In Exercises 19–30, solve each system by the addition method.

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Question

In Exercises 19–30, solve each system by the addition method. 3x - 4y = 11 2x + 3y = - 4

Exercise 27: Solve the system of equations 3x - 4y = 11 and 2x + 3y = -4 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, the system we're given is seven X minus six Y equals four and two X plus Y is equal to minus 13. Okay so a system of two equations. Now let's go ahead and label the equations. So the first equation is seven X minus six. Y equals four. An equation two is five, X plus Y equals minus 13. And labeling is a good practice. It just helps to make it clearer when you're doing the rest of the problem, you can follow through your work more easily. Alright. So we were asked to use the addition method and when we do the addition method, what we want is we want to choose either the extra for the Y term. So either variable, okay we want the coefficient to be the same but with an opposite sign so that when we add the equations together, that variable will be eliminated. Okay so in this case we have minus six Y. And we have plus one Y. Okay so if we multiply equation two by six then we'll have minus six Y plus six. Y. That's gonna eliminate that Y term. Okay you could choose to eliminate X as well if you wanted in this case it's just simpler to eliminate y. So we're gonna take equation one and we're gonna add equation two, we're gonna add equation two times six. So we have seven X minus six, Y equals four and then we have five X times 6, 30 X Y times six plus six Y. And on the right hand side minus 13 times six minus 78. And when you do the multiplication, when you're multiplying this equation, make sure you multiply every term. You can't just multiply the term that you want to change. You need to multiply the entire equation so that it remains consistent. Alright, so now our addition method, we add these together seven X plus 30 X. Is going to give us 37 X minus six Y plus six Y. Well that gives us zero. Okay? And that's why we chose to multiply by six. This is what we wanted and then four minus 78 minus 74. Okay, so now we only have an X term in a constant term. We've eliminated the Y term. Okay. And sometimes the addition method is called the elimination method for that reason. And now all we're left to do is to divide by 37 and we get X is equal to -2. Alright, so we've used our elimination method to find an X value that satisfies both equations. Now we need a corresponding y value. Okay? And again this is satisfying both equations. So you can use either equation to find the corresponding y value, We're gonna choose equation to. Okay because the Y already has a coefficient of one, so it'll be a little bit simpler but either equation will work. So let's go up here and we're gonna sub the X value. We found X equals minus two into equation two. So we have five times X, which is minus two plus Y is equal to minus 13. Yes we have minus 10 plus Y is equal to minus 13. Moving the 10 to the right hand side, we get y is equal to negative three, so that is our corresponding Y value. So now we have an X value, we have a Y value and we can write our solution set as the point minus two minus three that we found. So this is a solution to the system that's going to correspond with answer D. I hope this video helped. Thanks everyone for watching.

In Exercises 19–30, solve each system by the addition method. 3x - 4y = 11 2x + 3y = - 4

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