### Video Transcript

If π₯, π¦ is the solution to the system of equations negative three π₯ plus five π¦ equals two, nine π₯ plus three π¦ equals 66, what is the value of π¦?

In the system of equations, if we want to solve for the value of π¦, we need to try and get rid of the π₯ variables. We know that negative three π₯ plus five π¦ equals two. We can call this equation one. And then nine π₯ plus three π¦ equals 66 can be equation two. Because we want to get rid of that π₯ variable, we can solve this equation by elimination. We solve by elimination when we make the coefficients of the variable weβre using opposite one another. Since equation two has a positive coefficient and equation one has a negative coefficient with π₯, we need our first equation to have a coefficient of negative nine, since our second equation has a coefficient of positive nine.

How could we go from negative three π₯ to negative nine π₯? We can multiply by positive three. This means we need to multiply the whole equation by positive three. Weβll distribute that positive three. Positive three times negative three π₯ equals negative nine π₯. Positive three times five π¦ equals 15π¦, and three times two equals six. Weβve rearranged equation one so that it now says negative nine π₯ plus 15π¦ equals six. Weβll take equation one and add equation two. We need to check that all the variables are lined up together and they are. And then we add negative nine π₯ plus nine π₯, which cancel each other out, eliminating the π₯ variable. We add a 15π¦ and three π¦ to get 18π¦ and six plus 66 equals 72. To get π¦ by itself, we divide both sides by 18. 18π¦ divided by 18 is just π¦.

This tells us that π¦ equals 72 over 18. You might notice that both 72 and 18 are divisible by nine. 72 divided by nine is eight; 18 divided by nine is two. And we know that eight over two can be simplified to four. And so we found that π¦ equals four. By the elimination method, we found that π¦ equals four. Letβs take a look at how we would solve this problem if we had used substitution. We would have started in the same place with these two equations. To solve by substitution, weβll need to convert one of these equations into something that says π₯ equals. This is because, in order to solve for π¦, we need to substitute a value for π₯. If we do this, in the equation one, we subtract five π¦ from either side and then we see that negative three π₯ equals two minus five π¦. We divide all of our terms by negative three. And then we find that π₯ equals negative two-thirds plus five-thirds π¦.

After weβve rearranged equation one, weβll plug in what we found for π₯ into equation two. Weβll have a new statement that says nine times negative two-thirds plus five-thirds π¦ plus three π¦ equals 66. We have to multiply nine by two-thirds. Nine times negative two equals negative 18. And negative 18 divided by three equals negative six. Nine times five thirds π¦, nine times five equals 45. We have 45 over three π¦, which equals 15π¦. And then bring down your three π¦ and your 66. Weβre actually starting to see a pattern that we had in our elimination method. In our elimination method, we were also adding 15π¦ and three π¦. So we can say that 15π¦ plus three π¦ equals 18π¦, and negative six plus 18π¦ equals 66. We add six to both sides. 18π¦ equals 72. And again we see that π¦ equals 72 over 18 which we know reduces to four.

This was substitution where we solved our first equation for π₯. We could have solved our second equation for π₯. If we did that, we would have had π₯ equals 66 over nine minus π¦ over three. We could reduce 66 over nine to 22 thirds and then you would have to take this value and plug it into the first equation. At the end of that method, it would still yield π¦ equals four.