Using matices to do this
[ 1 -2 1] * [x] = [-5]
[ 2 -1 1] [y] [ 1]
[2 1 2] [z] [10]
AX = B
A*A^(-1)X = A^(-1)B
X = A^(-1)B
SO we need the inverse of A this can be worked out using a long and tedious method or just use a calculator
A^-1 = [ -.6 1 -.2]
[ -.4 0 2]
[.8 -1 .6]
A^-1B = [2}
[4}
[1]
so x =2 y=4 z=1
now you do the other one
Check
2 - (4*2) +1 = 2 - 8 +1 = -5
2006-11-11 03:02:32
·
answer #1
·
answered by Anonymous
·
0⤊
0⤋
Eliminate one unknown from each equation at a time.
x - 2y +z = 5 Multiply both sides of the equation by 2
2x - 4y + 2z = 10 Now take one of the other equations
2x + y + 2z = 10 Subtract leaving -5y = 0 i.e. y = 0
Use another equation 2x - y + z = 1
2x - 4y +2z = 10 Subtract
3y - z = -9, Substitute 0 for y and z = 9 Substitute these into
x - 2y + z = 5, i.e. x - 0 + 9 = 5 hence x = -4
Proof:- 2x - y + z =1 i.e. -8 - 0 + 9 = 1
Do the next one yourself :- x = y = z = 1.5
2006-11-11 03:22:21
·
answer #2
·
answered by Anonymous
·
0⤊
0⤋
Mulitply the first by -2 and add to the second:
-2x + 4y - 2z = 10
2x -y + z = 1
------------------------
3y - z = 11
z = 3y - 11
Do the same thing with 1 & 3:
-2x + 4y - 2z = 10
2x + y + 2z = 10
5y = 20
y = 4
Substitute this back in the previous equation:
z = 12 - 11 = 1
Then you can get x:
x - 8 + 1 = -5
x - 7 = -5
x = 2
Now, do the same thing for the second set. :-)
2006-11-11 03:00:20
·
answer #3
·
answered by Dave 6
·
0⤊
0⤋
For substitution, you resolve for on variable in terms of the different one, and plug that into the 2nd equation. for occasion: a million) 2x-5y=6 => 5y=2x-6 => y=(2x-6)/5, then plug in this "y" to the 2nd equation in this undertaking.. 3x + 4((2x-6)/5) = 32... 3x + (8x-24)/5 = 32 15x/5 + (8x-24)/5 = 32 15x + 8x - 24 = 32(5) = a hundred and sixty 23x = 184, or x = 8.. Then plug that x into the two equation to unravel for "y" 2(8) - 5y = 6 sixteen - 5y = 6 5y = sixteen - 6 = 10, so y = 2 comparable for #2
2016-12-28 18:47:07
·
answer #4
·
answered by para 3
·
0⤊
0⤋
1) x= -4 y=0 z= 9
Sorry, that's all I know. I'm only in 6th grade after all.
2006-11-11 18:11:05
·
answer #5
·
answered by ♫tweet75♫ 3
·
0⤊
0⤋
yep. a 3x3 Matrix.
2006-11-11 02:55:28
·
answer #6
·
answered by Halo 5
·
0⤊
0⤋
set up a matrix and solve it with a calculator.
2006-11-11 02:54:52
·
answer #7
·
answered by travis R 4
·
0⤊
0⤋
Get x in terms of y&z. Get y in terms of z. Get z
Substitute and get the value y and x
2006-11-11 03:25:02
·
answer #8
·
answered by openpsychy 6
·
0⤊
0⤋