Find a point in the xy plane that the first coordinate plus the second coordinate equals 3... HELP?

Question

Find a point in the xy plane that the first coordinate plus the second coordinate equals 3, and the square of the first coordinate minus the second coordinate equals -9.

Alright, basically I keep coming up with x=3, which would make y=0. Please help me, I'm really frustrated with this problem.

Answer 1

If the first coordinate plus the second coordinate equals 3, then we have the equation x + y = 3. (This is the line in the xy plane through the points (3,0) and (0,3)).

If the square of the first coordinate minus the second coordinate equals -9, then we have the equation x^2 - y = -9. (This is the equation of the parabola y = x^2 + 9.)
We want to find the intersection of the line and the parabola.

Rewrite the first equation as y = 3 - x and substitute it into the second equation; we now have x^2 - (3 - x) = -9, which can be written as x^2 + x - 3 = -9, or x^2 + x + 6 = 0. If we use the quadratic formula to solve this equation, we get
x = (-1 (+ or -) squareroot( 1^2 - (4)(6))) / 2 ,
or x = ( -1 + i squareroot(23)) /2 or x = ( -1 - i squareroot(23)) / 2

This means that the only solutions for x are imaginary; thus there are no points in the xy plane that satisfy the conditions you have given because the line and the parabola do not intersect.

Are you sure you have stated the problem correctly? Should you have said that the square of the first coordinate minus the square of the second coordinate equals -9?
Or maybe that the square of the first coordinate minus the second coordinate equals +9? If this is the case, the problem is simple because we can solve the equation x^2 + x -3 = 9. Rewrite it as x^2 + x - 12 = 0 and factor it like this:
(x + 4)(x - 3)= 0. Then x = -4 or x = 3.
If x = -4 then y = 3 - x = 3 - (-4) = 3 + 4 = 7.
Then -4 + 7 = 3 and (-4)^2 - 7 = 16 - 7 = 9
If x = 3 then y = 3 - x = 3 - 3 = 0.
Then 3 + 0 = 3 and 3^2 - 0 = 9 - 0 = 9

Recheck the problem. If you have stated it correctly, then there are no points in the xy plane that can solve it.

Answer 2

Sounds like you are correct, except you have your points backward I think. What would happen if you made the first coordinate 0, and the second one 3?