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 nonzero \(y\)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 (\(k\) 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 
\(y=kx\) or \(\displaystyle \frac{{{{y}_{1}}}}{{{{x}_{1}}}}=\frac{{{{y}_{2}}}}{{{{x}_{2}}}}\) 
The value of \(y\) varies directly with \(x\), \(y\) is directly proportional to \(x\)
Special Case: Direct Square variation: \(y=k{{x}^{2}}\) 
Inverse or Indirect Variation 
\(\displaystyle y=\frac{k}{x}\) or \(xy=k\) 
The value of \(y\) varies inversely with \(x\), \(y\) is inversely proportional to \(x\), \(y\) is indirectly proportional to \(x\)
Special case: Indirect Square variation: \(\displaystyle y=\frac{k}{{{{x}^{2}}}}\) 
Joint Variation 
Like direct variation, but involves more than one variable.
Example: \(y=kx{{z}^{2}}\) 
Example: \(y\) varies jointly with \(x\) and the square of \(z\) 
Combined Variation 
Involves a combination of direct variation or joint variation, and indirect variation.
Example: \(\displaystyle y=\frac{{kxw}}{{{{z}^{2}}}}\) 
Example: \(y\) varies jointly as \(x\) and \(w\) and inversely as the square of \(z\) 
Partial Variation 
Two variables are related by the sum of two or more variables (one of which may be a constant).
Example: \(y={{k}_{1}}x+{{k}_{2}}\) 
Example: \(y\) is partly constant and partly varies directly with \(x\) 
Direct or Proportional Variation
When two variables are related directly, the ratio of their values is always the same. If \(k\), the constant ratio is positive, the variables go up and down in the same direction. If \(k\) is negative, as one variable goes up, the other goes down. (\(k\ne 0\))
Think of linear direct variation as a “\(y=mx\)” line, where the ratio of \(y\) to \(x\) is the slope (\(m\)). With direct variation, the \(y\)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 \(y=mx+b\), which is our typical linear equation.)
Direct variation problems are typically written:
→ \(\boldsymbol {y=kx}\), where \(k\) is the ratio of \(y\) to \(x\) (which is the same as the slope or rate).
Some problems will ask for that \(k\) 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 \(x\) and \(y\) and you can simply set up a ratio to find the other value. I’m thinking the \(k\) 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 \(y=k{{x}^{2}}\). There is a word problem example of this here.)
Remember the example of making $10 an hour at the mall (\(y=10x\))? 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 \(({{x}_{1}},{{y}_{1}})\), or the second \(({{x}_{2}},{{y}_{2}})\).
\(\displaystyle \frac{{{{y}_{1}}}}{{{{x}_{1}}}}\,\,=\,\,\frac{{{{y}_{2}}}}{{{{x}_{2}}}}\)
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 \(x\) and \(y\).
Direct Variation Problem  Formula Method  Proportion Method 
The value of \(y\) varies directly with \(x\), and \(y=20\) when \(x=2\).
Find \(y\) when \(x=8\).
(Note that this may be also be written “\(y\) is proportional to \(x\), and \(y=20\) when \(x=2\). Find \(y\) when \(x=8\)“.) 
\(\begin{array}{l}y=kx\\20=k2\\k=10\end{array}\) \(\begin{array}{l}y=kx\\y=10x\\y=10(8)\\y=80\end{array}\)
Since \(x\) and \(y\) vary directly, we know that \(y=kx\). Since the problem was stated that \(y\) varies directly with \(x\), we place the \(y\) first.
Solve for \(k\), using the values of \(x\) and \(y\) that we know (\(x=2,\,\,y=20\)). We see that \(k=10\).
Now use \(y=10x\). We plug the new \(x\), which is 8. We get the new \(y=80\). 
\(\displaystyle \begin{align}\frac{{{{y}_{1}}}}{{{{x}_{1}}}}&=\frac{{{{y}_{2}}}}{{{{x}_{2}}}}\,\\\frac{{20}}{2}&=\frac{y}{8}\,\\\\2y&=160\\y&=80\end{align}\)
We can set up a proportion with the \(y\)’s on top, and the \(x\)’s on bottom (think of setting slopes equal to each other 🙂 )
When we see the word “when” in the original problem (“\(y=20\) when \(x=2\)”), it means that that \(x\) goes with that \(y\).
