Question

How do you find the exact solutions to the system `x^2/36-y^2/4=1` and `x=y`?

Answer 1

No solution.

Explanation:

To solve these systems of equations let us put `x=y` in `x^2/36-y^2/4=1`

and we get `x^2/36-x^2/4=1`

i.e. `x^2/36-(9x^2)/36=1`

or `(-8x^2)/36=1`

or `-2x^2=9` or `2x^2+9=0`

But for no real number we have `2x^2+9=0`

Hence, we do not have any solution.

Observe that while `x^2/36-y^2/4=1` represents hyperbola `x=y` is a line passing through origin with slope of `1` and the two do not intersect.
graph{(x-y)(x^2-9y^2-36)=0 [-20, 20, -10, 10]}

Answer 2

`x = y = +-(3sqrt(2))/2i`

Explanation:

This system only has Complex solutions...

Given:

`{ (x^2/36-y^2/4 = 1), (x=y) :}`

Substitute `y=x` in the first equation to get:

`x^2/36-x^2/4 = 1`

Multiply both sides by `36` to get:

`x^2-9x^2 = 36`

That is:

`-8x^2 = 36`

Divide both sides by `-8` to get:

`x^2 = -36/8 = -18/4 = ((3sqrt(2))/2i)^2`

Hence:

`x = y = +-(3sqrt(2))/2i`