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Lesson 4: Graphing Linear Equations and Inequalities Quiz 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 graphs.
1. Graph the following equations on separate graphs.
a) 2 6x y+ =
2 6
2 6
13
2
x y
y x
y x
+ =
= − +
= − +
slope = 1
2−
y-intercept is ( )0,3
b) 4y =
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 2 of 8
c) 4 2 8x y− =
4 2 8
2 4 8
2 4
x y
y x
y x
− =
− = − +
= −
slope = 2;
y-intercept is ( )0, 4−
d) 2x = −
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 3 of 8
2. Which of the following are solutions of 2x + y = 5? Answer yes or no.
a) (2, 1)
( )?
?
?
2 5
2 2 1 5
4 1 5
5 5
x y+ =
+ =
+ =
= �
Yes
b) (-3, 1)
( )?
?
?
2 5
2 3 1 5
6 1 5
5 5
x y+ =
− + =
− + =
− = �
No
3. Given the equation 5 7 10 3 20x y x− + = +
a) Write the equation in slope-intercept form.
5 7 10 3 20
7 2 10
2 10
7 7
x y x
y x
y x
− + = +
− = − +
= −
b) State the slope and y-intercept.
Slope is 2
7; y-intercept is
10
7− , or
100,
7
−
.
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 4 of 8
4. What are the x- and y-intercepts of 4 3 16x y− = − ?
x-intercept y-intercept 0y = 0x =
( )4 3 16
4 3 0 16
4 16
4
x y
x
x
x
− = −
− = −
= −
= −
( )4 3 16
4 0 3 16
3 16
16
3
x y
y
y
y
− = −
− = −
− = −
=
x-int is ( )4,0− ; y-int is 16
0,3
5. Find the slope of the line containing the points (4, -3) and (2, 6).
3 6
4 2
9
2
9
2
m− −
=−
−=
= −
Slope is 9
2−
6. Write the following forms of linear equations and describe important
variables.
a) slope-intercept
y mx b= + ; m is the slope of the line, b is the y-intercept of the
line.
b) standard
Ax By C+ = ; In this equation, there are no particularly important
variables. However, slope = A
B− and the y-intercept =
C
B
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 5 of 8
c) point-slope
( )1 1y y m x x− = − ; m is the slope of the line, and the point ( )1 1,x y
is a point on the line.
7. Write the equation of the line described in slope-intercept form.
a) passes through the points (4, 4) and (-2, -1)
( )( )
4 1
4 2
5
6
m− −
=− −
=
( )
( )
1 1
54 4
6
5 104
6 3
5 2
6 3
y y m x x
y x
y x
y x
− = −
− = −
− = −
= +
b) passes through the point (-3, 5) and is perpendicular to the line 3 4 7x y− =
3 4 7
4 3 7
3 7
4 4
x y
y x
y x
− =
− = − +
= −
The slope of the given line is 3
4, so the slope of the line
perpendicular to the given line is 4
3− .
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 6 of 8
( )45 3
3
5 4
1
y mx b
b
b
b
= +
= − − +
= +
=
4
3m = − , and 1b = The equation of the line is:
41
3
y mx b
y x
= +
= − +
8. Write the equation of the line described in standard form.
a) 2
3m = and passes through the point (6, 7)
( )27 6
3
7 4
3
y mx b
b
b
b
= +
= +
= +
=
( )
23
3
23 3 3
3
3 2 9
2 3 9
y mx b
y x
y x
y x
x y
= +
= +
⋅ = ⋅ +
= +
− + =
The equation in standard form is 2 3 9x y− + = or 2 3 9x y− = −
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 7 of 8
b) x-intercept = -5 and y-intercept = 3
The two points are ( )5,0− and ( )0,3 .
( )
3 0
0 5
3
5
m−
=− −
=
( )
33
5
35 5 3
5
5 3 15
3 5 15
y mx b
y x
y x
y x
x y
= +
= +
⋅ = ⋅ +
= +
− + =
The equation in standard form is 3 5 15x y− + = or 3 5 15x y− = −
9. Sketch the graphs of the following inequalities on the coordinate plane. Use separate graphs for each inequality.
a) 3x ≤
0
3 6 9
Lesson 4: Graphing Linear Equations and Inequalities Quiz solutions
Algebra 1
© 2009 Duke University Talent Identification Program
Page 8 of 8
b) 2 3 12x y− >
2 3 12
3 2 12
24
3
x y
y x
y x
− >
− > − +
< −
Slope is 2
3; y-intercept is ( )0, 4−
