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.

i need a picure

step-by-step explanation:

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column