We can then cross multiply to get the new \(y\).

It’s really that easy. Can you see why the proportion method can be the preferred method, unless you are asked to find the \(k\) constant in the formula?
Again, if the problem asks for the equation that models this situation, it would be “\(y=10x\)”.
Direct Variation Word Problem:
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? 
\(\begin{align}y&=kx\\2500&=k100\\k&=25\end{align}\) \(\begin{align}y&=25x\\y&=25(1000)\\y&=25000\end{align}\)
Since the amount of money is directly proportional (varies directly) to the number who attend, we know that \(y=kx\), where \(y=\) the amount of money raised and \(x=\) 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 \(y\)).
We need to fill in the numbers from the problem, and solve for \(k\). We see that \(k=25\). We have \(y=25x\). We plug the new \(x\), which is 1000.
We get the new \(y=25000\). If 1000 people attend, $25,000 would be raised! 
\(\displaystyle \begin{align}\frac{{\text{ }\!\!\$\!\!\text{ }\!\!\$\!\!\text{ }}}{{\text{attendees}}}&=\frac{{\text{ }\!\!\$\!\!\text{ }\!\!\$\!\!\text{ }}}{{\text{attendees}}}\\\frac{{2500}}{{100}}&=\frac{y}{{1000}}\\\\100y&=2500000\\y&=25000\end{align}\)
We can set up a proportion with the \(y\)’s on top (amount of money), and the \(x\)’s on bottom (number of attendees). We can then cross multiply to get the new amount of money (\(y\)).
We get the new \(y=25000\). 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? 
\(\displaystyle \begin{align}\frac{{{{y}_{1}}}}{{{{x}_{1}}}}&=\frac{{{{y}_{2}}}}{{{{x}_{2}}}}\\\frac{{10}}{1}&=\frac{y}{{20}}\\\\y&=200\end{align}\)
We can set up a proportion with the \(y\)’s on top (representing gallons), and the \(x\)’s on bottom (representing number of loads). Remember that “per load” means “for 1 load”.
We can then cross multiply to get the new \(y\). 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 \(y\) is proportional to the square of \(x\), or \(y=k{{x}^{2}}\). 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 \(y\) varies directly with the square of \(x\), and if \(y=4\) when \(x=3\), what is \(y\) when \(x=2\)? 
\(\displaystyle \begin{align}y&=k{{x}^{2}}\\4&=k\cdot {{3}^{2}}\,\,\\k&=\frac{4}{9}\,\end{align}\) \(\displaystyle \begin{align}y&=\frac{4}{9}{{x}^{2}}\\y&=\frac{4}{9}\cdot {{2}^{2}}\,\,\,\,\,\,\\y&=\frac{{16}}{9}\end{align}\)
Since \(y\) is directly proportional (varies directly) to the square of \(x\), we know that \(y=k{{x}^{2}}\). Plug in the first numbers we have for \(x\) and \(y\) to see that \(\displaystyle k=\frac{{4}}{9}\).
We have \(\displaystyle y=\frac{4}{9}{{x}^{2}}\). We plug the new \(x\), which is 2, and get the new \(y\), which is \(\displaystyle \frac{{16}}{9}\). 
\(\displaystyle \begin{align}\frac{{{{y}_{1}}}}{{{{{\left( {{{x}_{1}}} \right)}}^{2}}}}&=\frac{{{{y}_{2}}}}{{{{{\left( {{{x}_{2}}} \right)}}^{2}}}}\\\frac{4}{{{{3}^{2}}}}&=\frac{y}{{{{2}^{2}}}}\\y=\frac{{4\cdot {{2}^{2}}}}{{{{3}^{2}}}}&=\frac{{16}}{9}\end{align}\)
We can set up a proportion with the \(y\)’s on top, and \({{x}^{2}}\)’s on the bottom.
