Systems of equations with more than two variables

[Mbandi]

Solve for x, y, and z

2x - y + 6z = 1
x - y + z = 2
x + y + z = 1

 

[galactus]

Try adding the 3rd row to the 1st and 2nd rows. That'll eliminate y and you'll have two equations with 2 variables.

 

[Mbandi]

See Systems of equations for more than two variables on free math help website. This question is given as an explanatory example. I have missed out on the second step of the process. Check...

2x - y + 6z = 1 (equation 1)
x - y + z = 2 (equation 2)
x + y + z = 1 (equation 3)

Steps:

1. Solve equation 1 for y
It becomes y = 2x + 6z - 1

2. Plug value of y or (2x + 6z - 1) into equations 2 and 3 above.

Equation 2 : x - y + z = 2

x -(2x+6z-1) + z = 2

x -2x -6z +1 =2 At this stage +z in the original equation 2 seems to have disappeared. My question is where has it gone????


Again on page 2 0f 5 on the same example for equation 3: x+y+z=1

Plugging 2x +6z-1 into it is shown as

x + 2x + 6z - 1 + z = 1

i.e 3x + 7x - 1 =1

and then 3x +7z = -1+1 ( HOW IS THIS -1+1 COMING UP ????)

Kindly help out.

 

[skeeter]

did you take galactus' advice?

2x - y + 6z = 1 (equation 1)
x + y + z = 1 (equation 3)
--------------------------------
3x + 7z = 2

x - y + z = 2 (equation 2)
x + y + z = 1 (equation 3)
------------------------------
2x + 2z = 3

now you have two equations with two unknowns ...

3x + 7z = 2
2x + 2z = 3

use the method of elimination to solve the two variable system ...
multiply the 1st equation by 2, multiply the 2nd equation by -3 ... then add to eliminate the x terms

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

6x + 14z = 4
-6x - 6z = -9
-----------------
8z = -5
z = -5/8

use one of the previous equations of the two variable system to solve for x ...

2x + 2(-5/8) = 3
2x - (5/4) = 3
2x = 17/4
x = 17/8

finally, use any one of the three original equations to solve for y ...

x + y + z = 1
(17/8) + y + (-5/8) = 1
(3/2) + y = 1
y = -1/2

 

[Mbandi]

Thanks to Galactus and skeeter. Great help indeed. Cheers