Direct, Inverse, Joint and Combined Variation
This section covers:
Direct or Proportional Variation
Inverse or Indirect Variation
Joint and Combined Variation
Partial Variation
More Practice
When you start studying algebra, you will also study how two (or more) variables can relate to each other specifically. The cases
you’ll study are:
Direct Variation, where one variable is a constant multiple of another
Inverse or Indirect Variation, where when one of the variables increases, the other one decreases (their product is constant)
Joint Variation, where more than two variables are related directly
Combined Variation, which involves a combination of direct or joint variation, and indirect variation
Partial Variation, where two variables are related by a formula, such as the formula for a straight line (with a non-zero -
intercept)
These sound like a lot of fancy math words, but it’s really not too bad. Here are some examples of direct and inverse variation:
Direct: The number of dollars I make varies directly (or you can say varies proportionally) with how much I work ( is
positive).
Direct: The length of the side a square varies directly with the perimeter of the square.
Inverse: The number of people I invite to my bowling party varies inversely with the number of games they might get to play
(or you can say is proportional to the inverse of).
Inverse: The temperature in my house varies indirectly (same as inversely) with the amount of time the air conditioning is
running.
Inverse: My GPA may vary directly inversely with the number of hours I watch TV.
Partial (Direct): The total cost of my phone bill consists of a fixed cost per month, and also a charge per minute.
Here is a table for the types of variation we’ll be discussing:
Type of Variation Formula Example Wording
Direct or Proportional
Variation or
The value of varies directly with , is directly
proportional to
Special Case: Direct Square variation:
Inverse or Indirect
Variation or
The value of varies inversely with , is inversely
proportional to , is indirectly proportional to
Special case: Indirect Square variation:
Joint Variation
Like direct variation, but involves
more than one variable.
Example:
Example: varies jointly with and the square of
Combined Variation
Involves a combination of direct
variation or joint variation, and
indirect variation.
Example:
Example: varies jointly as and and inversely as
the square of
Partial Variation
Two variables are related by the
sum of two or more variables
(one of which may be a
constant).
Example:
Example: is partly constant and partly varies
directly with
Direct or Proportional Variation
When two variables are related directly, the ratio of their values is always the same. If , the constant ratio is positive, the variables
go up and down in the same direction. If is negative, as one variable goes up, the other goes down. ( )
Think of linear direct variation as a “ ” line, where the ratio of to is the slope ( ). With direct variation, the -intercept is
always 0 (zero); this is how it’s defined.
(Note that Part Variation (see below), or “varies partly” means that there is an extra fixed constant, so we’ll have an equation like
, which is our typical linear equation.)
Direct variation problems are typically written:
→ , where is the ratio of to (which is the same as the slope or rate).
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Some problems will ask for that value (which is called the constant ratio, constant of variation or constant of
proportionality – it’s like a slope!); others will just give you 3 out of the 4 values for and and you can simply set up a ratio to find
the other value. I’m thinking the comes from the word “constant” in another language.
(I’m assuming in these examples that direct variation is linear; sometime I see it where it’s not, like in a Direct Square Variation
where . There is a word problem example of this here.)
Remember the example of making $10 an hour at the mall ( )? This is an example of direct variation, since the ratio of how
much you make to how many hours you work is always constant.
We can also set up direct variation problems in a ratio, as long as we have the same variable in either the top or bottom of the
ratio, or on the same side. This will look like the following. Don’t let this scare you; the subscripts just refer to the either the first set
of variables , or the second .
Direct Variation Word Problem:
We can solve the following Direct Variation problem in one of two ways, as shown. We do these methods when we are given any
three of the four values for and .
Direct Variation Problem Formula Method Proportion Method
The value of varies
directly with , and
when .
Find when .
(Note that this may be also
be written “ is proportional
to , and when .
Find when “. )
Since and vary directly, we know that .
Since the problem was stated that varies
directly with , we place the first.
Solve for , using the values of and that we
know ( ). We see that .
Now use . We plug the new , which is 8.
We get the new .
We can set up a proportion with the ’s on
top, and the ’s on bottom (think of setting
slopes equal to each other )
When we see the word “when” in the original
problem (“ when ”), it means
that that goes with that .
We can then cross multiply to get the new .
It’s really that easy. Can you see why the proportion method can be the preferred method, unless you are asked to find the
constant in the formula?
Again, if the problem asks for the equation that models this situation, it would be “ ”.