We can plug in the numbers we have, and then cross multiply to get the new \(y\).
We then get the new \(\displaystyle y=\frac{{16}}{9}\).

Inverse or Indirect Variation
Inverse or Indirect Variation refers to relationships of two variables that go in the opposite direction (their product is a constant, \(k\)). 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 Speed of car (\(x\)) 
Time to get to school (\(y\)) (minutes) 
\(x\) times \(y\) 
25  10  \(25\times 10=250\) 
30  8.33  \(\displaystyle 20\times 8.33~\approx 250\) 
35  7.14  \(\displaystyle 35\times 7.14~\approx 250\) 
40  6.25  \(40\times 6.25=250\) 
(Note that \(\approx \) means “approximately equal to”).
Do you see how when the \(x\) variable goes up, the \(y\) goes down, and when you multiply the \(x\) with the \(y\), we always get the same number (Note that this is different than a negative slope, or negative \(k\) value, since with a negative slope, we can’t multiply the \(x\)’s and \(y\)’s to get the same number).
So the formula for inverse or indirect variation is:
→ \(\displaystyle \boldsymbol{y=\frac{k}{x}}\) or \(\boldsymbol{xy=k}\), where \(k\) is always the same number.
(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 \(\displaystyle y=\frac{k}{{{{x}^{2}}}}\text{ or }{{x}^{2}}y=k\).)
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 
\(\displaystyle y=\frac{k}{x}\text{ }\,\,\text{or }\,\text{ }xy=k\)
\(\displaystyle {{x}_{1}}{{y}_{1}}={{x}_{2}}{{y}_{2}}\)
In our case, \(k=250\)
\(\displaystyle xy=250\text{ }\,\,\text{ or }\,\,\text{ }y=\frac{{250}}{x}\) 
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 \(x\) and \(y\):
Indirect Variation Problem  Formula Method  Product Rule Method 
The value of \(y\) varies inversely (or indirectly) with \(x\), and \(y=4\) when \(x=3\).
Find \(x\) when \(y=6\).
The problem may also be worded like this: Let \({{x}_{1}}=3\), \({{y}_{1}}=4\), and \({{y}_{2}}=6\). Let \(y\) vary inversely as \(x\). Find \({{x}_{2}}\). 
\(\displaystyle \begin{align}{{y}_{1}}&=\frac{k}{{{{x}_{1}}}}\\\,4&=\frac{k}{3}\\\,k&=12\end{align}\) \(\displaystyle \begin{align}{{y}_{2}}&=\frac{{12}}{{{{x}_{2}}}}\\6&=\frac{{12}}{{{{x}_{2}}}}\\6{{x}_{2}}=&12;\,\,\,{{x}_{2}}=2\end{align}\)
Since \(x\) and \(y\) vary inversely, we know that \(xy=k\), or \(\displaystyle y=\frac{k}{x}\).
We first fill in the \(x\) and \(y\) values with \({{x}_{1}}\) and \({{y}_{1}}\) from the problem. Remember that the variables with the same subscript, such as \({{x}_{1}}\) and \({{y}_{1}}\), stay together. We then solve for \(k\), which is 12.
We then put the \({{y}_{2}}\) value in for \(y\). We then solve for \({{x}_{2}}\), which is 2. (If the \({{x}_{2}}\) value were given, you’d put that in for \(x\), and solve for \({{y}_{2}}\)).
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 \(k\), and the equation of variation, which is \(\displaystyle y=\frac{{12}}{x}\). 
\(\begin{array}{l}\,\,\,\,\,{{x}_{1}}{{y}_{1}}={{x}_{2}}{{y}_{2}}\\\\\left( 3 \right)\left( 4 \right)={{x}_{2}}\left( 6 \right)\\\,\,\,\,\,\,\,\,\,\,12=6{{x}_{2}}\\\,\,\,\,\,\,\,\,\,\,{{x}_{2}}=2\end{array}\)
We know that when you multiply the \(x\)’s and \(y\)’s (with the same subscript) we get a constant, which is \(k\). You can see that \(k=12\) 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, \({{x}_{2}}\).
