### 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 weβre 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 weβre 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, weβre gonna have to introduce something. And weβre going to introduce π, which is our proportionality constant. And when we do that, what weβre gonna have is an equation which is π¦ is equal to ππ₯π§ over π’. Now with these kind of questions, thereβs 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 youβre given. Then you form a new equation. And with this new equation, you then solve the problem by finding the missing value.

So letβs see if we got any information that can help us find π. Well, yes, we have because weβre 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 weβre 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 weβve done is substituted in two for π because weβve 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 weβre 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 weβve 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.