Direct Variation Word Problem:
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Here’s another:
Direct Variation Problem Formula Method Proportion Method
The amount of money raised
at a school fundraiser is
directly proportional to the
number of people who
attend.
Last year, the amount of
money raised for 100
attendees was $2500.
How much money will be
raised if 1000 people
attend this year?
Since the amount of money is directly
proportional (varies directly) to the number who
attend, we know that , where the
amount of money raised and the number of
attendees. (Since the problem states that the
amount of money is directly proportional to the
number of attendees, we put the amount of
money first, or as the ).
We need to fill in the numbers from the problem,
and solve for . We see that . We have
. We plug the new , which is 1000.
We get the new . If 1000 people attend,
$25,000 would be raised!
We can set up a proportion with the ’s on
top (amount of money), and the ’s on
bottom (number of attendees). We can then
cross multiply to get the new amount of
money ( ).
We get the new . If 1000 people
attend, $25,000 will be raised!
Direct Variation Word Problem:
Here’s another; let’s use the proportion method:
Direct Variation Problem Proportion Method
Brady bought an energy efficient
washing machine for her new
apartment.
If she saves about 10 gallons of water
per load, how many gallons of water
will she save if she washes 20 loads
of laundry?
We can set up a proportion with the ’s on top (representing gallons), and
the ’s on bottom (representing number of loads). Remember that “per
load” means “for 1 load”.
We can then cross multiply to get the new . Brady will save 200 gallons if
she washes 20 loads of laundry.
See how similar these types of problems are to the Proportions problems we did earlier?
Direct Square Variation Word Problem:
Again, a Direct Square Variation is when is proportional to the square of , or . Let’s work a word problem with this type
of variation and show both the formula and proportion methods:
Direct Square Variation
Problem Formula Method Proportion Method
If varies directly with the
square of , and if
when , what is
when ?
Since is directly proportional (varies directly)
to the square of , we know that . Plug in
the first numbers we have for and to see that
.
We have . We plug the new , which is 2,
and get the new , which is .
We can set up a proportion with the ’s on
top, and ’s on the bottom.
We can plug in the numbers we have, and
then cross multiply to get the new .
We then get the new .
Inverse or Indirect Variation
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Inverse or Indirect Variation refers to relationships of two variables that go in the opposite direction (their product is a constant,
). Let’s suppose you are comparing how fast you are driving (average speed) to how fast you get to your school. You might have
measured the following speeds and times:
Average Spee d of car (
)
Time to get to school ( )
(minutes) times
25 10
30 8.33
35 7.14
40 6.25
(Note that means “approximately equal to”).
Do you see how when the variable goes up, the goes down, and when you multiply the with the , we always get the same
number (Note that this is different than a negative slope, or negative value, since with a negative slope, we can’t multiply the ’s
and ’s to get the same number).
So the formula for inverse or indirect variation is:
→ or , where is always the same number.
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(Note that you could also have an Indirect Square Variation or Inverse Square Variation, like we saw above for a Direct Variation.
This would be of the form .)
Here is a sample graph for inverse or indirect variation. This is actually a type of Rational Function (function with a variable in the
denominator) that we will talk about in the Rational Functions, Equations and Inequalities section here.
Formula Graph
In our case,
Inverse Variation Word Problem:
We might have a problem like this; we can solve this problem in one of two ways, as shown. We do these methods when we are
given any three of the four values for and :
Indirect Variation Problem Formula Method Product Rule Method
The value of varies
inversely (or indirectly) with
, and when .
Find when .
The problem may also be
worded like this:
Let , , and .
Let vary inversely as .
Find .
Since and vary inversely, we know that
, or .
We first fill in the and values with and
from the problem. Remember that the variables
with the same subscript, such as and , stay
together. We then solve for , which is 12.
We then put the value in for . We then solve for
, which is 2. (If the value were given, you’d
put that in for , and solve for ).
The formula way may take a little more time, but
you may be asked to do it this way, especially if
you need to find , and the equation of variation,
which is .
We know that when you multiply the ’s
and ’s (with the same subscript) we get
a constant, which is . You can see that
in this problem.
We can just substitute in all the numbers
that we are given and solve for the
number we want – in this case, .
This way is easier than the formula
method, but, again, you will probably be
asked to know both ways.
Inverse Variation Word Problem:
Here’s another; let’s use the product method:
Inverse Variation Problem Product Rule Method
For the Choir fundraiser, the number of tickets
Allie can buy is inversely proportional to the
price of the tickets.
She can afford 15 tickets that cost $5 each.
How many tickets can Allie buy if each cost
$3?
