Solve the system by elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z+5 solve the system by substitution. -x-y-z=-8 -4x+4y+5z=7 2x+2z=4

Step-by-step explanation:

Given system is-

-2x+2y+3z=0  (1)

-2x-y+z=-3   (2)

2x+3y+3z=5   (3)

From (1)-(2) we get

3y+2z=3 (4)

Again from (2)+(3),

2y+4z=2 (5)

Now multiplying equation (4) by -2

-6y-4z=-6 (6)

Adding  (5) and (6)

-4y=-4

dividing by  -4

y=-4 / -4

 =1

So,  y=1 ,

using  y=1  in  (5)

2(1)+4z=2

or, 2+4z=2

or, 4z=2-2

or, 4z=0

or, z=0

using  y=1 and z=0   in (3)

2x+3(1)+3(0)=5

or, 2x+3+0=5

or, 2x=5-3

or, 2x=2

or, x=2/2

     =1

So, the solution is-

(x,y,z)=(1 , 1 , 0)