Formulas
Professor Tim Busken
Department of Mathematics
Grossmont College
January 29, 2013
Professor Tim Busken
Formulas
Formulas
Learning Objectives:
Solve a Formula with numerical replacements for all but one
of its variables.
Solve formulas for the indicated variable.
Solve basic percent problems by translating them into an
equation.
Professor Tim Busken
Formulas
Formulas
Definition
A formula in mathematics is an equation that contains more than
one variable.
Professor Tim Busken
Formulas
Formulas
Classroom Example A store selling art supplies finds that they
can sell x sketch pads each week at a price of p dollars each,
according to the formula x = 900 − 300p . What price should they
charge for each sketch pad if they want to sell 375 pads each
week?
Professor Tim Busken
Formulas
Formulas
Classroom Example A boat is traveling upstream against a
current. If the speed of the boat in still water is r and the speed of
the current is c , then the formula for the distance traveled by the
boat is d = (r − c ) · t , where t is the length of time. Find c if d = 60
miles, r = 18 miles per hour, and t = 5 hours.
Professor Tim Busken
Formulas
Square
Rectangle
Triangle
s
l
s
s
w
w
l
a
c
h
b
s
Perimeter = 4s
Area = s 2
Perimeter = 2l + 2w
Area =
Area = lw
Professor Tim Busken
Perimeter = a + b + c
Formulas
1
bh
2
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Given the formula p = 2l + 2w , solve for the
Professor Tim Busken
Formulas
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Solution:
Given the formula p = 2l + 2w , solve for the
We begin by adding −2w to both sides of the formula.
p + (−2w ) = 2l + 2w + (−2w )
Professor Tim Busken
Addn. Prop. of Equality
Formulas
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Solution:
Given the formula p = 2l + 2w , solve for the
We begin by adding −2w to both sides of the formula.
p + (−2w ) = 2l + 2w + (−2w )
Professor Tim Busken
Addn. Prop. of Equality
Formulas
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Solution:
Given the formula p = 2l + 2w , solve for the
We begin by adding −2w to both sides of the formula.
p + (−2w ) = 2l + 2w + (−2w )
Addn. Prop. of Equality
p − 2w = 2l + 0
Additive Inverse
2l = p − 2w
Professor Tim Busken
Additive Identity
Formulas
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Solution:
Given the formula p = 2l + 2w , solve for the
We begin by adding −2w to both sides of the formula.
p + (−2w ) = 2l + 2w + (−2w )
Addn. Prop. of Equality
p − 2w = 2l + 0
Additive Inverse
2l = p − 2w
We now multiply both sides by
Additive Identity
1
2
1
1
· 2 · l = · (p − 2w )
2
2
Professor Tim Busken
Multiplication Prop. of Equality
Formulas
Solving Formulas for an indicated Variable
Classroom Example
length, l .
Solution:
Given the formula p = 2l + 2w , solve for the
We begin by adding −2w to both sides of the formula.
p + (−2w ) = 2l + 2w + (−2w )
Addn. Prop. of Equality
p − 2w = 2l + 0
Additive Inverse
2l = p − 2w
We now multiply both sides by
Additive Identity
1
2
1
1
· 2 · l = · (p − 2w )
2
2
l=
Multiplication Prop. of Equality
(p − 2w )
Multiplicative Inverse
2
Professor Tim Busken
Formulas
Solving Formulas for an indicated Variable
Classroom Examples
Solve bx + 6 = cx + 2 for x .
Solve
y +9
= 5 for y
x −6
Professor Tim Busken
Formulas
Basic Percent Problems
Definition
Recall that percent means per one hundred. For example, 25%
25
, 0.25 and 14 .
is the same as 100
Professor Tim Busken
Formulas
Basic Percent Problems
Definition
Recall that percent means per one hundred. For example, 25%
25
, 0.25 and 14 . Similarly, 75% is equivalent to
is the same as 100
0.75.
Professor Tim Busken
Formulas
Basic Percent Problems
Definition
Recall that percent means per one hundred. For example, 25%
25
, 0.25 and 14 . Similarly, 75% is equivalent to
is the same as 100
0.75.
To change a decimal to a percent, we move the decimal point
two places to the right and write a % symbol.
