Lesson Objectives
• Learn how to solve direct variation problems
• Learn how to solve inverse variation problems
• Learn how to solve joint variation problems
• Learn how to solve variation word problems

## How to Solve Variation Problems

In this lesson, we will discuss the various types of variation problems (direct, inverse, and joint) that we will encounter. Let's begin with the most common type of variation problem, which is known as "direct variation".

### Solving Direct Variation Problems

We will see direct variation whenever y depends on a multiple of x. For example, the circumference of a circle is found by using the formula:
C = 2πr
The circumference "C" is always found by multiplying the radius "r" by the constant 2π (2 pi). Therefore, as the radius "r" increases, the circumference will increase and as the radius "r" decreases, the circumference will decrease. Because of this, we can say that the circumference of a circle varies directly with the radius.
In general, we can say that "y" varies directly with "x" if there is some real number "k" such that:
y = kx
or
k = y/x
y is also said to be proportional to x. The number k is called the constant of variation or the constant of proportionality.
When x > 0, as x increases, y increases, and when x decreases, y decreases.
If k > 0, then as x ↑ by 1 unit, y ↑ k units.
If k > 0, then as x ↓ by 1 unit, y ↓ k units.
Let's suppose at a local gas station, the cost per gallon is $7. We can say our total cost "y" is equal to the constant cost per gallon of "7" times the number of gallons purchased "x". This gives us the following equation: y = 7x y = 7x x = 0y = 0 x = 1y = 7 x = 2y = 14 x = 3y = 21 x = 4y = 28 x = 5y = 35 We can see that each time x or the number of gallons purchased increases by 1, y or the total cost increases by 7. When we solve a direct variation problem, we can use the following steps. • Write the variation equation: y = kx or k = y/x • Use substitution to find the value for k • Rewrite the variation equation: y = kx with the known value for k • Find the required answer using substitution Let's look at an example. Example 1: Solve each variation problem If y varies directly with x, and y is 60 when x is 5, find y when x is 20. Step 1) Write the variation equation: $$k=\frac{y}{x}$$ Step 2) Use substitution to find the value for k: $$k=\frac{60}{5}=12$$ Step 3) Rewrite the variation equation: y = kx with the known value for k: $$y=12x$$ Step 4) Find the required answer using substitution: $$y=12(20)$$ $$y=240$$ y is 240 when x is 20. ### Direct Variation as a Power We will also see direct variation as a power. We can say that "y" varies directly with the nth power of "x" if there is some real number "k" such that: $$y=kx^n$$ To solve a direct variation as a power problem, we use the same procedure. The only difference is we change up the variation equation to include our power. Let's look at an example. Example 2: Solve each variation problem If y varies directly with the square of x, and y is 27 when x is 3, find y when x is 11. Step 1) Write the variation equation: $$y=kx^{2}$$ $$k=\frac{y}{x^{2}}$$ Step 2) Use substitution to find the value for k: $$k=\frac{27}{3^{2}}=\frac{27}{9}=3$$ Step 3) Rewrite the variation equation: y = kx2 with the known value for k: $$y=3x^2$$ Step 4) Find the required answer using substitution: $$y=3(11)^2$$ $$y=3(121)=363$$ y is 363 when x is 11. ### Inverse Variation Similar to direct variation, we have inverse variation. Inverse variation occurs when y depends on some constant number divided by x. For example, we have all used the distance formula to solve a motion word problem. $$d=rt$$ Where d is the distance traveled, r is the rate of speed, and t is the amount of time traveled. Let's first solve this equation for t: $$t=\frac{d}{r}$$ Over a given distance, time varies inversely with the rate of speed. This should be fairly intuitive. If we travel a set distance, as we increase speed, the trip will be shorter. Similarly, as we decrease speed, our trip will take longer. As an example, suppose we take a day trip that is 200 miles. Let's set our first-rate of speed as 40 miles per hour. $$t=\frac{200}{40}=5$$ This trip would take 5 hours. What happens if we increase our speed to 100 miles per hour? $$t=\frac{200}{100}=2$$ Now we can see our same trip is only going to take 2 hours. We increased our speed and the time of our trip went down. What happens if we decrease our speed to 20 miles per hour? $$t=\frac{200}{20}=10$$ This trip will take 10 hours. This is longer because we decreased our speed. $$t=\frac{200}{r}$$ r = 10t = 20 r = 20t = 10 r = 40t = 5 r = 80t = 2.5 r = 160t = 1.25 r = 320t = 0.625 If we look at the table above, we can see that the time our trip takes decreases as we increase our speed and increases as we decrease our speed. In general, we can say that "y" varies inversely with "x" if there is some real number "k" such that: $$y=\frac{k}{x}$$ or $$k=xy$$ To solve an inverse variation problem, we can use the following steps. • Write the variation equation: y = k/x or k = yx • Use substitution to find the value for k • Rewrite the variation equation: y = k/x with the known value for