Solve the system of elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

x = 1 , y = 1 and z = 0

Step-by-step explanation:

-2x+2y+3z=0    (1)

-2x-y+z=-3     (2)

2x+3y+3z=5    (3)

To find the values of x,y and z

(3) - (2)⇒ 2y + 4z = 2

y + 2z = 1   (4)

(1) - (2)⇒ 3y +2z =3   ---(5)

(5) - (4)⇒ 2y = 2

y = 1

Substituting the value of y in (4) we get z =0

Substituting the value of y and z  in (1) we get x = 1

Therefore x = 1 , y = 1 and z = 0

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204