Solve the system by the addition method.

(2,1) (2,-1) (-2,1) (-2,-1)

Step-by-step explanation:

x^2 -3y^2 =1

3x^2 + 3y^2 = 15

4x^2 = 16

Divide by 4

4x^2/4 = 16/4

x^2 = 4

Take the square root

sqrt(x^2) = sqrt(4)

x = ±2

Let x= 2

x^2 - 3y^2 =1

4 - 3y^2 = 1

Subtract 4 from each side

-3y^2 = 1-4

-3y^2 = -3

Divide by -3

y^2 =1

Take the square root

y = ±1

(2,1) (2,-1) are the two solutions when x=2

Let x= -2

x^2 - 3y^2 =1

4 - 3y^2 = 1

Subtract 4 from each side

-3y^2 = 1-4

-3y^2 = -3

Divide by -3

y^2 =1

Take the square root

y = ±1

(-2,1) (-2,-1) are the two solutions when x=-2

idk lol

step-by-step explanation:

answer for #1 is c

answer for #2 is d

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