Solving Variation Problems Lesson

In this lesson, we will learn about direct, inverse, joint, and combined variation.

Direct Variation

We have previously studied the slope-intercept form of a line.

Slope-Intercept Form of a Line

$$y = mx + b$$ When b is equal to zero, the y-intercept occurs at (0, 0), and our linear equation can be expressed as: $$y = mx$$ Here, we can say that the dependent variable y will directly vary based on x. As a simple example, suppose a consignment store charges a commission of 15% on the sale of goods. How could we model their earnings? $$y = 0.15x, x ≥ 0$$ Here, y is the total commission earned and x is the total sales or value of all goods sold. We can see that y will vary directly based on x. This type of direct variation equation is a linear function with a y-intercept of (0, 0). When m > 0, then the slope of the line is positive. This means as x increases ↑, y increases ↑ and as x decreases ↓, y decreases ↓. More specifically, for our example above, as x increases by 1, y increases by 0.15. The graph below can help to visualize the idea behind direct variation. Desmos Link for More Detail Now that we understand the basics, let's give an official definition of direct variation. First, we can say that two quantities will vary directly when one quantity is a constant multiple of another quantity. If the quantities x and y are related by the equation: $$y = kx$$ For some nonzero real number k, then we say that y varies directly with x. The k in our equation above is known as the constant of variation or the constant of proportionality. Additionally, some books will phrase direct variation in a different way. You may see "y varies directly as x" or "y is directly proportional to x" in your textbook. Let's look at an example.
Example #1: Find the direct variation equation.
Jamie works as a cashier at a local grocery store. Her paycheck is directly proportional to the number of hours worked. She is paid $20 per hour and can choose to work from 10 hours per week up to 40 hours per week. Write an equation that models her gross weekly pay (pay before any deductions). $$y = 20x, 10 ≤ x ≤ 40$$ Here, y is the gross weekly pay and x is the number of hours worked, which is limited from 10 to 40 hours. In this scenario, the k or constant of variation is 20. Notice if we doubled the number of hours worked, the pay would also double. For example, working 10 hours for the week would give a gross pay of $200, whereas, working 20 hours for the week would give a gross pay of $400. $$y = 20(10) = 200$$ $$y = 20(20) = 400$$ Desmos Link for More Detail

Solving a Direct Variation Problem

• Write out the variation equation
  • $$y = kx$$
  • $$k = \frac{y}{x}$$
• Substitute the given values and find k
• Replace k in the equation from step 1
• Find the remaining unknown values

Let's look at an example.
Example #2: Solve each direct variation problem.
If y varies directly with x, and y = 7 when x = 5, find y when x = 10.
1) Write out the variation equation. $$y = kx$$ 2) Substitute the given values and find k.
We are told that y = 7 when x = 5. $$k = \frac{y}{x}$$ $$k = \frac{7}{5}$$ 3) Replace k in the equation from step 1. $$y = \frac{7}{5}x$$ 4) Find the remaining unknown values.
We are asked to find y when x = 10. $$\require{cancel}y = \frac{7}{5} \cdot 10 = \frac{7}{\cancel{5}} \cdot 2\cancel{10}= 14$$ We can say that y would be 14 when x is 10.

Partial Variation

Most textbooks don't cover partial variation but you may see it from time to time, so we will briefly cover it here. A simple example occurs when we have a line that does not pass through the origin. Let's go back to the slope-intercept form of the line. $$y = mx + b$$ For direct variation, we stated that b is zero, which means the line has a y-intercept at (0, 0). In the case of partial variation, we will now look at a line where b is not zero. We can write our partial variation as: $$y = kx + c$$ Where k is nonzero and represents the constant of variation. The c here is a nonzero constant. Let's look at an example.
Example #3: Solve the variation problem.
A local banquet hall has a fixed fee of $500 plus an additional fee per person. Suppose that the cost of a reception is quoted at $3100 for a total of 100 guests. Find the cost for 160 guests.
1) Write out the variation equation. $$y = kx + c$$ y is the total cost, k is the additional fee per person, and c is the fixed fee of $500. $$y = kx + 500$$ 2) Substitute the given values and find k.
We are told that y = 3100 when x = 100. $$y = kx + 500$$ $$3100 = k(100) + 500$$ $$3100 - 500 = 100k$$ $$100k = 2600$$ $$k = 26$$ 3) Replace k in the equation from step 1. $$y = 26x + 500$$ 4) Find the remaining unknown values.
We are asked to find the cost for 160 guests. $$y = 26(160) + 500 = 4660$$ So the cost for 160 guests would be $4660.

