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Sal checks whether (-1,7) is a solution of the system: x+2y=13 and 3x-y=-11. Created by Sal Khan and Monterey Institute for Technology and Education.
Video transcript
Is negative 1 comma 7 a solution for the system of linear equations below? And they give us the first equation is x plus 2y is equal to 13. Second equation is 3x minus y is equal to negative 11. In order for negative 1 comma 7 to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and y is equal to 7, need to satisfy both of these equations in order for it to be a Solution. So let's try it out. Let's try it out with the first equation. So we have x plus 2y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative 1, and y is equal to 7, will x plus 2y equals 13? So we have negative 1 plus 2 times 7-- y should be 7-- this needs to be equal to 13. And I'll put a question mark there because we don't know whether it does. So this is the same thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least, satisfy this first equation. This point does sit on the graph of this first equation, or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue. We have 3 times negative 1 minus y, so minus 7, needs to be equal to negative 11. I'll put a question mark here because we don't know whether it's true or not. So let's see, we have 3 times negative 1 is negative 3. And then we have minus 7 needs to be equal to negative 11-- I put the question mark there. Negative 3 minus 7, that's negative 10. So we get negative 10 equaling negative 11. No, negative 10 does not equal a negative 11. So x equaling negative 1, and y equaling 7 does not satisfy the second equation. So it does not sit on its graph. So this over here is not a solution for the system. So the answer is no. It satisfies the first equation, but it doesn't satisfy the second. In order to be a solution for the system, it has to satisfy both equations.