Systems of equations help?

Question

I dont understand how to do these and i have a sheet of them due on thursday
EX:
x+2y-z=-7
2x-2y-z=6
x+y-2z=-6
Thank you so much for your help.

Answer 1

x+2y-z= -7 ........................equation 1

2x-2y-z=6 ........................equation 2

x+y-2Z = -6 ......................equation 3
----------------------------------------------------------------------------
subtrating equation 3 from equation 1.

x+2y-z=-7
x+y-2z=-6
---------------
y +z = -1 ...........................equation 4


now multiply equation 3 with 2(number), the equation becomes

2x +2y-4z=-12 ....................equation 5

now subtracting equation 5 from equation 2

2x-2y-z=6
2x+2y-4z=-12
-------------------
-4y+3z=18 ..........................equation 6


now multiplying equation 4 with 4(number) and adding with equation 6

4y +4 z = -4
-4y+3z=18
------------------
7z=14

z=2 /////////////////////////////// Value of Z /////////////////////////////////

now put z=2 in equation 4

y+z=-1
y+2=-1

y= -3 //////////////////////// Value of Y /////////////////////////////

now put value of y and z in any equation to get value of x

x+2y-z=-7
x+2(-3)-2=-7

x= -7+2+6

x=1 ///////////////////////Value of X ////////////////////////////////////////////

Answer 2

One way would be to solve the first equation for "x", then substitute x into the second equation. Then solve the second equation for "z", then substitute z into the third equation. That will let you solve for y. Then substitute y back into the second equation ... and so on.

For example, from the first equation,
x=z-7-2y
so the second equation becomes
2(z-7-2y) -2y -z =6
giving
2z-14-4y-2y=2z-14-6y=6
and
z-7-3y=3
therefor
z=10+3y
and so on

Answer 3

Combine equations 1 and 2 to eliminate one variable.
(Add them together. Then 3x - 2z = -1)
Then combine equations 1 and 3 to eliminate the same variable.
(Multiply both sides of equation 3 by 2 and subtract from equation 1 to get -3x + 9z = 15)
Then you have 2 equations and 2 unknowns.
Combine them to find one unknown.
(Add them together and 7z=14, z=2)
Substitute this into one of the 2 equations (with 2 unknowns) to find the second unknown. (x=1)
Substitute these two into one of the first three equations to find the third.
x=1;y=-3;z=2
You can substitute these values into the original equations to show that it is the (one and only) solution. All other answers are incorrect. (Go back to Gallifrey doctor! Tom you are wrong too.)

Answer 4

Add 1st and 2nd together,
3x-2z= -1......(1)
multiply 3rd by 2 : 2x+2y-4z= -12......(2)
add 2nd and (2) together
4x-5z= -6......(3)
Now,4*(1) - 3*(3) ,
7z=14,
z=2;
Substitute z=2 into (1)
3x-2*2= -1
x=1;
Substitute x=1,z=2 into 3rd equation to find y= -3;
Therefore x=1 , y= -3 , z=2 .

Answer 5

equations:
x+2y-z=-7
2x-2y-z=6
x+y-2z=-6

solution:
use two equation to eliminate a variable
x+2y-z=-7
+2x-2y-z=6
=3x-2z=-1

use another equation to eliminat the same variable you eliminated from the first equation used
2x-2y-z=6
+2(x+y-2z)=2(-6)

2x-2y-z=6
+2x+2y-4z=-12
=4x-5z=-6

use the equations you obtained to find x and z
5(3x-2z)=5(-1)
+2(4x-5z)=2(-6)

15x-10z=-5
+ 8x-10z=-12
7x=7
x=1

substitute it to any equation you have
3x-2z=-1
3-2z=-1
-2z=-4
z=2

substitute x and z to any equation to solve for y
x=1
z=2
y=-3

Answer 6

Use Guassian elimination:

1 2 -1 -7
2 -2 -1 6
1 1 -2 -6

1 2 -1 -7
0 -6 1 20 second row - 2*(first row)
0 -1 -1 1 third row - first row

1 0 -3 -5 first row + 2*(third row)
0 0 -5 26 second row + 6*(third row)
0 1 1 -1 -1*(third row)

1 0 -3 -5
0 1 1 -1 exchange second and third row
0 0 -5 26


1 0 -3 -5
0 1 1 -1
0 0 1 -5.2 (third row)/(-5)

1 0 0 -20.6 (first row) + 3*(third row)
0 1 0 4.2 (second row)-(third row)
0 0 1 -5.2

So x = -20.6, y= 4.2 and z = -5.2

Answer 7

Can't you also use a Scientific calulator and do a 4*3 thingamagig? Anyone know what I'm talkin' bout?

Answer 8

answer= x=1, y=-1,z=2... try this x+2y-z-7,2x-2y-z=6,x+y-2z=-6/answers/.edu link -http://www.math.purdue.edu/~archava/mts2262.pdf#search='x2yz%3D7%2Fanswer%2F.edu'

Answer 9

there r two ways
1ellimination & substitution
2cross multiplication
in both cases answers will be same.