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

I need help figuring out what to do with this problem, please help if you can :))-example-1
User Lorro
by
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1 Answer

10 votes

Answer:

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

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

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

User Assaf Karmon
by
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