DIFFICULT MATH QUESTION!! HELP NEEDED PLEASE!!?

2x = 1 - 3y

x = 1/2 - 3/2 y

x = 0.5 - 1.5y ......... Eq. 1

3y² - x² = 2 ............. Eq. 2

Sub 0.5 - 1.5y from Eq. 1 for x in Eq. 2:

3y² - (0.5 - 1.5y) = 2

3y² - 0.5 + 1.5y = 2

3y² + 1.5y = 2 + 0.5

3y² + 1.5y = 2.5

3(y² + 0.5y) = 2.5

y² + 0.5y = 2.5 / 3

y² + 0.5y = 0.833

y² + 0.5y + 0.625 = 0.833 + 0.0625

(y + 0.25)² = 0.896

y + 0.25 = √0.896

y + 0.25 = ± 0.947

y = - 0.25 ± 0.947

If y = - 0.25 + 0.947,

y = 0.697

and

x = 0.5 - 1.5(0.697)

x = 0.5 - 8.034

x = - 7.534

One Solution Set (- 7.534, 0.697)

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If y = - 0.25 - 0.947,

y = - 3.733

and

x = 0.5 - 1.5(- 3.733)

x = 0.5 - (- 5.6)

x = 0.5 + 5.6

x = 6.1

Other Solution (6.1, - 3.733)

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2x = 1 - 3y <=== lets call this equation A

3y^2 - x^2 = 2 <=== equation B

first solve for a variable (x or y) for one of the equations.

2x = 1 - 3y <=== divide all terms by 2

x = 1/2 - (3/2)y

Plug in x = 1/2 - (3/2)y into equation B

3y^2 - ((3/2)y)^2 = 2

3y^2 - (9/4)y^2 = 2

(12/4)y^2 - (9/4)y^2 = 2

(3/4)y^2 = 2 <==== multiply both sides by 4/3

y^2 = 2*(4/3)

y^2 = 8/3 <==== square root both sides

y = ±sqrt(8/3)

Plug the value of y into equation A

2x = 1 - 3(sqrt8/3) <=== divide all terms by 2

x = (1/2) - (3/2)(sqrt8/3)

2x = 1 - 3y

Square both sides of this equation

4x^2 = 1 - 6y + 9y^2

You now have x^2 in terms of y, substitute this into 3y^2 - x^2 = 2 to give a quadratic in y. Solve for y, then find x from the first equation.