how do you solve this?

Question

2x-4y+2z=16
-2x+5y+2z=-34
x-2y+2z=4

find x, y, and z
show work.

Answer 1

Hi ;
first Take Two of functions ;

Step 1;
2x- 4y + 2z =16
x - 2y + 2z = 4 { multiply the second function by ' -2 '} so;

2x- 4y + 2z =16
-2x + 4y - 4z = -8
―――――――
-2z = 8
z = 8/-2
z = - 4

Step 2;
2x-4y+2z=16
-2x+5y+2z=-34 { Just ignore 'x' in here } so ;
―――――――
y+ 4z = -18
y + 4(-4) = -18
y -16 = -18
y = -18 + 16
y = -2

Step 3;
we have the value of ' y ' & ' z ' so just fill them in to one of the functions and get ' x ' .

x- 2y + 2z = 4
y = -2 & z = -4
x - ( 2*-2) + (2*-4) = 4
x + 4 - 8= 4
x -4 = 4
x = 4+4
x = 8

The Result ;
x = +8
y = -2
z = -4

Now checking;
2x-4y+2z=16 ;
(2 *8) - (4*-2) + (2* -4) = 16 ; 16 +8 -8 6 ;16+0=16 ; 16 =16

-2x+5y+2z=-34
(-2*8)+(5*-2) +(2*-4) = -34; -16-10-8 = -34; -16-18=-34; -34=-34

x-2y+2z=4
8 -(2*-2) +(2*-4) = 4 ; 8 +4 -8 = 4 ; 4+0=4; 4=4

It is the correct answer.
Good Luck

Answer 2

Here are your original equations. I numbered them:

(#1) 2x-4y+2z=16
(#2) -2x+5y+2z=-34
(#3) x-2y+2z=4


Add equations (#1) and (#2) together:
(2x-4y+2z=16) + (-2x+5y+2z=-34)

= y+4z=-18 (this new thing is now (#4))

Multiply (#3) by 2, and add it to equation (#2):
(2x-4y+4z=8) + (-2x+5y+2z=-34)

= y+6z=-26 (this new thing is (#5))

Multiply (#5) by -1. Then add (#4) and (#5) together:
(y+4z=-18) + (-y-6z=26)

= -2z=8 --------> z = -4

Plug z into equation (#4):
y + 4(-4) = -18 --------> y = -2

Plug y and z into equation (#1):
2x - 4(-2) + 2(-4) = 16

2x + 8 - 8 = 16 ------> x = 8
------------------------------------
x = 8, y = -2 and z = -4

Answer 3

it has more than one solution. one solution is:

i- Choose one of the equations to start with (it's better to choose the one that has an unknown with samaller number multiplied in it-- in this case, equation 3, x-2y+2z=4, because it has x multiplied by 1).
ii- x=4+2y-2z
iii- substitute x in first equation : 2(4+2y-2z) -4y+2z=16
8+4y-4z-4y+2z=16
8-2z=16
-2z=8
so z= -4
iv- now substitude the value of z in second and third equation:
-2x+5y-8=-34
x-2y+8=4
it became two equations with two unknowns.
again choose the initialing equation (here again x-2y+8=4 (the second equation))
x=4-8+2y
x= -4+2y
v- now substitute x in eq. 1 :
-2(-4+2y)+5y=-26
8-4y+5y=-26
-y=34
y=-34
vi- now substitute the value of y in the second eq.(x-2y+8=4) to obtaion the value of x:
x-2(-34)+8=4
x+68+8=4
x+76=4
x=-72


So z= -4 , y=-34 , x=-72 is the answer of this 3-equations with 3 unknowns.

Answer 4

2x-4y+2z= 16; x-2y+z= 8 ---eq. 4
from eq 3 u have x-2y = 4-2z -- eq 5

sub eq 5 into 4 and u get 4-2z+z = 8 ; -z=4; z = -4

subsitute z=-4 in any two equation and solve for x any y...u want me to do it... okj

x-2y = 12 --eq 6
-2x+5y= -26 -- eq 7

multiply eq 6 with 2 and add it to eq7
y=-2 and thus x = 16

Answer 5

add equations #1 & #2
y+4z=-18 (#4)
add #2 & twice #3
y+6z=-26 (#5)
subtract #4 from #5
2z=-8
z=-4
substituting in # 4
y+4(-4)=-18
y-16=-18
y=-2
substituting in #1
2x-4(-2)+2(-4)=16
2x+8-8=16
2x=16
x=8

Checking in #2
-2(8)+5(-2)+2(-4)=-34
-16-10-8=-34
checking in #3
8-2(-2)+2(-4)=4
8+4-8=4

Answer 6

2x-4y+2z=16
-2x+5y+2z=-34
adding
y+4z=-18
2x-4y+2z=16
2x-4y+4z=8
subtracting
-2z=8
z=-4
substituting in
y+4z=-18
y-16=-18
y=-2
substituting in
2x-4y+2z=16
2x+8-8=16
2x=16
x=8
x=8,y=-2,z=-4

Answer 7

da way to solve da equation is to solve da 1st and 2nd then da 2nd and 3rd then solve both da results wat u get but u wont get da anwer coz ur sum is wrong