Two orienteering courses are a represented on suitable axes as 3x - 2y = - 17 and 3y - 4x =24 the courses have one checkpoint in the common. Determinate, algebraically the coordinates of this checkpoint

3x - 2y = -17 -> 9x - 6y = -51

3y - 4x = 24 -> 6y - 8x = 48

9x - 6y = -51

6y - 8x = 48

+

x = -3

3x - 2y = -17

3(-3) - 2y = -17

-9 - 2y = -17

-2y = -8

y = 4

Coordinate : (-3, 4)

step-by-step explanation:

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

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)