### Video Transcript

What is the solution π₯, π¦ to the system of equations three π₯ minus two π¦ equals 10 and four π₯ minus π¦ equals 10?

Letβs consider two different ways to solve this problem. First, weβll use the elimination method. If we label the first equation one 3π₯ minus two π¦ equals 10 and our second equation four π₯ minus π¦ equals 10, to use the elimination method, weβll want either our π₯-coefficients or our π¦-coefficients to be the opposite of one another. Notice here we have negative two π¦. We want to see if we can get the second equation to say positive two π¦. We can do that by multiplying the whole second equation through by negative two. Negative two times four π₯ equals negative eight π₯. Negative two times negative π¦ is positive two π¦. And we canβt forget to multiply the other side of the equal sign. Negative two times 10 equals negative 20. Now our second equation looks a little bit different, but itβs still equivalent.

Now we bring down our first equation and write it so that the coefficients of the π₯- and π¦-variables are in-line with one another. And then, we add the two equations together. Negative eight π₯ plus three π₯ equals negative five π₯. Positive two π¦ minus two π¦ cancels out and negative 20 plus 10 equals negative 10. From there, we divide both sides by negative five. Negative five π₯ divided by negative five equals π₯ and negative 10 divided by negative five equals positive two. We found our value for π₯. We take this value and we plug it back in for π₯ in either equation one or equation two.

If we plug in two for the π₯-value in equation one, our statement would say three times two minus two π¦ equals 10. Three times two equals six. Six minus two π¦ equals 10. Subtract six from both sides. And we will then have negative two π¦ equals four. If negative two π¦ equals four, we divide both sides by negative two. And weβll see that π¦ equals negative two. Four divided by negative two equals negative two. Now we know that π₯ equals two and π¦ equals two. But remember, weβre solving for the coordinate form. So we want to write it like this: two, negative two.

Now, letβs consider the substitution method. Again, weβll write down both equations. But this time we want to solve for one of our variables. With this method, you could use either equation and solve for either variable. But letβs take the second equation and solve for π¦ because it has a coefficient of negative one. And that will be pretty easy to work with. To do this, weβll subtract four π₯ from both sides of the equation. When we do that, we get negative π¦ equals 10 minus four π₯. We want a positive π¦. So weβll multiply through by negative one. And we will then have π¦ equals negative 10 plus four π₯.

Now that we have π¦ in terms of π₯, for equation two, we can take what we know that π¦ is equal to and substitute it in for π¦ in equation one. We take negative 10π₯ plus four and plug it in for the value of π¦. And our statement one now says three π₯ minus two times negative 10 plus four π₯ equals 10. Now, we need to distribute this negative two. Negative two times negative 10 equals positive 20. Negative two times four π₯ equals negative eight π₯. And itβs all equal to 10. We can combine our like terms: three π₯ minus eight π₯. And then, we subtract 20 from both sides. 10 minus 20 equals negative 10.

And now, weβre back to a statement that we had in the elimination method. Negative five π₯ equals negative 10. Therefore, we can say that π₯ equals two. We plug in two for π₯ and the equation we found for what π¦ equals. So we say π¦ equals negative 10 plus four times two. Four times two is eight. Negative 10 plus eight equals negative two. π¦ equals negative two. Both methods show the solution to be two, negative two for this system of equations.