In Exercises 19–30, solve each system by the addition method. 4x + 3y = 15 2x - 5y = 1

Question

Exercise 25: Solve the system of equations 4x + 3y = 15 and 2x - 5y = 1 using the addition method.

Accepted Answer

Hey everyone welcome back in this problem we are asked to consider the given system of equations and solve using the addition method. Okay. The equations were given our nine x minus two equals 13 and minus three X plus Y equals minus five. And so that the system of two equations, let's get started just with labeling the equations we have equation +19 x minus two equals 13 and two is minus three X plus Y equals minus five K. Labeling the equations just makes it easier to follow through with your work and keep track of your equations. Alright so we're asked to use the addition method. Now when we're doing the addition method, what we want is we want to choose one of the variables so either X or y. We want them to have the same coefficient case or the same variable in the different equation you want to have the same coefficient but with the opposite sign. So that when you add the equations together that variable will cancel it will become zero. Okay so you can choose either X or Y. In this case let's go ahead and choose Y. Okay. And so in equation one we have minus two Y. So what we want is we want equation to to have the same coefficient but a different sign. So we want this to be plus two Y. Okay well how do we get that? We can multiply equation two by two and then add them together. So we're gonna take equation one and we're gonna add equation two multiplied by two. So we get nine X minus two. I equals 13 is our first equation. And now equation two multiplied by two minus three X times two minus six X plus y times two plus two Y. And minus five times two minus 10. Okay. Just make sure you multiply the entire equation. Not just the term that you want to change. Okay that keeps the equation consistent. Alright. And we add these together with our addition method nine X plus negative six X. Going to give us three X negative two Y plus two Y. That gives us zero and that's what we wanted now we've eliminated the variable. Why? Okay sometimes the addition method is called the elimination method. For this reason we've eliminated the variable Y. And on the right hand side 13 plus negative 10 is three. Okay so dividing by three we get X is equal to one. So that's great. We've done we've used our addition method and we found an X. Value of one. Now in order to find the solution we need a corresponding y value. Okay an X value alone is not a solution, we need a corresponding y value. So let's go up here and what we want to do is we want to substitute this X value into our equation to find that corresponding y value and because this is going to be a solution to both equations, we can choose either equation. Okay so let's go ahead and choose equation too. Um Because the y term just as a coefficient one, it'll be a little bit easier to simplify but you can choose either equation. So let's substitute our X value, X equals not minus one. X equals positive one into equation two. We're gonna get minus three times X. Which is one plus Y equals minus five is negative three plus Y equals negative five. Moving the three to the right hand side, we get y is equal to negative two. Okay so now we have our X value and our Y value and we can write our solution set as the point that we found one minus two. Okay. And so this right here is the solution to the system of equations. It is the set containing the .1 minus to answer a. That's it for this one. Thanks everyone for watching.

In Exercises 19–30, solve each system by the addition method. 4x + 3y = 15 2x - 5y = 1

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