Algebra 1
Chapter 3.7 Formulas &
Functions
Objective
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Students will solve formula’s for one of its
variables and rewrite equations in function
form.
Before we begin…
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In this lesson we will look at equations that
have more than one variable…
More specifically, we will solve a formula for
one of it’s variables and rewrite or transform
an equation in function form…
The key to being successful here is your
ability to analyze an equation and make
decisions that will help you solve the
problem…
Solving Formulas
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A formula is an algebraic equation that relates
two or more real-life quantities.
A formula usually contains 2 or more
variables…
You should already be familiar with some
formulas like perimeter, area, volume and
surface area.
In this lesson we will be transforming
(changing) the formula to isolate the unknown
variable…
Let’s look at an example…
Example #1
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Using the formula for the area of a rectangle A = ℓw.
Find ℓ in terms of A and w
Use the new formula to find the length of a rectangle that
has an area of 35 square feet and a width of 7 feet
The first step is to understand what the formula is
saying. In this instance it says Area = length ● width
When you are asked to find ℓ in terms of A and w, you
are being asked to isolate ℓ to one side of the equation
To do that, you undo the operations in the formula using
just the variables…
Example #1 (continued)
A = ℓw
In this instance we will begin by working on the right side of the equation.
Since ℓw means to multiply, we will undo the multiplication by dividing
both sides by w. Which will look like this
The w’s on the right
side cancel out
A = ℓw
leaving
This is the
transformed
formula with ℓ
isolated on the
right side
w
w
A =ℓ
w
ℓ
Example #1 (continued)
A =ℓ
w
Now that we have the transformed formula all we have to do is substitute the
given data into the formula and simplify to determine the length of the rectangle.
Use the new formula to find the length of a rectangle that
has an area of 35 square feet and a width of 7 feet
35 = ℓ
7
5 =ℓ
Comments
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Of course, in the first example we used a
simple formula so that you understand what
this strategy is trying to accomplish.
You can use this strategy for any
formula…This strategy also has application in
your science class with science formula’s
Let’s look at a more complex formula…
Example #2
In this example we will use the formula for converting temperature
from Celsius to Fahrenheit and solve for F
C
b g
5
F  32
9
The first step is deciding which side of the equation to work on.
Since F is on the right side we will begin there…
Second, I notice that the formula illustrates the distributive
property…
I can do one of 2 things here….I can distribute the 5/9
or I can multiply by the reciprocal of 5/9 to get rid of the
fraction on the right side of the equation…(we talked
about this in a previous lesson…)
My decision is to multiply by the reciprocal to get rid of the
fraction on the right side…
Example #2 (continued)
5
Original formula
C b
F  32g
9
F
IJ b g
G
HK
9
9 5
C
F  32
5
5 9
Multiply by the reciprocal of
5/9. The fractions on the right
side cancel out.
9
C  F  32
5
9
C  F  32
5
32
+ 32
To undo the -32 on the right add
32 to both sides of the equation
9
C  32  F
5
The transformed formula with F
isolated on one side of the equation
Comments
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In the previous example you were not given
any data…all you were asked to do was
solve for F, which means to isolate F on one
side of the formula…
Let’s look at another example where you
have to transform the formula and use data to
solve the problem…
In this example we will use the formula for
simple interest I=prt, where I = interest;
p=principal; r=rate; t=time.
Example #3
I = prt
A. Solve for r
B. Find the interest rate (r) for an investment of $1500 (p)
that earned $54 (I) in 1 year (t)
A: Transform the formula so that r is isolated to one side of the equation
To do that you have to
know that prt means
multiply p times r times t.
To undo multiplication
use division like this:
I = prt
Original Formula
I = prt
The p and the t on the
right side cancel out
leaving r
pt p t
I=r
pt
The transformed formula
r is isolated on one side
of the equation.
Example #3 (continued)
B. Find the interest rate (r) for an investment of $1500 (p) that earned
$54 (I) in 1 year (t)
I=r
The transformed formula
pt
54
Substitute the data into the formula
=r
1500(1)
54
Simplify
=r
1500
0.036 or 3.6%
=r
Solution
Equations in Function Form
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Continuing with the theme of isolating
variables…
A two-variable equation is written in Function
Form if one of its variables is isolated to one
side of the equation
The isolated variable is called the output and
it is a function of the other variable called the
input.
