Solve the system by the addition method.

Answer:
(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