MATH 60
UNIT 4.3
1
SOLVING EQUATIONS FOR SPECIFIED VARIABLES
Solve each equation for the specified variable.
3
4
Example 1:
x9
Multiply both sides by the reciprocal of whatever is multiplying x. In this case the
reciprocal of 34 is 43 .
43
3 4
x  9  34
x  91  43
 12
The same procedures apply if we are working with letters (variables).
Solve a 
Example 2:
F
m
for F
F is being divided by m, so multiply (to undo the division) both sides by
m a
1

ma 
m
1
.
F m
m 1
F m
m 1
ma  F
Solve A  h2 ( B  b) for B
Example 3:
When we are solving for a variable, we want to isolate it (get it on one side all by itself).
In this case h2 is multiplying both B and b, so let’s get rid of it first, by multiplying by the
reciprocal
2
h
.
h
2
2
h  A  2 ( B  b) h
2A  B  b
h
Next, subtract b from both sides.
2A
h
2A
h
b  B bb
b  B
©Cerritos College MLC
No part of this work may be reproduced without the prior written consent of the Cerritos College Math Learning Center.
MATH 60
UNIT 4.3
2
Since we should simplify our answers, we combine the terms over a common
denominator.
2 A  b  2 A  bh
h
h
h
 2 Ahbh
This is accomplished by multiplying b by
and then combining it with the
Our final answer is:
2Abh
h
2A
h
.
B
Problems:
Solve each of the following problems for the indicated variables.
1. x  y  7 for y
3. p  a  b  c for b
5. I  prt for r
7. E  MC 2 for M
9. p  2L  2W for W
11. V  a  gt for t
13.
2
3
2. 2 y  x  4 for y
4. x  by  c  1 for x
6. V  LWH for W
8. V   r 2 h for h
10. x  by  c  1 for y
12. t  a  bp for b
x  a for x
15. y 
k
x
14. 3x  c  b for x
16. p 
for k
k
v
for k
Answers:
1. y  7  x
2. y 
x4
2
3. b  p  a  c
4. x  1  by  c
5. r 
I
pt
6. W 
V
LH
7. M 
E
C2
8. h 
V
9. W 
P 2 L
2
10. y 
1 x  c
b
11. t 
V a
g
12. b 
t a
a t
or b 
p
p
13. x 
3
a
2
14. x 
bc
3
15. k  xy
 r2
16. k  pv
©Cerritos College MLC
No part of this work may be reproduced without the prior written consent of the Cerritos College Math Learning Center.
h
h