We know that when you multiply the ’s and ’s we get a constant,
which is . The number of tickets Allie can buy times the price of each
ticket is . We can let the ’s be the price of the tickets.
We can just substitute in all the numbers that we are given and solve for
the number we want. We see that Allie can buy 25 tickets that cost $3.
This makes sense, since we can see that she only can spend $75 (which
is !)
“Work” Inverse Proportion Word Problem:
Here’s a more advanced problem that uses inverse proportions in a “work” word problem; we’ll see more “work problems” here in
the Systems of Linear Equations Section and here in the Rational Functions and Equations Section.
“Work” Inverse Variation Problem Product Rule Method
If 16 women working 7 hours day can
paint a mural in 48 days, how many days
will it take 14 women working 12 hours a
day to paint the same mural?
(The three different values are inversely
proportional; for example, the more
women you have, the less days it takes
to paint the mural, and the more hours in
a day the women paint, the less days
they need to complete the mural.)
Since each woman is working at the same rate, we know that when we
multiply the number of women by the number of the hours a day by
the number of days they work , it should always be the same (a constant).
(Try it yourself with some easy numbers).
We can just substitute in all the numbers that we are given and solve for the
number we want (days). So we see that it would take 32 days for 14 women
that work 12 hours a day to paint the mural. In this case, our is 5376, which
represents the number of hours it would take one woman alone to paint the
mural.
Recognizing Direct or Indirect Variation
You mi ght be asked to l ook at functions (equations or points that compare ’s to unique ’s – we’ll discuss later in the Algebraic
Functions section) and determine if they are direct, inverse, or neither:
Function Direct, Inverse, or Neither Variation
Neither: Direct Variation line must go through .
Direct: This is the same as .
2 1 .5
4 8 16 Inverse: The product of the ’s and ’s is always 8; .
Inverse: This is the same as .
Neither: No in the function.
0 2 4
4 6 8
Neither: Even though this would be a line, there is no such that . Also,
direct variation line must go through .
Joint Variation and Combined Variation
Joint variation is just like direct variation, but involves more than one other variable. All the variables are directly proportional,
taken one at a time.
Let’s set this up like we did with direct variation, find the , and then solve for ; we need to use the Formula Method:
Joint Variation Problem Formula Method
Suppose varies jointly with and the
square root of .
When and , then .
Find when and .
Again, we can set it up almost word for word from the word problem. For the
words “varies jointly”, just basically use the “ ” sign, and everything else will
fall in place.
Solve for first by plugging in variables we are given at first; we get .
Now we can plug in the new values of and to get the new .
We see that . Really not that bad!
Joint Variation Word Problem:
We know the equation for the area of a triangle is ( base and height), so we can think of the area having a joint
variation with and , with . Let’s do an area problem, where we wouldn’t even have to know the value for :
Joint Variation Problem Math and Notes
The area of a triangle is jointly related to
the height and the base.
If the base is increased by 40% and the
height is decreased by 10%, what will be
the percentage change of the area?
Remember that when we increase a number by 40%, we are actually
multiplying it by 1.4, since we have to add 40% to the original amount.
Similarly, when we decrease a number by 10%, we are multiplying it by .9,
since we are decreasing the original amount by 10%.
Reduce the original values by the new values, and find the new “multiplier”;
we see that there will be a 26% increase in the area ( would be multiplied
by 1.26, or be 26% greater.)
You ca n p ut re al n umb ers to ver ify thi s, u sin g t he for mu la .
Joint Variation Word Problem:
Here’s another:
Joint Variation Problem Math and Notes
The volume of wood in a tree ( ) varies
directly as the height ( ) and the square
of the girth ( ).
If the volume of a tree is 144 cubic
meters ( ) when the height is 20
meters and the girth is 1.5 meters, what
is the height of a tree with a volume of
1000 and girth of 2 meters?
We can set it up almost word for word from the word problem. For the words
“varies directly”, just basically use the “ ” sign, and everything else will fall in
place. Solve for first; we get .
Now we can plug in the new values to get the new height.
The new height is 78.125 meters.
Combined Variation
Combined variation involves a combination of direct or joint variation, and indirect variation. Since these equations are a little more
complicated, you probably want to plug in all the variables, solve for , and then solve back to get what’s missing.
Let’s try a problem:
Combined Variation Problem Math and Notes
(a) varies jointly as and and inversely
as the square of . Find the equation of
variation when , , , and
.
(b) Then solve for when , , and
.
