Formulae

2.4. FORMULAE
81
31. At a minimum, we need to move each decimal point one place to the right
in order to clear the decimals from the equation. Consequently, we multiply
both sides of the equation by 10.
2.5x + 1.9 = 0.9x
10(2.5x + 1.9) = 10(0.9x)
25x + 19 = 9x
Original Equation.
Multiply both sides by 10.
Distribute the 10.
Note that the decimals are now cleared from the equation. Next, isolate all
terms containing the variable x on one side of the equation. Remove the term
9x from the right-hand side by subtracting 9x from both sides of the equation.
25x + 19 − 9x = 9x − 9x
16x + 19 = 0
Subtract 9x from both sides.
Simplify both sides.
Subtract 19 from both sides to remove the term 19 from the left-hand side of
the equation.
16x + 19 − 19 = 0 − 19
16x = −19
Subtract 19 from both sides.
Simplify both sides.
Finally, to “undo” multiplying by 16, divide both sides of the equation by 16.
−19
16x
=
16
16
19
x=−
16
2.4
Divide both sides by 16.
Simplify.
Formulae
1. We are instructed to solve the equation F = kx for x. Thus, we must isolate
all terms containing the variable x on one side of the equation. We begin by
dividing both sides of the equation by k.
F = kx
kx
F
=
k
k
F
=x
k
Original equation.
Divide both sides by k.
Simplify.
Note that this last equation is identical to the following equation
x=
F
k
Note that we have x =“Stuff”, where “Stuff” contains no occurrences of x, the
variable we are solving for.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
82
3. We are instructed to solve the equation E = mc2 for m. Thus, we must
isolate all terms containing the variable m on one side of the equation. We
begin by dividing both sides of the equation by c2 .
E = mc2
E
mc
= 2
2
c
c
E
=m
c2
2
Original equation.
Divide both sides by c2 .
Simplify.
Note that this last equation is identical to the following equation
m=
E
c2
We now have m =“Stuff”, where “Stuff” contains no occurrences of m, the
variable we are solving for.
5. We are instructed to solve the equation A = πr1 r2 for r2 . Thus, we must
isolate all terms containing the variable r2 on one side of the equation. We
begin by dividing both sides of the equation by πr1 .
A = πr1 r2
A
πr1 r2
=
πr1
πr1
A
= r2
πr1
Original equation.
Divide both sides by πr1 .
Simplify.
Note that this last equation is identical to the following equation
r2 =
A
πr1
Note that we have r2 =“Stuff”, where “Stuff” contains no occurrences of r2 ,
the variable we are solving for.
7. We are instructed to solve the equation F = ma for a. Thus, we must
isolate all terms containing the variable a on one side of the equation. We
begin by dividing both sides of the equation by m.
F = ma
ma
F
=
m
m
F
=a
m
Second Edition: 2012-2013
Original equation.
Divide both sides by m.
Simplify.
2.4. FORMULAE
83
Note that this last equation is identical to the following equation
F
m
Note that we have a =“Stuff”, where “Stuff” contains no occurrences of a, the
variable we are solving for.
a=
9. We are instructed to solve the equation C = 2πr for r. Thus, we must
isolate all terms containing the variable r on one side of the equation. We
begin by dividing both sides of the equation by 2π.
C = 2πr
C
2πr
=
2π
2π
C
=r
2π
Original equation.
Divide both sides by 2π.
Simplify.
Note that this last equation is identical to the following equation
C
2π
Note that we have r =“Stuff”, where “Stuff” contains no occurrences of r, the
variable we are solving for.
r=
11. We are instructed to solve the equation y = mx + b for x. Thus, we must
isolate all terms containing the variable x on one side of the equation. First,
subtract b from both sides of the equation.
y = mx + b
y − b = mx + b − b
Original equation.
Subtract b from both sides.
y − b = mx
Combine like terms.
Note that all the terms comtaining x, the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by m to complete the solution.
y−b
mx
=
m
m
y−b
=x
m
Divide both sides by m.
Simplify.
