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solve the system by elimination

-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

x = 1, y = 1, z = 0

Step-by-step explanation:

EQ 1: -2x + 2y + 3z = 0                  EQ 2: -2x  -   y +   z = -3

EQ 3: 2x + 3y + 3z = 5                  EQ 3:   2x + 3y + 3z = 5

                 5y + 6z = 5                                       2y + 4z = 2

Eliminate one of the variables for the new equations:

2(5y + 6z = 5) -->  10y + 12z = 10

-3(2y + 4z = 2) -->   -6y - 12z = -6

                                4y          = 4

                                          y = 1

Substitute y = 1 into one of the new equations to solve for z:

5(1) + 6z = 5

5   + 6z = 5

        6z = 0

          z = 0

Substitute y = 1 and z = 0 into one of the original equations to solve for x:

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

2x + 3 + 0 = 5

2x             = 2

             x = 1

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)