Download Literal Equations

Foundation Lesson III
Literal Equations
Manipulating Variables and Constants
A literal equation is one which is expressed in terms of variable symbols (such as d, v, and a) and
constants (such as R, g, and π). Often in science and mathematics you are given an equation and asked
to solve it for a particular variable symbol or letter called the unknown.
The symbols which are not the particular variable we are interested in solving for are called literals, and
may represent variables or constants. Literal equations are solved by isolating the unknown variable on
one side of the equation, and all of the remaining literal variables on the other side of the equation.
Sometimes the unknown variable is part of another term. A term is a combination of symbols such as
the products ma or πr2. In this case the unknown (such as r in πr2) must factored out of the term before
we can isolate it.
The following rules, examples, and exercises will help you review and practice solving literal equations
from physics and chemistry.
PROCEDURE
In general, we solve a literal equation for a particular variable by following the basic procedure below.
1. Recall the conventional order of operations, that is, the order in which we perform the operations of
multiplication, division, addition, subtraction, etc.:
a. Parenthesis
b. Exponents
c. Multiplication and Division
d. Addition and Subtraction
This means that you should do what is possible within parentheses first, then exponents, then
multiplication and division from left to right, then addition and subtraction from left to right. If
some parentheses are enclosed within other parentheses, work from the inside out.
2. If the unknown is a part of a grouped expression (such as a sum inside parentheses), use the
distributive property to expand the expression.
3. By adding, subtracting, multiplying, or dividing appropriately,
a. move all terms containing the unknown variable to one side of the equation, and
b. move all other variables and constants to the other side of the equation. Combine like terms
when possible.
4. Factor the unknown variable out of its term by appropriately multiplying or dividing both sides of
the equation by the other literals in the term.
5. If the unknown variable is raised to an exponent (such as 2, 3, or ½), perform the appropriate
operation to raise the unknown variable to the first power, that is, so that it has an exponent of one.
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Laying the Foundation in Physics
Foundation Lesson III
EXAMPLES
1. F = ma. Solve for a.
F = ma
Divide both sides by m:
F
= a
m
Since the unknown variable (in this case a) is usually placed on the left side of the equation, we can
switch the two sides:
a =
F
m
2. PV
1 1 = P2 V2 . Solve for V2.
PV
1 1 = P2 V2
Divide both sides by P2:
PV
1 1
= V2
P2
V2 =
PV
1 1
P2
d
. Solve for t.
t
Multiply each side by t:
3. v =
tv = d
Divide both sides by v:
t =
d
v
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Foundation Lesson III
4. PV = nRT . Solve for R.
PV = nRT
Divide both sides by n:
PV
= RT
n
Divide both sides by T:
PV
= R
nT
PV
R =
nT
5. R =
ρL
. Solve for L.
A
ρL
R =
A
Multiply both sides by A:
RA = ρ L
Divide both sides by ρ:
RA
ρ
L =
68
= L
RA
ρ
Laying the Foundation in Physics
Foundation Lesson III
6. A = h ( a + b ) . Solve for b.
Distribute the h:
A = ha + hb
Subtract ha from both sides:
A − ha = hb
Divide both sides by h:
A − ha
= b
h
A − ha
b =
h
7. P = P0 + ρ gh. Solve for g.
Subtract P0 from both sides:
P − P0 = ρ gh
Divide both sides by ρh:
P − P0
= g
ρh
P − P0
g =
ρh
1
QV . Solve for Q.
2
Multiply both sides by 2:
8. U =
2U = QV
Divide both sides by V:
2U
= Q
V
2U
Q =
V
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Foundation Lesson III
1 2
kx . Solve for x.
2
Multiply both sides by 2:
9. U =
2U = kx 2
Divide both sides by k:
2U
= x2
k
Take the square root of both sides:
2U
= x
k
x =
2U
k
L
. Solve for L.
g
Divide both sides by 2π:
10. T = 2π
T
=
2π
L
g
Square both sides:
L
T2
=
2
4π
g
Multiply both sides by g:
gT 2
= L
4π 2
gT 2
L =
4π 2
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Laying the Foundation in Physics
Foundation Lesson III
Gm1m2
. Solve for r.
r2
Multiply both sides by r2:
11. F =
Fr 2 = Gm1m2
Divide both sides by F:
r2 =
Gm1m2
F
Take the square root of both sides:
r =
12.
Gm1m2
F
hi
s
= − i . Solve for so.
ho
so
Cross-multiply:
hi s o = − ho si
Divide both sides by hi:
s0 = −
ho si
hi
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71
Foundation Lesson III
13.
1
1
1
1
+
+
. Solve for R3.
=
REQ
R1 R2 R 3
Subtract
1
1
from both sides:
+
R1 R2
1
1
1
1
−
−
=
REQ R1 R2
R3
Take the reciprocal of both sides:
1
= R3
1
1
1
−
−
REQ R1 R2
R3 =
1
1
1
1
−
−
REQ R1 R2
This equation could be solved further with several more algebraic steps.
14. F = qvB sin θ . Solve for θ.
Divide both sides by qvB:
F
= sin θ
qvB
Take the inverse sine of both sides:
⎡ F ⎤
θ = sin −1 ⎢
⎥
⎣ qvB ⎦
15. µ mg cos θ = mg sin θ . Solve for µ. .
Divide both sides by mgcos θ:
µ =
72
mg sin θ
sin θ
= tan θ
=
mg cos θ
cos θ
Laying the Foundation in Physics
Foundation Lesson III
Name _____________________________________
Period ____________________________________
Literal Equations
Manipulating Variables and Constants
EXERCISES
Directions: For each of the following equations, solve for the variable in bold print. Be sure to show
each step you take to solve the equation for the bold variable.
1. v = at
2. P =
F
A
3. λ =
h
p
4. F ( ∆t ) = m∆v
5. U =
Gm1m2
r
6. C =
5
( F − 32 )
9
7. v 2 = v0 2 + 2a∆x
8. K avg =
9. K =
3
kBT
2
1 2
mv
2
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Foundation Lesson III
3RT
M
10. vrms =
3k BT
11. vrms =
12. F =
µ
Kq1q2
4πε 0 r 2
1
13.
1 1
1
=
+
si so
f
14.
1
1
1
+
=
CEQ
C1 C2
15. V =
4 3
πr
3
16. P + Dgy +
1 2
Dv = C
2
17. P + Dgy +
1
Dv 2 = C
2
18. x = x0 + v0t +
1 2
at
2
19. n1 sin θ1 = n2 sin θ2
⎛ M + m⎞
20. mg sin θ = µ mg cos θ ⎜
⎟
⎝ m ⎠
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Laying the Foundation in Physics
Foundation Lesson III
Laying the Foundation in Physics
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