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? 
\(\begin{array}{c}{{x}_{1}}{{y}_{1}}={{x}_{2}}{{y}_{2}}\\\\\left( 5 \right)\left( {15} \right)={{x}_{2}}\left( 3 \right)\\75=3{{x}_{2}}\\{{x}_{2}}=25\end{array}\)
We know that when you multiply the \(x\)’s and \(y\)’s we get a constant, which is \(k\). The number of tickets Allie can buy times the price of each ticket is \(k\). We can let the \(x\)’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 \(k\)!)

“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.) 
\(\displaystyle \begin{array}{c}{{x}_{1}}{{y}_{1}}{{z}_{1}}={{x}_{2}}{{y}_{2}}{{z}_{2}}\\\\\left( {16} \right)\left( 7 \right)\left( {48} \right)=\left( {14} \right)\left( {12} \right){{z}_{2}}\\5376=168{{z}_{2}}\\{{z}_{2}}=32\end{array}\)
Since each woman is working at the same rate, we know that when we multiply the number of women \((x)\) by the number of the hours a day \((y)\) by the number of days they work \((z)\), 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 \(k\) is 5376, which represents the number of hours it would take one woman alone to paint the mural.

Recognizing Direct or Indirect Variation
You might be asked to look at functions (equations or points that compare \(x\)’s to unique \(y\)’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  
\(y=3x2\)  Neither: Direct Variation line must go through \((0,0)\).  
\(8y=x\)  Direct: This is the same as \(\displaystyle y=\frac{1}{8}x;\,\,\,\,\,\,k=\frac{1}{8}\).  

Inverse: The product of the \(x\)’s and \(y\)’s is always 8; \(k=8\).  
\(\displaystyle x=\frac{{\frac{4}{5}}}{y}\)  Inverse: This is the same as \(\displaystyle xy=\frac{4}{5};\,\,\,\,\,k=\frac{4}{5}\).  
\(y=40\)  Neither: No \(x\) in the function.  

Neither: Even though this would be a line, there is no \(k\) such that \(y=kx\). Also, direct variation line must go through \((0,0)\). 
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 \(k\), and then solve for \(y\); we need to use the Formula Method:
Joint Variation Problem  Formula Method 
Suppose \(x\) varies jointly with \(y\) and the square root of \(z\).
When \(x=18\) and \(y=2\), then \(z=9\).
Find \(y\) when \(x=10\) and \(z=4\). 
\(\begin{align}x&=ky\sqrt{z}\\18&=k\left( 2 \right)\sqrt{9}\\18&=6k\\k&=3\end{align}\) \(\begin{align}x&=ky\sqrt{z}\\x&=3y\sqrt{z}\\10&=3y\sqrt{4}\\10&=3y\left( 2 \right)\\y&=\frac{{10}}{{6}}=\frac{5}{3}\end{align}\)
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 \(k\) first by plugging in variables we are given at first; we get \(k=3\).
Now we can plug in the new values of \(x\) and \(z\) to get the new \(y\).
We see that \(\displaystyle y=\frac{5}{3}\). Really not that bad! 
Joint Variation Word Problem:
We know the equation for the area of a triangle is \(\displaystyle A=\frac{1}{2}bh\) (\(b=\) base and \(h=\) height), so we can think of the area having a joint variation with \(b\) and \(h\), with \(\displaystyle k=\frac{1}{2}\). Let’s do an area problem, where we wouldn’t even have to know the value for \(k\):
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?

\(\displaystyle \begin{array}{c}A=kbh\,\,\,\,\,\,\,\text{(original)}\\\,A=k\left( {1.4b} \right)\left( {.9h} \right)\,\,\,\,\,\,\text{(new)}\\\\\,A=k\left( {1.4} \right)\left( {.9} \right)bh\\A=k\left( {1.26} \right)bh\end{array}\)
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 (\(A\) would be multiplied by 1.26, or be 26% greater.)
You can put real numbers to verify this, using the formula \(\displaystyle A=\frac{1}{2}bh\).

