Solve the following systems. Whenever I have an equation with both decimals and fractions, my first thought might will be to get rid of all of the decimals and fractions. So I'm going to take this top equation and multiply it all the way through by two and I'm going to get 10X plus six y + C equaling 9. So those two equations are equivalents and I'm going to use 10X plus six y + Z equal 9 every time to work the problem. I'm going to have an equivalent for the second equation by multiplying it all the way 3 by 10. So five X -, 9 Y -2 Z equaling 4. And I'm going to take that third equation and multiply it also 3 by 10, so 30 X -20 four y + 4 Z equaling -4 So these are equivalents to the originals, but now they all have integers, so I'm going to use equation one and two. I'm going to eliminate the ZS. When I look here, I've got some that are positive, some that are negative, and they're all relatively small numbers that I can easily get by manipulating, multiplying through by something. So if I take one and two, I'm going to multiply equation one by two. So I'm going to get 20X plus 12Y +2 Z equal 18. I'm going to leave the second equation exactly the way it is. Five X -, 9 Y -2 Z equal 4. So I'm going to get a new equation when I combine those OF25X plus 3Y equaling 22. This is my new equation, let's call it four. Now I'm going to combine two and three, and I'm going to multiply equation 2 through by two. So I'm going to get ten X -, 18 Y -4 Z equaling 8 and I'm going to combine that with 30 X -20 four y + 4 Z equal -4. So when I combine this I get forty X -, 42 Y equaling 4. Now I can see that that entire equations divisible by A2. So an equivalent would be a twenty X -, 21 Y equaling 2. Just double checking, making sure we did all the math right, divided all three by two. So I'm going to call this my new equation 5. So now when I combine equations four and five, I can see that Y would be easily gotten rid of if I have a +3 Y and a -21 Y. If I multiply equation 4 through by A7, so 7 * 25 be 170, five X + 21 Y equal 154, and then I'm going to leave the second equation the way it was. Twenty X -, 21 Y is going to equal to, so we get 195 X equaling 156. If I then divide each side, I get 156 / 195. Let's see, does 156 / 195 reduce at all? What number comes out of both of those? Wow, 39 comes out of both of those. I used my calculator, I wasn't seeing that easily. 39 comes out of both 156 and 195, so that reduces to 4/5. Now if I have 4/5 for X, I'm going to substitute it into either equation four or five to get my Y. It doesn't matter which one I'm going to choose, four just because the coefficients were a little smaller on the Y. So 25 * 4/5 + 3 Y equal 22. Remember that 25 we could think of as over one, and five would then go into 25 five times. So 5 * 4 this first term would reduce to 3020. Sorry, five goes into 2555 times. 4 is 2020 + 3 Y equal 22. Three Y would equal 2Y would equal 2/3. Once I have my X and my Y, I'm going to substitute it in to one of the three equivalent equations looking for Z. And some things I might look for is in the XS. If anything has a coefficient that's evenly divisible by 5, and all three of them do. When I look at the YS, any of them that are evenly divisible by three, and all three of those also do. So then I'm going to look at the Z's and figure out which one's the easiest. And that's going to be equivalent equation one because the Z is by itself, it's a 1Z. So if I put this into equation one, I'm going to have 10 * 4/5 + 6 * 2/3 + Z equaling 9. This 10 is understood over 1 * 4 / 5 +6 is understood over one 2 / 3 + Z equal 95 goes into 10 twice 2 * 4 is 8/3 goes into six twice. 2 * 2 is 4, so 8 + 4 is 12. If we take the 12 to the other side, Z equal -3. So our ordered triplet, our X is 4/5, our Y is 2/3, and our Z is -3.