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MeganDarling
Group Title
Pleasee help x.x 1. Solve the system by elimination
2x+2y+3z=0
2xy+z=3
2x+3y+3z=5
2.Solve using substitution
xyz=8
4x+4y+5z=7
2x+2z=4
Thanks for any help :o
 2 years ago
 2 years ago
MeganDarling Group Title
Pleasee help x.x 1. Solve the system by elimination 2x+2y+3z=0 2xy+z=3 2x+3y+3z=5 2.Solve using substitution xyz=8 4x+4y+5z=7 2x+2z=4 Thanks for any help :o
 2 years ago
 2 years ago

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MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
Heya lol.. So any help?
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
Agh that doesn't help me at all x.x
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
I see that :o
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
O.o I most certainly cannot x.x
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
I'm dumb D:
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
Megan, your homework makes me sleepy
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
haha oh Hi hero x.x
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
I just need to be able to show all the correct steps with these 2 x.x they are worth a lot of points Dx
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
I'm willing to help you with the constraint and profit problems, but for the system of three equations, I'd rather use matrices.
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
I'm supposed to be learning how to do that too x.x
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
I'm not a teacher. I will post the full solutions to the problems, which will explain the steps that way. However, if you have any questions about anything, I can try to answer. What I refuse to do is "guide" student to an answer since, in a way, it wastes time.
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
8x8y=16 6x9y=108 With this problem I'm supposed to solve using matrices :P
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
haha I guess so
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
Well, before you do anything, reduce both first.
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
Divide the first equation by 8 and the second by 3
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
oh oops the first one =16 my bad there.
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
I'm just going to to respond to one of the previous questions you posted.
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
er okay :o
 2 years ago

Henry.Lister Group TitleBest ResponseYou've already chosen the best response.3
With Question 1. A system of elimination requires to to add or subtract two equations together to eliminate a variable. Since you have three equations, you will want to bring that down to two equations with two variables. If you minus the second equation from the first equation: 2x+2y+3z=0 () 2xy+z=3 (=) 3y+2z=3 Now we want to create a second equation that only has the two variables y and z. We also want to create a second equation that is different from the first one. To do this we need to involve the 3rd equation. In this case we will add the first and the last equations together. 2x+2y+3z=0 (+) 2x+3y+3z=5 (=) 5y+6z=5 Now we have two equations to solve together by adding or subtracting. But in this case we will have to alter equation 2 (multiply all of it by 3) so that we can subtract one from 2. 3y+2z=3 (x3) 9y+6z=9 Now we can subtract. 5y+6z=5 () 9y+6z=9 (=) 4y=4 y=1 Now that we have a value of y we can substitute that back into our equations to get other values. 3(1)+2z=3 3+2z=3 z=0 Now sub z=0 and y=1 back into equation 1. 2x+2(1)+3(0)=0 2x+2=0 x=1
 2 years ago

MeganDarling Group TitleBest ResponseYou've already chosen the best response.1
Thank you so much Henry (: Your awesome
 2 years ago
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