What is the solution to x+y-z=1 x+2z=0 2x+2y=4

EQ. 1: x+y-z = 1

EQ 2: x +2z =0

EQ. 3: 2x+2y=4

Solve for X in EQ. 2: x +2z=0, subtract 2Z from each side:

x = -2Z

Replace X in EQ. 1 with solution of EQ 2 and solve for Y:

-2z + y -z =1

Simplify:

-3z +y = 1

Add 3z to each side::

y =3z +1

Replace X and Y in EQ.3 and solve for Z:

2(-2z) +2(3z+1) = 4

Distributive property:

-4z + 6z +2 = 4

Simplify:

2z +2 = 4

Subtract 2 from each side:

2z = 2

Divide both sides by 2:

z = 1

Now replace z with 1 and solve:

X = -2z = -2(1) = -2

Y = 3z +1 = 3(1) +1 = 3+1 = 4

Z = 1

Solution is (-2,4,1)

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