We need to get additive inverses. So when we look at this top one, we want to figure out how to get 2R and 5R to be additive inverses. If I multiply the top equation by a five, that would give me ten r -, 25 S equaling -145. Because I need additive inverses, I'm going to multiply the second one through by a -2 so -10 R -4 S equal -58. That was a four. So when I add these, the 10 R -10 R are going to cancel and we get -29 S equaling negative 203. And if we divide -2 O 3 by -29, we're going to get negative oh, positive 7. And if I have +7 for the S. Now we need to go back and figure out how to get rid of the s s in order to solve for R. So the same concept. We're going to do the additive inverse. So I'm going to take this top equation and I'm going to multiply it by two. If I multiply the top equation 3 by two, we get 4S4 R -10 S equal -58. I'm going to take that second equation and multiply it by 5. So we get 20 five r -, + 10 S equal 145. So we get 29 R equaling 145 and -58 is going to give us 87. And then if we divide by 29, we're going to get R equaling 3, so our points going to be 3, 7 because we always list them in alphabetical order and R comes before us. The fact that we have at least one solution means it's consistent, and because it's exactly 1 solution means it's in dependent.