y
x y
x y
x y
x
x
+ =
− − =
+ =
− − =
+ =
− − =
=
=
If 5x = , then
2 3
2 5 3
10 3
7
x y
y
y
y
+ =
⋅ + =
+ =
= −
Solution is ( )5, 7−
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 8 of 9
c) 2 5 6
20 8 24
x y
y x
− =
− = −
Using elimination,
( )
2 5 6
20 8 24
4 2 5 6
8 20 24
8 20 24
8 20 24
0 0
x y
y x
x y
x y
x y
x y
− =
− = −
− =
− + = −
− =
− + = −
=
There are an infinite number of solutions. The solutions to this system
of equations is the set of ordered pairs ( ),x y that satisfy the equation
2 5 6x y− = .
10. Solve the following system graphically.
2 2
5 3
x y
x y
− <
+ ≤ −
2 2
2 2
11
2
x y
y x
y x
− <
− < − +
> −
5 3
5 3
x y
y x
+ ≤ −
≤ − −
See graph on next page. The maroon area is the solution to this
system of equations.
Unit 2: Lessons 3, 4, and 5 Unit 2 Exam solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 9 of 9
