# Video: Combined Variation and Its Applications

If π¦ is varied jointly with π₯ and π§ and inversely with π’ and π¦ = 4 when π₯ = 2, π§ = 1, and π’ = 1, find the value of π¦ when π₯ = 1, π§ = 2 and π’ = 2.

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### Video Transcript

If π¦ is varied jointly with π₯ and π§ and inversely with π’ and π¦ equals four when π₯ equals two, π§ equals one, and π’ equals one, find the value of π¦ when π₯ equals one, π§ equals two, and π’ equals two.

Okay, so the first thing we need to do with this problem is set up an equation. And what we got here is a proportionality problem. So weβre told that π¦ is varied jointly with π₯ and π§. So what this means is directly. So there is direct variation. So we can say that π¦ is proportional to π₯π§. And weβre also told that π¦ is inversely proportional to π’. So we could also say this is π¦ is proportional to one over π’.

So, great, now what we need to do is put this together to form an equation. And to enable us to do that, weβre gonna have to introduce something. And weβre going to introduce π, which is our proportionality constant. And when we do that, what weβre gonna have is an equation which is π¦ is equal to ππ₯π§ over π’. Now with these kind of questions, thereβs usually a general format. And that is that first of all, you need to find π, which is the proportionality constant. And you do that using the information youβre given. Then you form a new equation. And with this new equation, you then solve the problem by finding the missing value.

So letβs see if we got any information that can help us find π. Well, yes, we have because weβre told that π¦ equals four when π₯ equals two, π§ equals one, and π’ equals one. So what we can do is substitute these values in. So when we do, what weβre gonna get is four is equal to π multiplied by two multiplied by one over one. So therefore, we can say that four is equal to two π. So then if we divide both sides of the equation by two, we can say that two is equal to π.

So, great, we found π. We found our proportionality constant. So what we can do is the next step that I mentioned, which is to form a new equation. And this new equation is π¦ equals two π₯π§ over π’ because what weβve done is substituted in two for π because weβve worked out what π is. So is this useful? Well, it is, because what this is gonna enable us to do is find the value of π¦ when π₯ equals one, π§ equals two, and π’ equals two.

So what we do is substitute those values in. And when we do that, what weβre gonna have is π¦ is equal to two multiplied by one multiplied by two over two. Well, we can divide the numerator and the denominator by two. So weβve got π¦ equals two multiplied by one. So therefore, π¦ is equal to two. So we can say that the value of π¦ when π₯ equals one, π§ equals two, and π’ equals two is two.