$${\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1\,600}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{40\,000}}$$
Solve for each individual variable set first. Since 'y' is first, I will do that.
25y^2 = 40000
Divide by 25 on both sides.
y^2 = 1600
y = 40
Now, solve for 'x'
-1600x^2 = 40000
Divide by -1600
x^2 = -25
x = +- 5i (positive or negative)
I am going to try something else, so hang on a second; I'm not necessarily sure if this is what you meant, but this is solving for each individual variable as itself.
NUMBER 2
So first divide the whole equation by 25.
y^2 - 64x^2 = 1600
Add 64x^2 to both sides.
y^2 = 64x^2 + 1600
Take the root.
y = 8x + 40
Now, plug in (8x + 40) for 'y'.
25(8x + 40)^2 - 1600x^2 = 40000
Even the whole thing out now. This means taking the '^2' of (8x + 40) and distributing it to said term.
25 * 64x^2 + 1600 - 1600x^2 = 40000
Add [-1600x^2] to both sides.
25 * (64x^2 + 1600) = 1600x^2 + 40000
Take the square root of the whole thing.
$${\sqrt{{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{64}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,600}} = {\mathtt{1\,600}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40\,000}}}}$$
This works out unbelieveably nicely.
5 * (8x + 40) = 40x + 200
Distribute the 5 to the (8x + 40)
40x + 200 = 40x + 200
Interestingly enough, the whole equations cancels itself out.
Infinite solutions.