Let’s look at an example…
Example #4
Rewrite the equation 3x + y = 4 so that y is a function of x
In this example you are being asked to isolate y to one side of the
equation. Once you do that the equation will be in function form
To transform this equation we will continue with the concepts that
we have been working with, which is inverse operations
The first step is to decide which side of the equation you will be working
on….In this case since y is on the left side I will begin there
On the left side is a +3x. To undo the +3x subtract 3x from both sides
of the equation like this:
3x + y = 4
Since -3x and 4
-3x
-3x
The 3x’s on
are not like terms
the left cancel
you cannot
y = - 3x + 4
out leaving y
combine them
Example #4 (continued)
This is the transformed equation:
y = - 3x + 4
This equation represents y as a function of x
What that means is that the value of y
(output) is determined by whatever
number is input for x
For example:
if you input 1 for x, y must then equal 1
If you input 2 for x, y must then equal -2
If you input 3 for x, y must then equal -5
Example # 5
A. Rewrite the equation 3x + y = 4 so that x is a function of y
B. Use the result to find x when y = -2, -1, and 0
First decide which side of the equation to work on. Since x is on the left
begin there
On the left you have a +y. To undo the +y subtract y from both sides of
the equation
3x + y = 4
Since 4 and –y
The y’s on the
- y -y
are not like
left cancel out
terms they
3x = 4 – y
leaving 3x
cannot be
combined
3x = 4 – y is NOT in function form.
To get the equation in function form x
has to be all by itself.
Example #5 (continued)
3x
=4–y
To get x all by itself you have to undo the multiplication of 3 times x by dividing
both sides by 3.
Be careful here!...when you divide both sides by 3 you have to
divide the whole right side by 3!
3x
The 3’s on
the left
cancel out
leaving x
3
=4–y
3
x
=4–y
3
This is the equation in
function form with x
isolated on the left side
of the equation
Example #5 (continued)
B. Use the result to find x when y = -2, -1, and 0
x
=4–y
3
Now that you have the equation in function form, simply substitute the
value of y (input) into the equation to determine the value of x (output)
Input
Substitute
y = -2
x
= 4 – (-2)
x=2
x
3
= 4 – (-1)
x = 5/3
y = -1
Output
3
y=0
x
=4–0
3
x = 4/3
Comments
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Transforming equations and formulas is relatively
simple…
The key to being successful here is being
organized, laying out your problem and solving stepby-step.
This concept will be utilized extensively when we
plot linear equations on the coordinate plane later in
the course…
Make no mistake about it…at the level you are
required to know how to transform equations and
formulas….
Comments
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On the next couple of slides are some practice
problems…The answers are on the last slide…
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Do the practice and then check your answers…If
you do not get the same answer you must question
what you did…go back and problem solve to find
the error…
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If you cannot find the error bring your work to me
and I will help…
Your Turn
Solve for the indicated variable:
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1.
2.
3.
A = ½ bh
V = ℓwh
V = πr2h
A = ½ h(b1 + b2)
Solve for b
Solve for h
Solve for h
Solve for b2
Your Turn
Rewrite each equation so that y is a function of x
5.
6.
7.
8.
9.
10.
2x + y = 5
13 = 12x – 2y
9 – y = 1.5x
y – 7 = -2x
5
-3x + 4y – 5 = -14
1 (25 – 5y) = 4x – 9y + 13
5
Your Turn Solutions
1.
2.
3.
4.
b = 2a
h
h= V
ℓw
h= V
πr2
b2 = 2a – b1
h
5.
6.
7.
8.
9.
10.
y = 5 – 2x
y = 6x – 13/2
y = -1.5x + 9
y = -10x + 35
y = ¾ x – 9/4
y=½x+1
Summary
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A key tool in making learning effective is being able to
summarize what you learned in a lesson in your own
words…
In this lesson we talked about Formulas and
Functions Therefore, in your own words summarize
this lesson…be sure to include key concepts that the
lesson covered as well as any points that are still not
clear to you…
I will give you credit for doing this lesson…please see
the next slide…
Credit
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I will add 25 points as an assignment grade for you working on this
lesson…
To receive the full 25 points you must do the following:
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Have your name, date and period as well a lesson number as a
heading.
Do each of the your turn problems showing all work
Have a 1 paragraph summary of the lesson in your own words
Please be advised – I will not give any credit for work submitted:
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Without a complete heading
Without showing work for the your turn problems
Without a summary in your own words…