For this case, what we must do is fill squares in all the expressions until we find the correct result.
We have then:

x2 + y2 − 4x + 12y − 20 = 0 x2 + y2 − 4x + 12y = 20
x2 − 4x + y2 + 12y = 20
x2 − 4x + (12/2)^2 + y2 + 12y + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
x2 − 4x + (6)^2 + y2 + 12y + (-2)^2 = 20 + (6)^2 + (-2)^2
x2 − 4x + 36 + y2 + 12y + 4 = 20 + 36 + 4
(x − 2)2 + (y + 6)2 = 60

3x2 + 3y2 + 12x + 18y − 15 = 0
x2 + y2 + 4x + 6y − 5 = 0
x2 + y2 + 4x + 6y = 5
x2 + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2
x2 + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2
x2 + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9
(x + 2)2 + (y + 3)2 = 18

2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 − 12x − 8y − 4 = 0
x2 + y2 − 12x − 8y = 4
x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
(x − 6)2 + (y − 4)2 = 56

x2 + y2 + 2x − 12y − 9 = 0
x2 + y2 + 2x - 12y = 9
x2 + 2x + y2 - 12y = 9
x2 + 2x + (2/2)^2 + y2 - 12y + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
x2 + 2x + (1)^2 + y2 - 12y + (-6)^2 = 9 + (1)^2 + (-6)^2
x2 + 2x + 1 + y2 - 12y + 36 = 9 + 1 + 36
(x + 1)2 + (y − 6)2 = 46

7 0

3 years ago

Solve:

Misha Larkins [42]

Answer:

convert 3 1/3 to an improper fraction

3*3 = 9

9+1 = 10

10/3

multiply the numerator by 4

(10/3) * 4 = 40/3

40/3 = 13 1/3