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MeganDarling
Pleasee help x.x 1. Solve the system by elimination- -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5 2.Solve using substitution- x-y-z=-8 -4x+4y+5z=7 2x+2z=4 Thanks for any help :o
Heya lol.. So any help?
Agh that doesn't help me at all x.x
O.o I most certainly cannot x.x
Megan, your homework makes me sleepy
haha oh Hi hero x.x
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
I'm willing to help you with the constraint and profit problems, but for the system of three equations, I'd rather use matrices.
I'm supposed to be learning how to do that too x.x
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.
-8x-8y=16 6x-9y=-108 With this problem I'm supposed to solve using matrices :P
Well, before you do anything, reduce both first.
Divide the first equation by 8 and the second by 3
oh oops the first one =-16 my bad there.
I'm just going to to respond to one of the previous questions you posted.
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 (-) -2x-y+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
Thank you so much Henry (: Your awesome