I need help figuring out what to do with this problem, please help if you can :))

x = 3; y = 3/2; z = -1

Step-by-step explanation:

The diameter of the circle is 41, so the radius is 41/2.

DA, DB, and DC are radii of the circle, so their lengths are equal and they are half of the diameter.

DA = DB = DC = 41/2

Set each expression equal to 41/2.

Then you have a system of 3 equations in 3 variables.

8x + 2z - 3/2 = 41/2

6x + y + 1 = 41/2

11y - 2z + 2 = 41/2

Multiply both sides of all equations by 2 to get rid of the fractions.

16x + 4z - 3 = 41

12x + 2y + 2 = 41

22y - 4z + 4 = 41

Move all constants to the right side.

16x + 4z = 44     (I'll name this equation Equation 1.)

12x + 2y = 39      (This is Equation 2.)

22y - 4z = 37

Line up the variables.

16x         + 4z = 44

12x + 2y         = 39

      22y - 4z = 37

We will use the elimination method.

Add the first and third equations to eliminate z.

Write the second equation below it.

16x + 22y = 81

12x + 2y = 39

Now you have a system of 2 equations in 2 unknowns.

We will now eliminate y. Multiply he second equation by -11 and write it below the first equation. Then add them.

         16x + 22y = 81

(+)   -132x - 22y = -429

        -116x         = -348

Divide both sides by -116.

x = 3

Plug in 3 for x in Eq. 1 and solve for z.

16x + 4z = 44    Eq. 1

16(3) + 4z = 44

48 + 4z = 44

4z = -4

z = -1

Plug in 3 for x in Eq. 2 and solve for y.

12x + 2y = 39      Eq. 2

12(3) + 2y = 39

36 + 2y = 39

2y = 3

y = 3/2

x = 3; y = 3/2; z = -1