Joint Variation Word Problem:
Here’s another:
Joint Variation Problem  Math and Notes 
The volume of wood in a tree (\(V\)) varies directly as the height (\(h\)) and the square of the girth (\(g\)).
If the volume of a tree is 144 cubic meters (\({{m}^{3}}\)) 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? 
\(\begin{array}{l}V=k\text{(height)(girth}{{\text{)}}^{2}}\\V=kh{{g}^{2}}\\\\144=k(20){{(1.5)}^{2}}=45k\\144=45k;\,\,k=3.2\\\\V=kh{{g}^{2}};\,\,\,\,1000=3.2h\cdot {{2}^{2}}\\h=78.125\end{array}\)
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 \(k\) first; we get \(k=3.2\).
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 \(k\), and then solve back to get what’s missing.
Combined Variation Problem  Math and Notes 
(a) \(y\) varies jointly as \(x\) and \(w\) and inversely as the square of \(z\). Find the equation of variation when \(y=100\), \(x=2\), \(w=4\), and \(z=20\).
(b) Then solve for \(y\) when \(x=1\), \(w=5\), and \(z=4\). 
\(\begin{align}{l}y&=\frac{{kxw}}{{{{z}^{2}}}}\\100&=\frac{{k(2)(4)}}{{{{{(20)}}^{2}}}}=\frac{{8k}}{{400}}\\8k&=100(400)\\k&=\frac{{(100)(400)}}{8}=5000\end{align}\) \(\begin{align}y&=\frac{{5000xw}}{{{{z}^{2}}}}\text{ }\,\,\,\,\,\text{ (answer to a)}\\\\y&=\frac{{5000(1)(5)}}{{{{4}^{2}}}}\\y&=\,\,\,\frac{{25000}}{{16}}=\,\,\,1562.5\text{ }\,\,\,\,\text{ (answer to b)}\end{align}\)
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 \(k\). We know that “the square of \(z\)” is a fancy way of saying \({{z}^{2}}\).
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 \(k\) on the top!
Now that we have the \(k\), we have the answer to (a) above by plugging it in the original equation.
We can get the new \(y\) when we have “new” \(x\), \(w\), and \(z\) values.
For the second part of the problem, when \(x=1\), \(w=5\), and \(z=4\), \(y=1562.5\). (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 \(\boldsymbol {k}\) 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! 
\(\displaystyle \begin{align}C&=\frac{{k\left( {{{P}_{1}}} \right)\left( {{{P}_{2}}} \right)}}{{{{d}^{2}}}}\\200000&=\frac{{k\left( {1500000} \right)\left( {1200000} \right)}}{{{{{400}}^{2}}}}\\k&=\frac{{\left( {200000} \right){{{\left( {400} \right)}}^{2}}}}{{\left( {1500000} \right)\left( {1200000} \right)}}=.01778\\C&=\frac{{.01778\left( {{{P}_{1}}} \right)\left( {{{P}_{2}}} \right)}}{{{{d}^{2}}}}\,\,\,\,\,\,\leftarrow \text{ answer to (a)}\end{align}\)
\(\displaystyle \begin{align}134000&=\frac{{.01778\left( {1500000} \right)\left( {1700000} \right)}}{{{{d}^{2}}}}\\134000{{d}^{2}}&=.01778\left( {1500000} \right)\left( {1700000} \right)\\d&=581.7 \, \text{miles}\,\,\,\,\,\,\,\,\leftarrow \text{ answer to (b)}\end{align}\)
We can set it up almost word for word from the word problem. Remember to put everything on top for “jointly proportional” (including \(k\)) since these are direct variations, and everything on bottom for “inversely proportional”.
Solve for \(k\) first; we get \(k=.01778\).
Now we can plug in the new values to get the distance between the cities (\(d\)). We can actually cross multiply to get \({{d}^{2}}\), and then take the positive square root get \(d\).
The distance between Charlotte and Indianapolis is about 581.7 miles. 