Note that this last equation is identical to the following equation
y−b
Simplify.
m
Note that we have x =“Stuff”, where “Stuff” contains no occurrences of x, the
variable we are solving for.
x=
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
84
13. We are instructed to solve the equation F = qvB for v. Thus, we must
isolate all terms containing the variable v on one side of the equation. We
begin by dividing both sides of the equation by qB.
F = qvB
F
qvB
=
qB
qB
F
=v
qB
Original equation.
Divide both sides by qB.
Simplify.
Note that this last equation is identical to the following equation
v=
F
qB
Note that we have v =“Stuff”, where “Stuff” contains no occurrences of v, the
variable we are solving for.
1
15. We are instructed to solve the equation V = πr2 h for h. Thus, we must
3
isolate all terms containing the variable h on one side of the equation. First,
clear the fractions by multiplying both sides of the equation by the common
denominator, 3.
1 2
πr h
3
1 2
πr h
3(V ) = 3
3
V =
3V = πr2 h
Original equation.
Multiply both sides by 3.
Simplify. Cancel 3’s.
Note that all the terms comtaining h, the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by πr2 to complete the solution.
πr2 h
3V
=
πr2
πr2
3V
=h
πr2
Divide both sides by πr2 .
Simplify.
Note that this last equation is identical to the following equation
h=
3V
πr2
Note that we have h =“Stuff”, where “Stuff” contains no occurrences of h, the
variable we are solving for.
Second Edition: 2012-2013
2.4. FORMULAE
85
V
for R. Thus, we must
17. We are instructed to solve the equation I =
R
isolate all terms containing the variable R on one side of the equation. First,
clear the fractions by multiplying both sides of the equation by the common
denominator, R.
I=
V
R
R(I) = R
V
R
Original equation.
Multiply both sides by R.
RI = V
Simplify. Cancel R’s.
Note that all the terms comtaining R, the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by I to complete the solution.
V
RI
=
I
I
V
R=
I
Divide both sides by I.
Note that we have R =“Stuff”, where “Stuff” contains no occurrences of R,
the variable we are solving for.
kqQ
19. We are instructed to solve the equation F = 2 for q. Thus, we must
r
isolate all terms containing the variable q on one side of the equation. First,
clear the fractions by multiplying both sides of the equation by the common
denominator, r2 .
F =
kqQ
r2
r2 (F ) = r2
r2 F = kqQ
kqQ
r2
Original equation.
Multiply both sides by r2 .
Simplify. Cancel r2 ’s.
Note that all the terms comtaining q, the variable we are solving for, are already
isolated on one side of the equation. We need only divide both sides of the
equation by kQ to complete the solution.
r2 F
kqQ
=
kQ
kQ
2
r F
=q
kQ
Divide both sides by kQ.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
86
Note that this last equation is identical to the following equation
q=
r2 F
kQ
Note that we have q =“Stuff”, where “Stuff” contains no occurrences of q, the
variable we are solving for.
21. We are instructed to solve the equation P = 2W + 2L for W . Thus, we
must isolate all terms containing the variable W on one side of the equation.
First, subtract 2L from both sides of the equation.
P = 2W + 2L
P − 2L = 2W + 2L − 2L
Original equation.
Subtract 2L from both sides.
P − 2L = 2W
Combine like terms.
Note that all the terms comtaining W , the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by 2 to complete the solution.
2W
P − 2L
=
2
2
P − 2L
=W
2
Divide both sides by 2.
Simplify.
Note that this last equation is identical to the following equation
W =
P − 2L
2
Note that we have W =“Stuff”, where “Stuff” contains no occurrences of W ,
the variable we are solving for.
1
23. We are instructed to solve the equation A = h(b1 + b2 ) for h. Thus, we
2
must isolate all terms containing the variable h on one side of the equation.
First, clear the fractions by multiplying both sides of the equation by the
common denominator, 2.
1
h(b1 + b2 )
2
1
h(b1 + b2 )
2(A) = 2
2
A=
2A = h(b1 + b2 )
Second Edition: 2012-2013
Original equation.
Multiply both sides by 2.
Simplify. Cancel 2’s.
