Find the solution to this system of equations.
x+y=1
2x-y+z=1
x+2y+z=8/3

x=
y=
z=

Step-by-step explanation:

x + y = 1x = 1 - y

2x - y + z = 1

x + 2y + z = 8/3I am changing this equation...multiplying everything by 3this gets rid of the fractionchanges to : 3x + 6y + 3z = 8

now sub 1 - y in for x in both of the other equations

2x - y + z = 1                                     3x + 6y + 3z = 8

2(1 - y) - y + z = 1                              3(1 - y) + 6y + 3z = 8

2 - 2y - y + z = 1                                3 - 3y + 6y + 3z = 8

2 - 3y + z = 1                                      3 + 3y + 3z = 8

-3y + z = 1 - 2                                     3y + 3z = 8 - 3

-3y + z = -1                                          3y + 3z = 5

now we have :

-3y + z = -1

3y + 3z = 5

add

4z = 4

z = 4/4

z = 1 <

now sub 1 in for z in either of the resulting equations we got

3y + 3z = 5

3y + 3(1) = 5

3y + 3 = 5

3y = 5 - 3

3y = 2

y = 2/3 <===

so we know z = 1 and y = 2/3so sub those in to either of the beginning equations to find x

x + y = 1

x + 2/3 = 1

x = 1 - 2/3

x = 3/3 - 2/3

x = 1/3 <===

we need to check this now...

2x - y + z = 1

2(1/3) - 2/3 + 1 = 1

2/3 - 2/3 + 1 = 1

1 = 1 (correct)

so it checks outx = 1/3 , y = 2/3, and z = 1

i do not know this sorry : (

answer: 414,377,008

step-by-step explanation:

just multiply it