Solve the system of equations below algebraically.

5x^2 + y^2 - x + 20y - 48 = 0

x-2y+3=0

A) (0,0) and (1, 2)

B) (0, 2)

C) (1, - 3)

D) (-3,0) and (1, 2)

D) (-3, 0) and (1, 2).

Step-by-step explanation:

5x^2 + y^2 - x + 20y - 48 = 0

x - 2y + 3 = 0

From the second equation:

x = 2y - 3, so we substitute this in the first equation:

5 (2y - 3) ^2 + y^2 - (2y - 3) + 20y - 48 = 0

5 (4y^2 - 12y + 9) + y^2 - 2y + 3 + 20y - 48 = 0

20y^2 - 60y + 45 + y^2 - 2y + 3 + 20y - 48 = 0

21y^2 - 42y = 0

21y (y - 2) = 0

y = 0, 2.

So when y = 0 x = 2 (0) - 3 = - 3

and when y = 2, x = 2 (2) - 3 = 1.