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Solving systems of linear equations — Basic example

Watch Sal work through a basic Solving systems of linear equations problem.

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Video transcript

- [Instructor] We're told the system of equations above has solution X, Y. What is the value of X? Pause this video and have a go at it before we work through it together. All right, now there's several ways that you could approach it, but the way I like to think about it right when I look at it is if in one of the equations I've already explicitly solved for a variable. So I have Y is equal to two X. Well, in the other equation I could substitute for that variable that has been solved for. So if Y is equal to two X, wherever I see Y in the other equation, I can replace it, I can substitute that with a two X. And then what does that give us? Well, this second equation here becomes three X plus, instead of Y I can write two X because I'm substituting, and then that is equal to 30. And then what's three X plus two X? Well, that of course is equal to five X is equal to 30. I can divide both sides by five, since that's the coefficient on the X term. And I get X is equal to six. And I am done. Now, another way we could approach this is through elimination. If I subtract Y from both sides of this top equation, I will get zero is equal to two X minus Y, or of course, I could write that as two X minus Y is equal to zero. And then I'll rewrite this second equation three X plus Y is equal to 30. And then notice, if I add the left-hand side to the left-hand side and the right-hand side to the right-hand side, the Ys are going to cancel out. I'll get two X plus three X is five X, negative Y plus Y, that's zero Y. They just cancel out. And then zero plus 30 is equal to 30. And so we get back to this step. You solve for X. You get X equals six again.