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

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{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1