Now this looks really complicated, and you may get “word problems”
like this, but all we do is fill in all the variables we know, and then solve
for . We know that “the square of ” is a fancy way of saying .
Remember that what follows the “varies jointly as” is typically on the
top of any fraction (this is like a direct variation), and what follows
“inversely as” is typically on the bottom of the fraction. And always put
on the top!
Now that we have the , we have the answer to (a) above by plugging it
in the original equation.
We can get the new when we have “new” , , and values.
For the second part of the problem, when , , and ,
. (Just plug in).
Combined Variation Word Problem:
Here’s another; this one looks really tough, but it’s really not that bad if you take it one step at a time:
Combined Variation Problem Math and Notes
The average number of phone calls per day
between two cities has found to be jointly
proportional to the populations of the
cities, and inversely proportional to the
square of the distance between the two
cities.
The population of Charlotte is about
1,500,000 and the population of Nashville is
about 1,200,000, and the distance between
the two cities is about 400 miles. The
average number of calls between the cities
is about 200,000.
(a) Find the and write the equation
of variation.
(b) The average number of daily
phone calls between Charlotte and
Indianapolis (which has a population of
about 1,700,000) is about 134,000. Find
the distance between the two cities.
In reality, the distance between these two
cities is 585.6 miles, so we weren’t too far
off!
We can set it up almost word for word from the word problem. Remember to put
everything on top for “jointly proportional” (including ) since these are direct
variations, and everything on bottom for “inversely proportional”.
Solve for first; we get .
Now we can plug in the new values to get the distance between the cities ( ). We
can actually cross multiply to get , and then take the positive square root get .
The distance between Charlotte and Indianapolis is about 581.7 miles.
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𝑧
√
=𝑘(2)
9
‾
√
= 6𝑘
=−3
𝑥
𝑥
10
10
𝑦
=𝑘𝑦
𝑧
√
=−3𝑦
𝑧
√
=−3𝑦
4
‾
√
=−3𝑦(2)
= = −
10
−6
5
3
=
𝑘
𝑘=−3
𝑥
𝑧
𝑦
𝑦=−
5
3
𝐴=𝑏ℎ
1
2
𝑏=ℎ=
𝑏 ℎ
𝑘=
1
2
𝑘
𝐴=𝑘𝑏ℎ (original)
𝐴=𝑘(1.4𝑏) (.9ℎ) (new)
𝐴=𝑘(1.4) (.9) 𝑏ℎ
𝐴=𝑘(1.26) 𝑏ℎ
𝐴
𝐴=𝑏ℎ
1
2
𝑉
ℎ
𝑔
𝑚
3
𝑉=𝑘(height)(girth
)
2
𝑉=𝑘ℎ
𝑔
2
144 = 𝑘(20) = 45𝑘
(1.5)
2
144 = 45𝑘;𝑘= 3.2
𝑉=𝑘ℎ ; 1000 = 3.2ℎ⋅
𝑔
2
2
2
ℎ= 78.125
=
𝑘 𝑘 = 3.2
𝑘
𝑦
𝑥
𝑤
𝑧
𝑦= 100
𝑥= 2 𝑤= 4
𝑧= 20
𝑦
𝑥= 1
𝑤= 5
𝑧= 4
𝑙𝑦
100
8𝑘
𝑘
=
𝑘𝑥𝑤
𝑧
2
= =
𝑘(2)(4)
(20)
2
8𝑘
400
= 100(400)
= = 5000
(100)(400)
8
𝑦
𝑦
𝑦
= (answer to a)
5000𝑥𝑤
𝑧
2
=
5000(1)(5)
4
2
= = 1562.5 (answer to b)
25000
16
𝑘
𝑧 𝑧
2
𝑘
𝑘
𝑦
𝑥
𝑤 𝑧
𝑥= 1
𝑤= 5
𝑧= 4
𝑦= 1562.5
𝑘𝑘
𝐶
200000
𝑘
𝐶
=
𝑘( ) ( )
𝑃
1
𝑃
2
𝑑
2
=
𝑘(1500000) (1200000)
400
2
= = .01778
(200000)
(400)
2
(1500000) (1200000)
=← answer to (a)
.01778 ( ) ( )
𝑃
1
𝑃
2
𝑑
2
134000
134000
𝑑
2
𝑑
=
.01778 (1500000) (1700000)
𝑑
2
= .01778 (1500000) (1700000)
= 581.7 miles ← answer to (b)
𝑘
𝑘 𝑘 = .01778
𝑑
𝑑
2
𝑑
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12/28/19, 11:07 AM
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