Asked by Samantha
Can someone show me how to solve the system by elimination with?
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
Answers
Answered by
Scott
add the 3rd equation to the 1st and then the 2nd to eliminate x
1 & 3 ... 5y + 6z = 5 ... (A)
2 & 3 ... 2y + 4z = 2 ... (B)
... multiply by 3/2 ... 3y + 6z = 3 ... (C)
subtract (C) from (A) to eliminate z
... 2y = 2
substitute the y-value back to find x and z
1 & 3 ... 5y + 6z = 5 ... (A)
2 & 3 ... 2y + 4z = 2 ... (B)
... multiply by 3/2 ... 3y + 6z = 3 ... (C)
subtract (C) from (A) to eliminate z
... 2y = 2
substitute the y-value back to find x and z
Answered by
Samantha
So it would go like
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.
Answered by
Steve
you have found y=1
You also have
5y+6z=5
so, 5+6z = 5, so z=0
Now you have any of the original equations, such as
2x+3y+3z=5
2x+3+0=5
x=1
You also have
5y+6z=5
so, 5+6z = 5, so z=0
Now you have any of the original equations, such as
2x+3y+3z=5
2x+3+0=5
x=1
Answered by
Samantha
So does that mean for “B” I was right? Or did you just go with mine?...and I’m still confused on how you found x, for everything else I understand now. Thank you
Answered by
Samantha
So it would go like
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.
-2x+2y+3z=0
2x+3y+3z=5
Equal
0x+5y+6z=5?
-2x-y+z=-3
2x+3y+3z=5
Equal
0x+3y+3z=5
But if I multiply 3/2 with what you got (aka) 2 & 3 ... 2y + 4z = 2 ... (B)
3/2*2=3
3/2*4=6
3/2*2=3
So I’d plug in the variables and get
3y+6z=3
And then subtract
5y + 6z = 5
From
3y + 6z = 3
And get
2y+0z=2
So basically if our variable has 0 it drops (aka) it’s dead? As in we don’t use it?
So I’d get 2y=2... but you said substitute y back to get x and z but how do I do that????
(So basically I have 2 problems with how I just solved this problem #1 I have no clue how you got “B” because I guess I solved it wrong??? And I have no clue on how to substitute y back to find x and z).... I know all I did is somewhat retyped what you said do, but I was showing you why i needed help instead of just being given the answers I needed someone to help explain and guide me step by step so I’d be able to look back and try doing it on my own.