- 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

### 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 = 0 | y = 0 |

x = 1 | y = 7 |

x = 2 | y = 14 |

x = 3 | y = 21 |

x = 4 | y = 28 |

x = 5 | y = 35 |

- 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

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 = kx

^{2}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 = 10 | t = 20 |

r = 20 | t = 10 |

r = 40 | t = 5 |

r = 80 | t = 2.5 |

r = 160 | t = 1.25 |

r = 320 | t = 0.625 |

- 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

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/x

^{3}or k = yx

^{3}: $$k=yx^3$$ 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/x

^{3}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. $$I=\text{prt}$$ 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 = .01 | t = 5 | I = 250 |

r = .02 | t = 5 | I = 500 |

r = .02 | t = 10 | I = 1000 |

r = .2 | t = 2 | I = 2000 |

r = .45 | t = 1 | I = 2250 |

r = .9 | t = .5 | I = 2250 |

- 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

Example 5: Solve each variation problem.

If y varies jointly with x

^{2}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/x

^{2}z: $$k=\frac{y}{x^2z}$$ 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 = kx

^{2}z 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.

First, let's think about this as:

y = kx

^{3}

where y is the amount in dollars of raffle tickets sold

x is the amount in dollars of anticipated raffle winnings

We know how to find the constant of variation:

$$k=\frac{y}{x^3}$$ 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.

#### Skills Check:

Example #1

Solve each problem.

If y varies directly with the square of x, and y = 9/4 when x = 3, find y when x = 12.

Please choose the best answer.

Example #2

Solve each word problem.

The frequency of a vibrating string varies inversely with its length. A piano string that is 4 feet long vibrates at 125 cycles per second. What frequency would a string that is 5 feet long have?

Please choose the best answer.

Example #3

Solve each word problem.

The time required to empty a tank varies inversely with the rate of pumping. If a pump can empty a tank in 2 hours and 15 minutes at the rate of 200 gallons per minute, how long will it take the pump to empty the same tank at a rate of 1,000 gallons per minute?

Please choose the best answer.

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