Transcript Solving Equations for a Specific Variable

Solving Equations for
a Specific Variable

When solving for a specific variable, follow
the same rules as when you are solving
“regular” equations.
 Begin
by identifying the variable you are
solving for.
 Then, SIMPLIFY the equation.

Remember – get rid of fractions, parentheses or
like terms.
 Next,
ISOLATE that variable by “undoing” the
operations around that variable.

Get all of the same variables on the same side of
the equal sign.
 Then,
SOLVE for the variable.
Like this……
1. Solve for d:
•
T  5c 2 d
2
2
5c 5c
•
T
d
2
5c
2. Solve for n:
R  3m n
3m2 3m2
2
R

n
2
3m
Mark the variable you
are solving for.
“Undo” the operation
that “connects” the ‘d’
to the other numbers or
variables.
•
•
What operation connects
the 5 to the c2 and to the
d?
Multiplication – so what
is it’s opposite?
•
•
Division
So divide both sides by
everything in front of the
d.


If I’m solving for “w” I want all
of my “w’s” on the LEFT side
of the equal sign and all the
“u’s” on the RIGHT…..so
ISOLATE

Now SOLVE for “w”.
3. Solve for w:
5u  w  u  5w
5w
5w
5u 4w  u
5u
5u
4w  4u
4
4
w  u



If I’m solving for “u” I want all
of my “u’s” on the LEFT side
of the equal sign and all the
“w’s” on the RIGHT…..so
ISOLATE
Now SOLVE for “u”.
4. Solve for u:
5u  w  u  5w
u
u
4u w  5w
w w
4u  4w
4
4
u  w
5. Solve for A:
5
B  ( A  8)
7
5 (7) 40
(7)B (7)
A

7
7
7B  5 A 40
40
40
7 B  40  5 A
5
5
7 B  40  A
5
6. Solve for A:



Simplify –
get rid of the
parentheses
then the
fraction in
this case.
Now mark
your variable
and
ISOLATE.
And SOLVE
for “A”
7
B  ( A  3)
8
7 (8) 21
(8)B (8)
A 

8
8
8B  7 A 21
21
21
8B  21  7 A
7
7
8 B  21  A
7
7. Solve for c:
8. Solve for b:
1
Simplify –
A  h(b  c )
get rid of the
2
parentheses
(2)1 (2) 1
(2)
A  hb  hc
A  hb  hc
then the
2
2
2
2
fraction in
this case.
 Now mark
2 A  hb  hc
2 A  hb  hc
your variable hc
 hb  hb
hc
and
2 A  hb  hc
2 A  hc  hb
ISOLATE.
h
h
h
h
 And SOLVE
for “A”
2 A hb  c
2 A hc  b


2
2
h
h
h
h
c  Ab
b  Ac
h
h
1
A  h(b  c )
2
(2) (2)1 (2) 1

9. Solve for h:
(2) 1(2)
A  h(b  c )
2
2 A  h(b  c)
(b  c) (b  c)
2A
h
bc
10. Solve for L:
( L) A  W ( L)
L
A  WL
W
W
A L
W