Can someone me with this question? i seem to be getting a negative answer every time : ( - if (x, y) is a solution to the system of equations above and x> 0, what is the value of x?

Substitute 4 for y. Then -3x^2 + 16 = 52.

Solve for x. Subtract 16 from both sides, obtaining -3x^2 = 36.

Divide both sides by -3, obtaining x^2 = -12. This last result makes no sense, as no square of a real number could be negative. Probably this is where you're ":getting a negative answer."

If imaginary answers were allowed, then x = i*√12 = i*2√3 or x = -i*2√3.

x =

You get a negative value for x² because there are no real-number solutions to the set of equations. (The graphs do not intersect.)

Solve for x^2:

... x^2 = (y^2 -52)/3 = (16-52)/3) = -12

... x = 2√(-3) = ±i·2√3

Complex numbers cannot be compared to zero. (x > 0 has no meaning in this context.) It seems possible you want the complex solution with the positive imaginary part: x = i·2√3

idk

step-by-step explanation: