25y^2 - 1600x^2 = 40000

What are you doing GoldenLeaf?

Some of what you did made some sense.

First I'll simplify it by dividing everything by 25

$$25y^2 - 1600x^2 = 40000 \\

y^2-64x^2=1600$$

Now if x=0 then yes y=+or-40

If y=0 there are no solutions for x which means that the graph will not cross the x axis.

There is not going to be only 1 solution because there are 2 variables and only 1 equation.

you would need two eqations in order to have only one solution.

So yes there will be an infinite number of solutions .

Every point on this graph will be a solution.

I hope that helps!

$${\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{1\,600}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{40\,000}}$$

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.

thank you very much

Okay Anon, I got the second one for you, and that seems much more likely than the second one to be honest.

Hope I helped!