how do I solve for x,y and z?

Question

sorry again for posting so many math questions, but this one might even stump some of you.

x+2y+z=2
5x=z
y=-2x

Yeah seriously, though maybe it's not as hard as I'm making it out to be

Answer 1

x+2y+z=2 ....1
5x=z .....2
y=-2x ......3

from 2, you get the value of x in z
x = z/5 ..4

substitute 4 into 3 and you get the value of y in z
y = -2 (z/5)= -2z/5 ..5

substitute 4 & 5 into 1

z/5+2(-2z/5)+z=2
z/5-4z/5+z=2
-3z/5+5z/5=2
2z/5=2
2z=10
z=5

substitute the value of z to 2
5x=5
x=1

substitute this to 3
y=-2.1=-2

x=1
y=-2
z=5

Answer 2

You have x+2y+z=2, 5x=z and y=-2x.
Now, 5x=z => 5x - z = 0 and y=-2x => 2x +y = 0
You can write this problem in matrix form as
[ 1 2 1 ; 5 0 -1 ; 2 1 0] * [ x y z] = [ 2 0 0]'

Which implies
[ x y z] = inv([ 1 2 1 ; 5 0 -1 ; 2 1 0] )*[ 2 0 0]'

Which implies x = 1, y = -2, z = 5

[Here, I put ; to separate rows of matrix and ' to represent transform, I hope it helps you]
All the best

Answer 3

in 1st equation subtitute z by 5x and y by -2x ( from 2nd and 3rd equation) u'll get

x+2(-2x)+5x=2
x-4x+5x=2
6x-4x=2
2x=2
x=2/2=1
x=1 put the value of x in 3rd equation u get y=-2x=y=-2*1=-2
y=-2
put value of x in 2nd equation 5x=z 5*1=z
z=5

Answer 4

substisute equation 2 and 3 in the first one...and that would be

x+2(-2x)+(5x)=2
=> x-4x+5x=2
=> 2x=2
=> x=1

Substituting value of x =1 in 2nd and 3rd equation we get
z= 5
y= -2

(x,y,z)=(1,-2,5))

Answer 5

adding equations 1 and 2
6x+2y+z=z+2
cancellingout the z
6x+2y=2
4x+2y=0 (from eqn. 3)
subtracting2x=2
x=1sub in eqn.
y=-2
sub in eqn.2
z=5
so the soln.set{1,-2,5}

Answer 6

i subtituted z=5x and y=-2x into the first eqn.

then you can solve for x,
x+2(-2x)+5x=2
x-4x+5x=2
2x=2
x=2/2=1

then just substitute x into the 2nd eqn to get z,
5(1)=z
z=5

and 3rd eqns to get y,
y=-2(1)=-2

(x,y,z)=(1,-2,5)