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

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.