Algebra 1: Unit 5 Problem Set A
Name
Mr. Chamberlain
Date
Period
_
Equations come in many forms. Using algebraic properties, we should now feel
comfortable converting any equation into "y=" form. Notice the first re-write step in the
example problems below, where both terms on the right side are split apart in the
division step. This creates a specific coefficient for the x term. This coefficient can be
negative or fractional. Use this same technique whenever you convert to ''y='', which
we will call slope-intercept form, which is written as: Y
= mx + b
where m is
the slope and b is the y-intercept.
Example 1: Solve for slope-intercept form:
«this
4x+2y=-10
-4x
-4x
2y
-4x-10
=---
2
2
-4x
10
2
2
y=---
is standard form»
«
subtract 4x from both sides »
«
divide both sides by 2»
«re-write
Y = -2x-5
«final
("pull apart") the fraction»·
re-write to get "mx+b">
l.
Graph 4x + 2y = -10 using intercepts
form (aka the "thumb cover-up" method)
x y
Graph y = -2x-5 using your knowledge
of slope-intercept form (y = mx + b ) to
start at the y-intercept and move "up-andover" in either direction using the slope.
o
~."l,t; 0
\
4.0
6.0
3,0
~6.0
8.0
4.0
\
-4.0
2.0
'It:2.O
4.0
·2.0
5JJ
Hmmm ... notice any similarities between these graphs, Einsteinr?
Example 2: Solve for slope-intercept form
= 12
2x-3y
-2x
--
-2x
--
«subtract
-3y
-2x+12
-3
-3
-3y
-2x
«
12
divide both sides by -3 »
«re-write
--=--+-3
-3 -3
2
y=-x-4
3
2x from both sides»
("pull apart") the fraction).»>
final re-write (to get the "mx + hI!) »
«
2.
Graph 2x - 3y = 12 using intercepts form
(aka the "thumb cover-up" method)
Graph y = ~ x - 4 using your knowledge
3
of slope- intercept form (y = mx + b ) to
start at the y-intercept and move "up-andover" in either direction using the slope.
8.1)
8.0
6:0
6.0
4.0
4.0
2.0
-8.0
-5.0
-4.0
-2.0
:2.0
4))
-:3.0
-5.0
-4.0
,2..0
2JJ
-2.0
/,)-#"
/'
4.0
-6.0
-6.0
·8.0
-8.0
Based on the graphs above, what observation can you make about the equations
2x-3y
= 12
and
2
y = -x-4
3
??
IJ.....~rf)
V\.f' \.,...
3.
Convert both of the standard form equations below into slope-intercept form.
Graph both equations on the coordinate plane provided .
~
.
a)
~
I
~~
1[0
t5\
40\
20
~!3~O ~5.0
b)
-4,0
\
2.0
I
I
.1
I
6.0
8.0
13.0
8.0
4x-3y=12
t'
/
I
,/
t?
4.
-5,1)
-8,0
Convert each ofthe standard form equations below into slope-intercept form.
Graph both equations on the coordinate plane provided.
a)
3x+2y=8
_
2.v\ ':~ "'"
8.0
.....
6.0
~
ti."o?;'firt:''1';!f,1;i~'
"'"'))
""","','.,
.0
~.
·3.0
b)
·5,0
-4.0
-2.0
/'
I
u
U
~
2,1)
5x-3y = 15
V
"
·15.0
5.
Find the slope of the lines through the following pairs of points using the slope formula.
Graph and label the lines below.
- -y-~
-----:::$""- t3)
c)
(6,0)
V'n -
-~
-\2
(0,-4)
d)
- C'f- 0
0-- b
(4,5)
(4, -7)
Vep/'Jr., c,.t:. \
\ t t"tL
y
-'
----to
~
rec "ct
/J
u . t,
Ot
iJ
t i
\
),
6.
Find the slope of the lines through the following pairs of points using the slope formula.
Graph and label the lines below.
slope
= m = Y2 -
Yl
X2 -Xl
(4,-3)
e)
(-5, -3)
[+7)V-\ w'n+-~{
g)
f)
(lY\L
(-8, 8) (8, -8)
h)
,.<c.,.-~:<.--.~-...,...n-"-.~ -:"
_""",_.'-':i;"" ••
(i .
~
·6.0
·3.0
(-8, -8) (8, 8)
7.
Convert both of the standard form equations below into slope-intercept form.
Graph both equations on the coordinate plane provided.
a)
7x+2y=1O
..,.-.
~- IX+(O
b)
8.
4x-3y=12
Convert both of the standard form equations below into slope-intercept form.
Graph both equations on the coordinate plane provided.
a)
3x+2y=18
bO
-'2y•
ik:
Jr- i~ \ u
ud
b)
5x-3y
-2.-)(
2-.
+-q .
= 15
-
/'
-S)l~ \S
~
~