Combined Variation Word Problem:
Here’s another:
Combined Variation Problem  Math and Notes 
\(y\) varies jointly with \({{x}^{3}}\) and \(z\), and varies inversely with \({{r}^{2}}\).
What is the effect on \(y\) when \(x\) is doubled and \(r\) is halved?

Since we want \(x\) to double and \(r\) to be halved, we can just put in the new “values” and see what happens to \(y\). Make sure to put them in parentheses, and “push the exponents through”:
\(\displaystyle \begin{align}y&=\frac{{k{{x}^{3}}z}}{{{{r}^{2}}}}\,\,\,\,\,\,\,\,\text{(original)}\\y&=\frac{{k{{{\left( {2x} \right)}}^{3}}z}}{{{{{\left( {\frac{1}{2}r} \right)}}^{2}}}}\,\,\,\,\,\,\,\,\text{(new)}\end{align}\) \(\displaystyle y=\frac{{k8{{x}^{3}}z}}{{\frac{1}{4}{{r}^{2}}}}=\frac{8}{{\frac{1}{4}}}\frac{{k{{x}^{3}}z}}{{{{r}^{2}}}}=\left( {\frac{8}{1}\cdot \frac{4}{1}} \right)\frac{{k{{x}^{3}}z}}{{{{r}^{2}}}}=32\frac{{k{{x}^{3}}z}}{{{{r}^{2}}}}\)
We can set it up with everything on top for “varies jointly” (including \(k\)) since these are direct variations, and everything on bottom for “varies inversely”.
Now substitute “\(2x\)” for \(x\), since \(x\) is doubled, and “\(\displaystyle \frac{1}{2}r\)” for \(r\), since \(r\) is halved.
Simplify to see that we have a 32 in front of the old variation. (Don’t forget to flip and multiply when we divide by a fraction.) So the effect on \(y\) would be 32 times greater, or multiplied by 32.

One word of caution: I found a variation problem in an SAT book that stated something like this: “If \(x\) varies inversely with \(y\) and varies directly with \(z\), and if \(y\) and \(z\) are both 12 when \(x=3\), what is the value of \(y+z\) when \(x=5\)”. I found that I had to solve it setting up two variation equations with two different \(k\)’s (otherwise you can’t really get an answer). So watch the wording of the problems. 🙁
Here is how I did this problem:
Variation Problem  Math and Notes 
If \(x\) varies inversely with \(y\) and varies directly with \(z\), and if \(y\) and \(z\) are both 12 when \(x=3\), what is the value of \(y+z\) when \(x=5\) 
\(\displaystyle \begin{align}x&=\frac{{{{k}_{1}}}}{y}\\3&=\frac{{{{k}_{1}}}}{{12}}\\{{k}_{1}}&=36\end{align}\) \(\displaystyle \begin{align}x&=\frac{{36}}{y}\\5&=\frac{{36}}{y}\\y&=\frac{{36}}{5}\end{align}\) \(\displaystyle \begin{align}x&={{k}_{2}}z\\3&={{k}_{2}}12\\{{k}_{2}}&=\frac{1}{4}\end{align}\) \(\displaystyle \begin{align}x&=\frac{1}{4}z\\5&=\frac{1}{4}z\\z&=20\end{align}\) \(y+z=27.2\)
Set up 2 variation equations, the first using \({{k}_{1}}\) and the second \({{k}_{2}}\) as constants. The first equation is inverse variation, and the second equation is direct variation.
Now we can solve for \({{k}_{1}}\) and \({{k}_{2}}\) separately, using the fact that \(y\) and \(z\) are both 12 when \(x=3\). We get \({{k}_{1}}=36\) and \(\displaystyle {{k}_{2}}=\frac{1}{4}\).
We can then put the constants back in the equation and solve for \(y\) and \(z\) when \(x=5\). Cross multiply to see that \(\displaystyle y=\frac{{36}}{5}=7.2\) and \(\displaystyle z=20\), so \(y+z=27.2\).

Partial Variation
You don’t hear about Partial Variation or something being partly varied or part varied very often, but it means that two variables are related by the sum of two or more variables (one of which may be a constant). An example of part variation is the relationship modeled by an equation of a line that doesn’t go through the origin.