Direct Variation with Powers

In some cases, we will have problems where y varies directly with a power of x. If n is a positive real number, then y varies directly with the nth power of x if there exists some nonzero real number k such that: $$y = kx^n$$ Let's look at an example.
Example #4: Solve each direct variation problem.
When air resistance is not taken into account, the distance a body falls from rest varies directly with the square of the time it falls. Suppose that a skydiver falls 256 feet in 4 seconds, how far will they fall in 7.5 seconds?
1) Write out the variation equation. Here y will represent the distance the skydiver has fallen in feet after x seconds. $$y = kx^2, x ≥ 0$$ 2) Substitute the given values and find k.
We are told that y = 256 when x = 4. $$k = \frac{y}{x^2}$$ $$k = \frac{256}{4^2} = \frac{256}{16} = 16$$ 3) Replace k in the equation from step 1. $$y = 16x^2, x ≥ 0$$ 4) Find the remaining unknown values.
We want to find how far they will fall in 7.5 seconds. This means we want to find y when x = 7.5. $$y = 16(7.5)^2 = 16(56.25) = 900$$ The skydiver will fall 900 feet in 7.5 seconds. Desmos Link for More Detail

Inverse Variation

Another scenario, known as inverse variation, occurs when y increases as x decreases. In this situation, the relationship can be expressed as a rational function. If the quantities x and y are related by the equation: $$y = \frac{k}{x}$$ For some nonzero real number k, then we can say that y varies inversely with x. You may see this in your textbook as "y is inversely proportional to x" or "y varies inversely as x". Let's look at an example.
Example #5: Find the inverse variation equation.
The distance from Irvine to Baton Rouge is 1800 miles. The time that it takes to drive from Irvine to Baton Rouge depends on the rate of speed. The car must follow the posted speed limits of 45 mph to 75 mph for the entirety of the trip. Write an equation that models the time it takes to drive from Irvine to Baton Rouge. $$y = \frac{1800}{x}, 45 ≤ x ≤ 75$$ Here, y represents the time in hours that it takes to travel from Irvine to Baton Rouge, which is a total distance of 1800 miles. The x represents the average rate of speed in miles per hour. We can see that as the rate of speed increases, the time for the drive is going to decrease, whereas, if the rate of speed decreases, the time for the drive is going to increase. For example, if we drive at an average speed of 50 mph, the trip will take 36 hours, however, if we drive at an average speed of 75 mph, the trip will take 24 hours. Again, this type of inverse variation equation is a rational function. When k > 0, as x increases ↑, y decreases ↓, and as x decreases ↓, y increases ↑. Desmos Link for More Detail

Solving an Inverse Variation Problem

• Write out the variation equation
  • $$y = \frac{k}{x}$$
  • $$k = xy$$
• Substitute the given values and find k
• Replace k in the equation from step 1
• Find the remaining unknown values

Let's look at an example.
Example 6: Solve each inverse variation problem.
If y varies inversely with x, and y = 3 when x = 15, find y when x = 9.
1) Write out the variation equation. $$y = \frac{k}{x}$$ 2) Substitute the given values and find k. We are told that y = 3 when x = 15. $$k = xy$$ $$k = (3)(15) = 45$$ 3) Replace k in the equation from step 1. $$y = \frac{45}{x}$$ 4) Find the remaining unknown values. We are asked to find y when x = 9. $$y = \frac{45}{9} = 5$$ We can say that y would be 5 when x is 9.

