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)

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!