Chapter
8
–
Solving
Linear
Equations
Section
8.1
–
Solve
ax=b,
x/a=b,
an
a/x=b
An
equation
is
a
statement
that
2
mathematical
expressions
have
the
same
value
(
are
equal
).
Eg:
3x‐2
=
4
x=variable,
3=numerical
coefficient,
‐2&4
=
constant
terms
An
opposite
operation
(or
inverse
operation)
undoes
another
operation.
Add
–
subtract
multiply
–
divide
Eg
#1:
Solve
2x
=
¾
using
a
number
line.
(ax=b)
3/4
l
l
l
l
l
0
=2x
1
The
length
of
the
curly
bracket
is
2x,
so
½
of
this
length
would
be
1x
or
x.
*change
the
number
line
to
eighths.*
¾
or
6/8
l
l
l
l
l
l
l
l
l
0
=2x
1
x
or
3/8
x
or
3/8
**the
number
line
shows
each
x=3/8
Eg
#2:
Solve
m/3
=
‐2/3
algebraically.
(x/a
=
b)
x3
(m/3)
=(
‐2/3)x3
m
=
‐1.2
or
‐6/5
or
‐1
1/5
check:
Left
side
Right
side
=m/3
=‐2/5
=‐1.2/3
=
‐2/5
=
‐.4
=
‐.4
∴
m
=
1.2
since
the
left
side
is
equal
to
the
right
side.
Eg
#3
Solve
2x
=
.12
with
a
diagram
model.
Use
cups
with
paperclips
for
the
x’s.
X
X
=
10
1
1
Replace
each
paperclip
with
the
coins
it
represents.
(value
not
weight)
=
10
1
1
5
5
1
1
∴
x
=
.06
or
6
cents
Eg
#4
Solve
125/x
=
5
with
algebra
(a/x
=
b)
**we
need
to
get
our
variable
out
the
denominator**
xX(125/x)
=
(5)xX
125
=
5x
5
5
25
=
x
Some
common
formulas:
D
=
vt
or
D
=
st
or
D
=
rt
D
=
distance
v/s/r
=
velocity/speed/rate
(all
the
same)
t
=
time
(usually
in
hours)
Manipulate
to
solve
for
velocity:
D
=
vt
t
t
D
=
v
T
The
%
formula
p/w
=
%/100
**memorize**
P
=
part
w
=
whole
%
=
the
%
used
and
not
the
decimal
equivalent
Eg
#5
Coats
are
on
sale
at
25%
off.
If
the
sale
price
is
$176.25,
what
was
the
original
price?
176.25
75
W
100
75W
=
17525
75
75
W
=
235
∴
The
original
price
was
$235.00
Section
8.2
Solve
ax+b=c
and
x/a+b=c
Eg
#1
Solve
2x+5
=
17
‐5
‐5
2x
=
12
2
2
x
=
6
or
rationalizing
the
denominator
x12(k/3)
(–
½)x12
=
(1
¾)x12
Eg
#2
Solve
k/3
–
1/2
=
1
¾
+1/2
+1/2
4k
–
6
=
‐21
(reduced
to
lowest
terms)
+6
+6
K/3
=
1
¼
x3
x3
4k
=
‐15
K
=
‐3
¾
4
4
k
=
‐3.75
6
step
methods
for
word
problems
1) write
a
“let
x”
statement
(usually
the
smallest
unknown
value
or
the
coin
of
lesser
value)
2) write
a
word
diagram
3) write
an
equation
4) solve
equation
5) check
answer
6) write
an
English
statement
Eg
#3
A
telephone
plan
costs
5
cents
per
minute
plus
a
monthly
fee
of
$4.95.
If
your
bill
was
$18.75,
how
many
minutes
did
you
use?
1)
Let
x
represent
the
number
of
minutes
2)
Minutes
+
fixed
cost
=
total
cost
3)
.05x
+
4.95
=
18.75
‐4.95
‐4.95
4)
.05x
=
18.75
.05
.05
x
=
276
5)
Check:
.05(276)
+
4.95
=
18.75
13.80
+
4.95
=
18.75
18.75
=
18.75
6)
∴
You
used
276
minutes
that
month.
Eg
#4
A
jar
of
nickels
and
dimes
has
$4.75
in
it.
There
are
3
times
as
many
nickels
as
there
are
dimes.
How
many
dimes
are
there?