Professor Tim Busken
Formulas
Basic Percent Problems
Definition
Recall that percent means per one hundred. For example, 25%
25
, 0.25 and 14 . Similarly, 75% is equivalent to
is the same as 100
0.75.
To change a decimal to a percent, we move the decimal point
two places to the right and write a % symbol.
To change from a percent to a decimal, we drop the % symbol
and move the decimal point two places to the left.
Professor Tim Busken
Formulas
Basic Percent Problems
Recognizing key words in a percent problem is helpful in writing
the problem as an equation. Three key words in the statement of a
percent problem and their meanings are:
Professor Tim Busken
Formulas
Basic Percent Problems
Recognizing key words in a percent problem is helpful in writing
the problem as an equation. Three key words in the statement of a
percent problem and their meanings are:
of means multiplication (·)
is means equal (=)
what (or some equivalent) means the unknown, x
Professor Tim Busken
Formulas
Basic Percent Problems
Example :
Solution:
3 is what percent of 12?
3 is what percent of 12?
|
↓↓
3=
{z
↓
x
}
Professor Tim Busken
↓ ↓
· 12
Formulas
Basic Percent Problems
Example :
Solution:
3 is what percent of 12?
3 is what percent of 12?
|
↓↓
3=
3·
1
1
= x · 12 ·
12
12
{z
↓
x
}
↓ ↓
· 12
Multiplication Prop. of Equality
Basic Percent Problems
Example :
Solution:
3 is what percent of 12?
3 is what percent of 12?
|
↓↓
3=
3·
{z
↓
x
}
1
1
= x · 12 ·
12
12
↓ ↓
· 12
Multiplication Prop. of Equality
3
=x
12
1
x = = 0.25 = 25%
4
Professor Tim Busken
Formulas
Basic Percent Problems
Example :
Solution:
3.2 is 60% of what number?
3.2 is 60% of what
number
| {z
}?
↓ ↓ ↓
↓
3.2 = 60% ·
3.2 = 0.60 · x
Professor Tim Busken
↓
x
Convert % to a Decimal
Formulas
Basic Percent Problems
Example :
Solution:
3.2 is 60% of what number?
3.2 is 60% of what
number
| {z
}?
↓ ↓ ↓
↓
3.2 = 60% ·
3.2 = 0.60 · x
Professor Tim Busken
↓
x
Convert % to a Decimal
Formulas
Basic Percent Problems
Example :
Solution:
3.2 is 60% of what number?
3.2 is 60% of what
number
| {z
}?
↓ ↓ ↓
↓
3.2 = 60% ·
3.2 = 0.60 · x
Professor Tim Busken
↓
x
Convert % to a Decimal
Formulas
Basic Percent Problems
Example :
Solution:
3.2 is 60% of what number?
3.2 is 60% of what
number
| {z
}?
↓ ↓ ↓
↓
3.2 = 60% ·
3.2 = 0.60 · x
1
1
· 3.2 =
· 0.60x
0.60
0.60
↓
x
Convert % to a Decimal
Multiplication Prop. of Equality
3.2
= 1x
0.60
x=
Multiplicative Inverse
16
= 5.3
3
Professor Tim Busken
Formulas
Basic Percent Problems
Example :
Solution:
What number is 20% of 0.07
What
number
|
{z
} is 20% of 0.07?
↓
x
↓ ↓ ↓ ↓
= 20% · 0.07
x = (0.20) · (0.07)
Professor Tim Busken
Convert % to a Decimal
Formulas
Basic Percent Problems
Example :
Solution:
What number is 20% of 0.07
What
number
|
{z
} is 20% of 0.07?
↓
x
↓ ↓ ↓ ↓
= 20% · 0.07
x = (0.20) · (0.07)
Professor Tim Busken
Convert % to a Decimal
Formulas
Basic Percent Problems
Example :
Solution:
What number is 20% of 0.07
What
number
|
{z
} is 20% of 0.07?
↓
x
↓ ↓ ↓ ↓
= 20% · 0.07
x = (0.20) · (0.07)
Convert % to a Decimal
x = 0.014
Multiply
Don’t forget how to multiply these decimals correctly by hand!
Professor Tim Busken
Formulas