### Video Transcript

Given that π₯, π, π, π¦ is an arithmetic sequence, evaluate π¦ minus π₯ over π minus π.

Because we know these four variables form an arithmetic sequence, we can say that π is halfway between π₯ and π. We can also say that π is halfway between π and π¦. We can say this because in an arithmetic sequence, each consecutive term is one common difference away from the previous term. But another way to say this would be that π is the arithmetic mean of π₯ and π.

If we know that the arithmetic mean is the sum of values divided by the number of values and weβve already said that π was the arithmetic mean of π₯ and π, then π equals π₯ plus π divided by two. And since π is the arithmetic mean between π and π¦, π is equal to π plus π¦ divided by two.

To solve π₯ minus π¦ over π minus π would be to rewrite these equations so that π₯ is by itself and π¦ is by itself. Weβll move our π-equation over to give ourselves a little bit more room. For equation one, we multiply both sides of the equation by two, which gives us two π equals π₯ plus π. And when we subtract π from both sides, we see that two π minus π equals π₯.

For our second equation, weβll solve for π¦. By multiplying both sides of the equation by two, we get two π equals π plus π¦. And then, we subtract π from both sides, which gives us two π minus π equals π¦. This means weβre ready to plug in what we found for π₯ and for π¦ into our equation. π¦ is two π minus π, and our numerator is π¦ minus π₯. π₯ is two π minus π, and the denominator would stay the same, π minus π.

To solve, our first step will be to distribute the subtraction, which gives us two π minus π minus two π plus π in the numerator. We can combine our like terms. We have two π plus π equals three π. And negative π minus two π equals negative three π, which will give us three π minus three π over π minus π. In our numerator, both the π and the π have a factor of three. We can undistribute that three so that the numerator becomes three times π minus π, which is all over π minus π. π minus π divided by π minus π equals one. So, π¦ minus π₯ over π minus π equals three.

Before we finish, letβs consider one other method for solving this question. When weβre working with an arithmetic sequence, we know that there is a common difference between each consecutive terms. And this means we could rewrite π in terms of π₯, its previous term, and the common difference π. We could say that π equals π₯ plus π. And when we think about our third term, itβs equal to our second term π plus the common difference.

However, we can also write π in terms of π₯ and π. We can say that π will be equal to the first term π₯ plus two π. So, we have π equals π₯ plus two π. And of course, this means that we could write the fourth term as π₯ plus three π. We could say that π¦ equals π₯ plus three π. And then, we take what we found for π¦ and plug it in to our equation. Our numerator becomes π₯ plus three π minus π₯, and our denominator is π minus π. We plug in π₯ plus two π for π and π₯ plus π for π.

In our denominator, we need to distribute this subtraction. Minus π₯ plus π is the same thing as minus π₯ minus π. And we see in our numerator that π₯ minus π₯ equals zero. And in our denominator, π₯ minus π₯ equals zero. In our denominator, we have two π minus π which will just leave us with one π; we can write that as π. And in the numerator, we have three π. Three π over π simplifies to three and confirms that π¦ minus π₯ over π minus π equals three.