Here a few examples:
Partial Variation Problem 
Solution 
\(y\) is partly constant and partly varies (directly) with \(x\).
When \(y=4,\,x=2\), and when \(y=16,\,x=4\).
Find an equation connecting \(y\) and \(x\), and find \(y\) when \(x=6\). 
Since \(y\) varies partly with a constant and directly with \(x\), we have \(y={{k}_{1}}x+{{k}_{2}}\). We’ll used a system to solve for \({{k}_{1}}\) and \({{k}_{2}}\) by plugging in what we know:
\(\begin{align}y&={{k}_{1}}x+{{k}_{2}}\\4&=2{{k}_{1}}+{{k}_{2}}\\16&=4{{k}_{1}}+{{k}_{2}}\end{align}\) Use elimination to solve: \(\begin{array}{l}4=2{{k}_{1}}{{k}_{2}}\\\underline{{\,16=\,\,\,\,4{{k}_{1}}\,+{{k}_{2}}}}\\12\,=\,\,\,\,2{{k}_{1}};\,\,\,\,{{k}_{1}}=6\\{{k}_{2}}=164\left( 6 \right)=8\end{array}\) Now we have \(\,y=6x8\). When \(x=6,\,\,y=6\left( 6 \right)8=28\). 
\(y\) partly varies directly with \(x\) and also partly varies inversely with \(x\).
When \(y=2,\,x=2\), and when \(y=7,\,x=4\).
Find an equation connecting \(y\) and \(x\), and Find \(y\) when \(x=8\). 
Since \(y\) partly varies directly with \(x\) and partly varies inversely with \(x\), we have \(\displaystyle y={{k}_{1}}x+\frac{{{{k}_{2}}}}{x}\). We’ll used a system to solve for \({{k}_{1}}\) and \({{k}_{2}}\):
\(\displaystyle \begin{align}\,\,\,\,y&={{k}_{1}}x+\frac{{{{k}_{2}}}}{x}\\\,2&=2{{k}_{1}}+\frac{{{{k}_{2}}}}{{2}}\\\,\,7&=4{{k}_{1}}+\frac{{{{k}_{2}}}}{{4}}\end{align}\) Use substitution to solve: \(\displaystyle \begin{align}2&=2{{k}_{1}}+\frac{{{{k}_{2}}}}{{2}}\\{{k}_{2}}&=44{{k}_{1}}\\7&=4{{k}_{1}}+\frac{{44{{k}_{1}}}}{{4}}\\28&=16{{k}_{1}}44{{k}_{1}}\\{{k}_{1}}&=2;\,\,\,{{k}_{2}}=4\end{align}\)
Now we have \(\displaystyle \,y=2x+\frac{4}{x}\). When \(\displaystyle x=8,\,\,\,y=2\left( 8 \right)+\frac{4}{8}=\frac{{31}}{2}\). 
The cost of attending a fair consists of a fixed entrance cost \(f\) of $10, and a charge for riding rides, which is proportional to the number of rides ridden.
If the cost of attending the fair is $24 when 7 rides are ridden, find the cost of riding 10 rides. 
Since \(y\) varies partly with a constant \(f\) and directly with the number of rides ridden, say \(x\), we have \(y=cx+10\), where \(c\) (cost of each ride) is a constant of variation. Let’s plug in what we know:
\(\displaystyle \begin{align}y&=cx+10\\24&=c\left( 7 \right)+10\\c&=\frac{{14}}{7}=2\end{align}\)
The cost of riding 10 rides is \(y=2x+10=2\left( {10} \right)+10=\$30\). 
We’re doing really difficult problems now – but see how, if you know the rules, they really aren’t bad at all?
For Practice: Use the Mathway widget below to try a Variation problem. Click on Submit (the blue arrow to the right of the problem) and click on Find the Constant of Variation to see the answer.
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).
Learn these rules, and practice, practice, practice!
On to Introduction to the Graphing Display Calculator (GDC). I’m proud of you for getting this far! You are ready!