1) Let
x
represent
the
number
of
dimes
2) dimes
+
nickels
=
total
cost
3) x
+
3x
=
4.75
4) 10x
+
3(5)x
=
475
turn
them
all
to
pennies
25x
=
475
25
25
x
=
19
5)
Check:
dimes
=
x
=
19(.10)
=
$1.90
Nickels
=
3x
=
3(19)
=
57(.05)
=
$2.85
$1.90
+$2.85
$4.75
5) ∴
There
are
19
dimes
in
the
jar.
Section
8.3
Solve
a(x+b)
=
c
**Note:
distributive
property
is
a(b+c)
=
ab
+
ac**
A
fraction
bar
acts
to
group
values
(usually
those
above
(or
below)
it),
it
is
also
a
division
symbol.
Eg:
z
–
1
can
be
written
as
(z
–
1)
÷
5
or
1
(z
–
1)
5
5
Solve
the
following:
Eg
#1
2(x+1.5)
=
7.6
or
x(x+1.5)
=
7.6
(mult
through
brackets)
2
2
2x
+
3
=
7.6
x
+
1.5
=
3.8
‐3
‐3
‐1.5
‐1.5
2x
=
4.6
x
=
2.3
2
2
x
=
2.3
Eg
#2
or
N
+
1
=
‐3
N
+
1
=
‐3
2
4
2
4
=2
x4
N
+
1
=
‐3
x4
x2
N
+
1
=
‐3
x2
2
=1
4
2
4
2N
+
2
=
‐3
(we
multiplied
the
line
N
+
1
=
‐6
above
2(N+1)
and‐3
was
left
on
the
right)
4
‐2
‐2
‐1
‐1
2N
=
‐5
N
=
‐10
or
‐5
or
‐2
½
or
‐2.5
2
2
4
2
N
=
‐5
or
‐2.5
2
Eg
#3
If
the
average
daily
temperature
is
‐13.2°C
and
the
low
was
‐18.1°C,
what
was
the
high?
**Note:
the
daily
average
is
the
average
of
the
high
and
low
temperature.
ie:
high
+
low
2
high
‐18.1
=
‐13.2
2
x2
high
‐18.1
=
‐13.2
x2
2
high
–
18.1
=
‐26.4
+18.1
+18.1
high
=
‐8.3°C
Section
8.4
Variables
on
both
sides
of
the
equation
Basic
Step:
1)
When
necessary
multiply
through
the
brackets
and
/or
rationalize
the
denominator.
Group
like
terms.
2)
Move
the
x’s
to
the
side
with
the
most
(using
regular
algebra
rules)
Eg
#1
Solve
with
algebra:
Eg
#2
2(x+3)
–
3
=
8
–
3x
4(m‐2)
‐1(m+3)
=
m
–
1
2x
+6
‐3
=
8
–
3x
4m
‐8
–m
‐3
=
m
‐1
2x
+
3
=
8
–
3x
3m
–
11
=
m
‐1
+3x
+3x
‐m
‐m
5x+3
=
8
2m
‐11
=
‐1
+11
+11
‐3
‐3
2m=
10
2
2
5x
=
5
5
5
x
=
1
m
=
5
Eg
#3
In
your
hand
you
have
30
less
quarters
than
dimes.
If
the
value
of
coins
in
each
hand
is
the
same,
how
many
quarters
and
dimes
do
you
have?
Let
x
represent
quarters
Quarters
=
dimes
check:
x
=
x+30
quarters:
20
x
.25
=
$5.00
25x
=
10(x+30)
dimes:
(20+30)
x
.10
=
$5.00
25x
=
10x
+
300
‐10x
‐10x
15x
=
300
∴
there
are
20
quarters
and
50
dimes.
15
15
x
=
20
Eg
#4:
Zane
has
$35.50
and
saves
$4.25/week.
Jaxon
has
$24.23
and
saves
$5.50/week.
In
how
many
weeks
will
they
have
the
same
amount
of
money?
Let
x
represent
number
of
weeks
check:
Zane
=
Jaxon
Zane:
35.50
+
4.25(9)
=
$73.75
35.50
+
4.25x
=
24.25
+
5.50x
Jaxon:
24.25
=
5.50(9)
=
$73.75
‐4.25
‐4.25
35.50
=
24.25
+
1.25x
∴In
9
weeks
they
will
have
the
same
amount
‐24.25
‐24.25
of
money.
11.25
=
1.25x
1.25
1.25
9
=
x