1. -x-y-z=-8
-4x+4y+5z=7
2x+2z=4
solve by substitution.

2. -2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
solve by elimination.

3.
constraints:
x≥0
y≥0
y≤1/3x+3
5≥x+y

objective function: c=6x-4y
given the system of constraints, name all vertices of the feasible region. then find the maximum value of the given objective function.

1) x =19

y = -13

z = 2

2) x= 1

y= 1

z = 0

Step-by-step explanation:

1)

-x-y-z=-81

-2x-y+z=-3 2

2x+3y+3z=53

From eqn 1,

x= 8 - y-z

Put x in eqn 2 and 3

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

-16+2y+2z-y+z = -3

y+ 3z = 13 4

2(8-y-z)+3y+3z=5

16-2y-2z+3y +3z = 5

y+z=-11 5

Using elimination method for eqn 4 and 5

2z = 24

z = 2

y = -11-2 = -13

Put x and y into eqn 1

x = 8- -13-2 =19

2)

-2x+2y+3z=01

-2x-y+z=-32

2x+3y+3z=53

Subtract eqn 2 from eqn 1

2y - -y +2z = -3

3y +2z = 34

Add eqn 2 and eqn 3

2y +4z = 25

Multiply eqn 4 by 2 and eqn 5 by 3

6y + 4z =6 6

6y + 12z = 67

Subtract eqn 7 from eqn 6

-8z = 0

z = 0

2y +4z = 2

2y + 4*0=2

2y = 2

y = 1

Putting y = 1 and z = 0 in eqn 1

-2x+2*1+3*0=0

-2x + 2+ 0 = 0

-2x = -2

x = 1

the answer for this question is 38

w'x is 0.

hope it