###### Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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# Inverse Variation - Problem 1

Alissa Fong
###### Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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In inverse variations, the product of each x and y pair should equal the same constant, k. In other words, xy=k. Multiply the x and y values together in each pair. If the product is the same each time, the relationship varies inversely, or the relationship represents an inverse variation

Here I'm given tables of values and asked to tell whether or not they represent inverse relations. So here is something I'm going to keep in mind before I approach these problems. I know that for inverse variation the products of the x value and the y value should always be equal to the same number.

That's how you do this problem, I'm going to say it again. You want to find the products of each x value with it's paired y value and see if you get the same constant over and over. So let's check it out.

1 times 12 gives me 12. 2 times 6 also gives me 12 yeah. -3 times -4 gives me 12 so yes this table, part A, does represent an inverse relationship because every time I multiply my x and y values I get the same constant.

Let's try this table the same process. Is it true that 1 times 4 is equal to 3 times 12 oh-oh! I can already tell this is not going to be an inverse relationship because I have different products. Let's just verify with this last guy see if maybe I made an error, no. Those are all different products therefore this relation is not an inverse relationship.

These problems can be pretty easy if you guys just remember what inverse variation means. It means the product of your x and y values always gives you the same k or constant of variation.