Me i've been on this problem for 20 mins

Hello from MrBillDoesMath!

  x < 3

Discussion:

Let's solve  (1/3)x + 2x  > 3x - 2. Note that "x" occurs in three terms. Let's combine them by bringing all x's to one side of the equation. Subtract (1/3)x from the both sides:

(1/3)x - (1/3)x + 2x > 3x - 2  - (1/3)x  or

2x > 3x - (1/3)x -  2.

Subtract 2x from both sides

2x - 2x > 3x - 2x - (1/3)x - 2 or

0 >  x - (1/3)x - 2.

Add 2 to both sides:

0 + 2 >  x - (1/3)x -2 + 2 or

2 >  x - (1/3)x or

2 > x(2/3).

Multiply both sides by 3/2 to get:

2 * (3/2) > x (2/3) * (3/2) or

3 >  x (1)

Thank you,

MrB

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)