When we look at the systems of equations to solve, I look for coefficients that have some that are positive, some that are negative, possible ones throughout coefficients that are already the same. So in this problem, I'm going to get rid of the YS because I've got a couple that are positive and well, one positive and a couple negative. And I can see that I have a three Y and a negative 33 Y already when I identify the equations as 1-2 and three. So I'm going to take one and combine it with two. When I take one and two and add them together, I'm going to get a new equation that's 5X plus 3Y3Z. The Y is canceled equaling 15. Now because one is positive and three is negative, I'm going to combine one and three, and when I combine one and three, if I'm trying to get rid of the YS, I'm going to take that top equation and multiply it all through by two. I have 8X plus six y + 2 Z equaling 2. I'm going to take equation 3 and multiply it through by three. So I'm going to get 33 X -6 Y plus 9Z. If I multiply by three, I get 189. When I add those together, 33 and eight is going to give me 41 X the YS canceled plus 11 Z equal 191. So we have a equation 4 from combining one and two and an equation 5 from combining one and three. So now I'm going to look at five X + 3 Z equaling 15 and 40. One X + 11 Z equaling 191 are equations 4:00 and 5:00. So when I look at 4:00 and 5:00, I'm going to see that I don't have A1 coefficient in either one. I don't have positive negatives in either one. So I'm going to just choose the variable with the smaller coefficients. This time if I have 3:00 and 11:00. So I'm going to take equation 4 and I'm going to multiply it through by 11. So 55X plus 33Z equal 165. And I'm going to grab my calculator and just check that 11 * 15 is 165. Now I'm going to take the bottom equation to multiply it all through by a three. So I get 120 three X + 33 Z equaling 191 times 3573. I need one of these to be positive and one to be negative to be able to add them. So I'm going to look and look. I'm going to look and think about when I add them. I want my X's to be positive. It's just a little easier to divide by a positive. So I'm going to take the smaller X and multiply that equation through by a negative all the way through. So when I add these 123 and -55 is going to give me 68 X. And over here when I add 573 and -165 is going to give me 408. If I divide 408 by 68 I get six. So when X is 6 I'm now going to be able to find ZI think I'll substitute it into four. So 5 * 6 + 3 Z equaling 15 30 + 3 Z equal 15 three Z equal -15 by subtracting 30. So Z equal -5. So now I have X and I have Z going back to one of the originals. I'm going to choose #2 just because it's got the smallest coefficients. And I have 6 - 3 Y plus two Z. So 2 * -5 equaling 14. So 6 - 3 Y -10 equal 14 six and -10 by combining like terms -4 - 3 Y equal 14. I'm going to add 4 so -3 Y equal 18. Divide by -3 and we get -6. So the ordered triplet is 6, negative 6 and -5.