k • Find the required answer using substitution Let's look at an example. Example 3: Solve each variation problem If y varies inversely with x, and y is 25 when x is 5, find y when x is 10. Step 1) Write the variation equation: k = yx $$k=yx$$ Step 2) Use substitution to find the value for k: $$k=25(5)=125$$ Step 3) Rewrite the variation equation: y = k/x with the known value for k: $$y=\frac{125}{x}$$ Step 4) Find the required answer using substitution: $$y=\frac{125}{10}=\frac{25}{2}$$ y is 25/2 or 12.5 when x is 10. ### Inverse Variation as a Power We will also see inverse variation as a power. We say that "y" varies inversely with the nth power of "x" if there is some real number "k" such that: $$y=\frac{k}{x^n}$$ To solve an inverse variation as a power problem, we use the same procedure. The only difference is we change up the variation equation to include our power. Let's look at an example. Example 4: Solve each variation problem If y varies inversely with the cube root of x, and y is 1/40 when x is 10, find y when x is 15. Step 1) Write the variation equation: y = k/x3 or k = yx3: $$k=yx^3$$ Step 2) Use substitution to find the value for k: $$k=\frac{1}{40}\cdot 10^3$$ $$k=\frac{1}{40}\cdot 1000$$ $$k=\frac{1000}{40}=25$$ Step 3) Rewrite the variation equation: y = k/x3 with the known value for k: $$y=\frac{25}{x^3}$$ Step 4) Find the required answer using substitution: $$y=\frac{25}{15^3}=\frac{25}{3375}$$ $$y=\frac{1}{135}$$ y is 1/135 when x is 15. ### Joint Variation In some cases, we will have one variable that depends on several others. For example, we can think about the formula which is used to solve simple interest word problems. $$I=prt$$ For a given principal (amount invested) "p", the amount of simple interest earned "I" varies jointly with the interest rate "r" and the time "t". Let's suppose we have 5000 dollars to invest in a savings account that pays annual simple interest. If we invest our money at a rate of 3% for 6 years: $$I=5000(.03)(6)=900$$ This means we would earn 900 dollars in simple interest over a period of 6 years. What happens if we increase the rate to 7%, using the same time period of 6 years? $$I=5000(.07)(6)=2100$$ This means we would earn 2100 dollars in simple interest over a period of 6 years. We can see that as the interest rate increased, the amount of simple interest earned increased. Let's suppose we again use an interest rate of 7% but decrease our time period to 3 years. $$I=5000(.07)(3)=1050$$ We can see as the time period decreased from 6 years to 3 years, the simple interest earned decreased. I = 5000rt r = .01t = 5I = 250 r = .02t = 5I = 500 r = .02t = 10I = 1000 r = .2t = 2I = 2000 r = .45t = 1I = 2250 r = .9t = .5I = 2250 From the table above, we can see that increasing rate or time increases simple interest earned, while a decrease in rate or time results in a decrease in simple interest earned. In general, we can say that "y" varies jointly with "x" and "z" if there is some real number "k" such that: $$y=kxz$$ or $$k=\frac{y}{xz}$$ To solve an inverse variation problem, we can use the following steps. • Write the variation equation: y = kxz or k = y/xz • Use substitution to find the value for k • Rewrite the variation equation: y = kxz with the known value for k • Find the required answer using substitution Let's look at an example. Example 5: Solve each variation problem If y varies jointly with x2 and z, and y is 816 when x is 4 and z is 3, find y when x is 5 and z is -1. Step 1) Write the variation equation: k = y/x2z: $$k=\frac{y}{x^2z}$$ Step 2) Use substitution to find the value for k: $$k=\frac{816}{4^2 \cdot 3}$$ $$k=\frac{816}{16 \cdot 3}$$ $$k=\frac{816}{48}=17$$ Step 3) Rewrite the variation equation: y = kx2z with the known value for k: $$y=17x^2z$$ Step 4) Find the required answer using substitution: $$y=17(5)^2(-1)$$ $$y=17(25)(-1)$$ $$y=-425$$ y is -425 when x is 5 and z is (-1). ### Solving Variation Word Problems To solve a variation word problem, we first read and interpret our problem. We can then set up and solve the problem using the methods discussed earlier in the lesson. Let's look at an example. Example 6: Solve each variation word problem A local church sells raffle tickets to supplement revenue. The total revenue from raffle tickets sold varies directly with the cube of the anticipated winnings. If the raffle sold$5,000,000 worth of tickets when the anticipated winnings were $100, find the revenue from raffle tickets sold when the anticipated winnings are$25.
$$k=\frac{y}{x^3}$$ Let's plug in for x and y, we are told that the raffle sold $5,000,000 worth of tickets (y) when the anticipated winnings were$100 (x). $$k=\frac{5,\hspace{-.1em}000,\hspace{-.1em}000}{100^3}$$ $$k=\frac{5,\hspace{-.1em}000,\hspace{-.1em}000}{1,\hspace{-.1em}000,\hspace{-.1em}000}$$ $$k=5$$ Now we can set up our variation equation with the known value for k: $$y=5x^3$$ We want to find the revenue from raffle tickets sold when the anticipated winnings are $25, let's plug in 25 for x: $$y=5(25)^3$$ $$y=5(15,\hspace{-.1em}625)$$ $$y=78,\hspace{-.1em}125$$ When the anticipated winnings are$25, the church will sell \$78,125 worth of raffle tickets.