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In Exercises 7 - 10, determine whether each ordered pair is a solution of the system of equations

$ \left\{\begin{array}{l}- \log x + 3 = y\\-\dfrac{1}{9}x + y = \dfrac{28}{9}\end{array}\right. $

(a) $ (9 , \dfrac{37}{9}) $

(b) $ (10 , 2) $

(c) $ (1 , 3) $

(d) $ (2 , 4) $

a) $\left(9, \frac{37}{9}\right)$ is not a solution

b) $(10,2)$ is a solution

c) $(1,3)$ is a solution

d) (2, 4) is not a solution

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Numerade Educator

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

So for this question, were given the system of equations on Cem ordered pairs or us to the side of thes ordered pairs or actually solutions the system so we can go ahead and look at the first order pair work given, um, it's nine and 37 nines and then plug it into our top equation to see if it gives us, um uh ah. See if it gives us thea answers. Ever looking for to have a negative log of nine plus three is maybe equal to 37 9th Um, so I know right off the bat, this isn't true, because the negative log of nine is some really mess. See, decimal and our answer on the right side is really neat fraction. So I'm going to go ahead and say that this isn't true. And if you were to plug it into your calculator, you would see that the left side does not equal the right side. Okay, so, um, with that, let's keep going. Go to our next order pair, which is, um, 10 and two. So if we plug in a 10 groups, you need to be black. If we plug in a 10 to get the negative wog of 10 plus three. My vehicle to, um, two s o. The negative log of 10 is just negative. One plus three is equal to two. And so that's true. But we can't say that this ordered pairs a solution, because we need to also make sure that it works for the bottom equation. If we plug it into the bottom equation, we'll get that 1/9. Um, Times 10 plus two is supposedly equal to 28 9th Um, so this will become negative. 19th plus two is supposedly equal to 28 9th And then if we move this to the other side, we'll get that, too, is equal to 38 9th which is definitely not true. So we can say that 10 to is not an order pair either. A solution. Moving onto the next order pair, we have, um, one and three. So the negative log of 10 plus three is equal to three. And that's obviously true. Once again, we need to also check that it equals that it makes sense for the other equation as well. So we have negative 1/9 times. Um, one plus three is he go to 28? Ninth said this would become negative one night plus three. Does he go to 28 9th and then you'll get that this is 30 is equal to 29 9th which is not true. So we can't say that one and three are solutions either. Ah, finally, our last order pair, which is two and four. So if I plug in a too with for our first equation with the negative log of two plus three is supposedly equal to four. Well, I also know this isn't true for the same reasons. As I knew nine and 37 9th were true because the negative log of two it's really complicated decimal and adding a whole number two. A complicated decimal won't make it a whole number. So I know this isn't a solution either. So we looked at each of these ordered pairs and we've concluded that actually, none of them are solutions to our system.