### Video Transcript

Find the values of π₯ and π¦ given the arithmetic sequence nine π₯, four π¦ plus one, two π¦ minus six, four π¦ plus two.

Weβve been given four terms in an arithmetic sequence, and all four of these terms are given as expressions with a variable, either π₯ or π¦. In an arithmetic sequence, the terms are such that the difference between any two consecutive terms is the same amount, and we call that amount the common difference. This means if we take the first term and add the common difference, we get the second term. If we take the second term and add the common difference, we get the third term and so on. We could use this fact to set up some equations. For example, we could say that nine π₯ plus π equals four π¦ plus one. This is a true statement.

However, we have actually added an unknown variable. We now have three unknowns π₯, π, and π¦. We want to consider a strategy that would not require us to introduce a third variable. To do this, we can use what we know about arithmetic means. The terms between π sub one and π sub π in an arithmetic sequence are arithmetic means. This means that four π¦ plus one is an arithmetic mean. And two π¦ minus six is an arithmetic mean. Four π¦ plus one falls between π sub one and π sub three. And since four π¦ plus one is the arithmetic mean between π sub one and π sub three, then we combine π sub one and π sub three and divide by two to find π sub two. We say nine π₯ plus two π¦ minus six divided by two equals four π¦ plus one. And since two π¦ minus six is also an arithmetic mean, it falls between π sub two and π sub four.

We can create a second equation that says four π¦ plus one plus four π¦ plus two divided by two equals two π¦ minus six. In our first equation, we have two variables π₯ and π¦. In our second equation, we only have a π¦-variable, so itβs probably helpful for us to start there and solve for π¦. To do that, we can combine like terms in our numerator, which will give us eight π¦ plus three divided by two equals two π¦ minus six. We multiply both sides of our equation by two to give us eight π¦ plus three equals four π¦ minus 12. In our next step, we can subtract four π¦ from both sides and also subtract three from both sides of the equation. On the left, we will have our π¦-terms. Eight π¦ minus four π¦ equals four π¦.

And on the right side, weβll have the constants negative 12 minus three equals negative 15. We divide both sides by four, which leaves us with π¦ equals negative 15 over four. We can take what we found for π¦ and plug that into our first equation. But our calculations will be easier if we simplify before we plug in negative fifteen-fourths for π¦. Our first step to get that two out of the denominator, we multiply both sides of the equation by two, and we get nine π₯ plus two π¦ minus six equals eight π¦ plus two. We know weβre going to solve for π₯ here, so we want to get π¦ on the other side of the equation. To do that, weβll subtract two π¦ from both sides of the equation and weβll add six to both sides of the equation.

On the left side, weβre left with nine π₯. And on the right side, we have six π¦ plus eight. Then we substitute negative fifteen-fourths in for π¦. Six times negative fifteen-fourths plus eight will equal negative 29 over two. Dividing both sides of this equation by nine, we get that π₯ equals negative 29 over 18. We could go back and plug these values in to find the exact values of every term in this arithmetic sequence. However, this question was only asking us to identify the values of π₯ and π¦, which we have done here. π₯ equals negative 29 over 18, and π¦ equals negative fifteen-fourths.