2.4. FORMULAE
87
Note that all the terms comtaining h, the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by b1 + b2 to complete the solution.
2A
h(b1 + b2 )
=
b1 + b2
b1 + b2
2A
=h
b1 + b2
Divide both sides by b1 + b2 .
Simplify.
Note that this last equation is identical to the following equation
h=
2A
b1 + b2
Note that we have h =“Stuff”, where “Stuff” contains no occurrences of h, the
variable we are solving for.
25. We are instructed to solve the equation y − y0 = m(x − x0 ) for m. Thus,
we must isolate all terms containing the variable m on one side of the equation.
We begin by dividing both sides of the equation by x − x0 .
y − y0 = m(x − x0 )
y − y0
m(x − x0 )
=
x − x0
x − x0
y − y0
=m
x − x0
Original equation.
Divide both sides by x − x0 .
Simplify.
Note that this last equation is identical to the following equation
m=
y − y0
x − x0
Note that we have m =“Stuff”, where “Stuff” contains no occurrences of m,
the variable we are solving for.
GM m
for M . Thus, we must
27. We are instructed to solve the equation F =
r2
isolate all terms containing the variable M on one side of the equation. First,
clear the fractions by multiplying both sides of the equation by the common
denominator, r2 .
GM m
r2
GM m
r2 (F ) = r2
r2
F =
r2 F = GM m
Original equation.
Multiply both sides by r2 .
Simplify. Cancel r2 ’s.
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
88
Note that all the terms comtaining M , the variable we are solving for, are
already isolated on one side of the equation. We need only divide both sides
of the equation by Gm to complete the solution.
GM m
r2 F
=
Gm
Gm
r2 F
=M
Gm
Divide both sides by Gm.
Simplify.
Note that this last equation is identical to the following equation:
M=
r2 F
Gm
Note that we have M =“Stuff”, where “Stuff” contains no occurrences of M ,
the variable we are solving for.
29. We are instructed to solve the equation d = vt for v. Thus, we must isolate
all terms containing the variable v on one side of the equation. We begin by
dividing both sides of the equation by t.
d = vt
vt
d
=
t
t
d
=v
t
Original equation.
Divide both sides by t.
Simplify.
Note that this last equation is identical to the following equation
v=
d
t
Note that we have v =“Stuff”, where “Stuff” contains no occurrences of v, the
variable we are solving for.
31. Start with the area formula and divide both sides by W .
A = LW
LW
A
=
W
W
A
=L
W
Hence:
L=
Second Edition: 2012-2013
A
W
2.4. FORMULAE
89
To determine the length, substitute 1073 for A and 29 for W and simplify.
1073
29
L = 37
L=
Hence, the length is L = 37 meters.
33. Start with the area formula, then divide both sides by h.
A = bh
A
bh
=
h
h
A
=b
h
Hence:
b=
A
h
Next, substitute 2418 for A, 31 for h, and simplify.
2418
31
b = 78
b=
Hence, b = 78 feet.
35. Start with the area formula, then clear the equation of fractions by multiplying both sides of the equation by 2.
1
bh
2
1
2[A] =
bh 2
2
A=
2A = bh
Divide both sides by h.
2A
bh
=
h
h
2A
=b
h
Second Edition: 2012-2013
CHAPTER 2. SOLVING LINEAR EQUATIONS
90
Hence:
b=
2A
h
Next, substitute 1332 for A, 36 for h, and simplify.
2(1332)
36
2664
b=
36
b = 74
b=
Hence, b = 74 inches.
37. We need to isolate terms containing W on one side of the equation. Start
with the perimeter equation and subtract 2L from both sides.
P = 2W + 2L
P − 2L = 2W + 2L − 2L
P − 2L = 2W
Divide both sides by 2 and simplify.
P − 2L
2W
=
2
2
P − 2L
=W
2
Hence:
W =
P − 2L
2
Substitute 256 for P , 73 for L, then simplify.
256 − 2(73)
2
110
W =
2
W = 55
W =
Hence, the width is W = 55 meters.
Second Edition: 2012-2013