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3x - y - 2z = 11
x – 2y + 3z = 12
x + y - 2z = 5

2007-08-24 19:40:11 · 5 answers · asked by Splendidy 1 in Science & Mathematics Mathematics

5 answers

3x - y - 2z = 11
x + y - 2z = 5--------ADD

4x - 4z = 16
x - z = 4

x - 2y + 3z = 12
2x + 2y - 4z = 10----ADD

3x - z = 22

x - z = 4
- 3x + z = - 22----ADD
- 2x = - 18
x = 9
z = 5

x + y - 2z = 5
9 + y - 10 = 5
y = 6

x = 9 , y = 6 , z = 5

2007-08-24 19:52:07 · answer #1 · answered by Como 7 · 1 0

Simultaneous linear equations: If they are independent, you solve by substitution.

For example, the last equation becomes:

x = 2z - y- 5.

Plug this into the equation above it:

x - 2y + 3z = 12 becomes 2z - y - 5 - 2y + 3z = 12

Rearrange:

5z - 3y = 17

Now, this means that 5z = 3y + 17

So, you plug this and the previous equation into the third (as yet unused) equation and you are left with an expression that is in z only.

Once you have z, compute y

Once you have y, compute x

2007-08-25 02:50:24 · answer #2 · answered by Anonymous · 0 0

1st equation (finding x):
3x - y - 2z = 11
3x = 11 + y + 2z
x = (11 + y + 2z) / 3
x = 11/3 + 1/3y + 2/3z

2nd equation (finding y, substitute x):
11/3 + 1/3y + 2/3z - 2y + 3z = 12
- 1 2/3y + 3 2/3z = 12 - 11/3
- 1 2/3y = 25/3 - 3 2/3z
y = - 5 + 11/5z

1st equation (substitute y):
x = 11/3 + 1/3(- 5 + 11/5z) + 2/3z
x = 11/3 - 5/3 + 11/15z + 2/3z
x = 2 + 7/5z

3rd equation (finding z, substitute x and y):
(2 + 7/5z) + (- 5 + 11/5z) - 2z = 5
2 + 7/5z - 5 + 11/5z - 2z = 5
8/5z = 8
z = 5

1st equation (finding x, substitute z)
x =2 + 7/5(5)
x = 2 + 7
x = 9

2nd equation (substitute z):
y = - 5 + 11/5(5)
y = - 5 + 11
y = 6

Answer: x = 9; y = 6; z = 5

Proof:
1st equation:
3(9) - 6 - 2(5) = 11
27 - 6 - 10 = 11

2007-08-25 04:34:45 · answer #3 · answered by Jun Agruda 7 · 3 1

get rid of y in the first and third equation
3x - y - 2z = 11
x + y - 2z = 5
-------------------
4x - 4z = 16

get rid of y in the first and second equation
-2(3x - y - 2z = 11)
x – 2y + 3z = 12

-6x + 2y + 4z = -22
x – 2y + 3z = 12
---------------
-5x + 7z = -10


4x - 4z = 16
-5x + 7z = -10

5(4x - 4z = 16)
4(-5x + 7z = -10)

20x - 20z = 80
-20x + 28z = -40
-----------------------
8z = 40
z = 5

-5x + 7(5) = -10
-5x + 35 = -10
-5x = -45
x = 9

9 + y - 2(5) = 5
y = 6

2007-08-25 03:05:00 · answer #4 · answered by      7 · 0 1

Actually wat u want...................
The answer for the question or the method to solve.......
if answer,,,it is
x=9
y=6
z=5

2007-08-25 02:51:16 · answer #5 · answered by vanitha 1 · 0 0

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