Inverse Variation with Powers

Just like we saw with direct variation with powers, we also have inverse variation with powers. Here, y varies inversely with a power of x. If n is a positive real number, then y varies inversely with the nth power of x if there exists some nonzero real number k such that: $$y = \frac{k}{x^n}$$ Let's look at an example.
Example #7: Solve each inverse variation problem.
Tyco Lights sells lamps that can vary in brightness with an included lightbulb that starts at 5 watts and can go up to 30 watts. Tyco has determined the lifespan of their lightbulbs in hours varies inversely with the square of the power in watts used by the lightbulb. If the lightbulb lasts 1600 hours when 5 watts is used, how long would the lightbulb last when 20 watts is used?
1) Write out the variation equation. Here, y will represent the lifespan of the lightbulb in hours, and x will represent the amount of power used in watts. $$y = \frac{k}{x^2}, 5 ≤ x ≤ 30$$ 2) Substitute the given values and find k. We are told that y = 1600 when x = 5. $$k = x^2y$$ $$k = (5^2)(1600) = 40{,}000$$ 3) Replace k in the equation from step 1. $$y = \frac{40{,}000}{x^2}, 5 ≤ x ≤ 30$$ 4) Find the remaining unknown values. We want to know how long the lightbulb will last when 20 watts is used. This means we are asked to find y when x = 20. $$y = \frac{40{,}000}{20^2} = 100$$ The lightbulb would last 100 hours when 20 watts is used. Desmos Link for More Detail

Joint Variation

With joint variation, a variable will depend on the product of two or more other variables. We can say that y varies jointly with x and z if there exists a nonzero real number k such that: $$y = kxz$$ Here using the word "and" implies multiplication, not addition. You may see "y varies jointly as x and y" or "y is jointly proportional to x and y" in your textbook. Additionally, we might also see joint variation as a power. If m and n are real numbers, and k is a nonzero real number we can modify our above equation: $$y = kx^nz^m$$ If x, y, and z are related by the equation above, then y varies jointly with the nth power of x and the mth power of y.

Solving a Joint Variation Problem

• Write out the variation equation
  • $$y = kxz$$
  • $$y = kx^nz^m$$
• Substitute the given values and find k
• Replace k in the equation from step 1
• Find the remaining unknown values

Let's look at an example.
Example #8: Solve each joint variation problem.
The area of a triangle varies jointly with the lengths of the base and the height. A triangle with a base of 50 inches and a height of 30 inches has an area of 750 square inches. Find the area of a triangle with a base of 10 inches and a height of 22 inches.
1) Let's write out our variation equation. Here we have the A or area that will depend on base (b) and the height (h). $$A = kbh$$ 2) Substitute the given values and find k. $$750 = k(50)(30)$$ $$k = \frac{750}{1500} = \frac{1}{2}$$ 3) Replace k in the equation from step 1. $$A = \frac{1}{2}bh$$ 4) Find the remaining unknown values. $$A = \frac{1}{2}(10)(22) = 110$$ The area of a triangle with a base of 10 inches and a height of 22 inches is 110 square inches.

Combined Variation

We may also see a combination of all the variation types we went through in our lesson. These are the most challenging problems as you have to really be careful about how your equation is set up. For example, you may see something such as: y varies directly with x and inversely with z. $$y = k \cdot \frac{x}{z}$$ Let's look at an example.
Example #9: Solve each combined variation problem.
A skateboard park has found that the monthly rentals of their skateboards vary directly with their ad campaign and inversely with the price of the skateboard rental. When $100,000 is spent on their ad campaign and the price to rent a skateboard is $25, there are 32,000 skateboard rentals. Find the monthly rentals of skateboards when $50,000 is spent on their ad campaign and the price of a skateboard is $20.
Let's create an equation to model the situation. $$y = \frac{kx}{z}$$ Here y is the monthly rentals of skateboards, x is the money spent on their ad campaign, and z is the price of the skateboard rental. Let's now plug in what we know. $$32{,}000 = \frac{100{,}000k}{25}$$ $$32{,}000 = 4000k$$ $$k = \frac{32{,}000}{4000}$$ $$k = 8$$ Now that we know the value for k, let's plug into the model. $$y = \frac{8x}{z}$$ Now, we can plug in to answer our question. $$y = \frac{8(50{,}000)}{20} = 20{,}000$$ When $50,000 is spent on their ad campaign and the price of a skateboard is $20, the monthly rentals of skateboards is 20,000.

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