Question

How do you solve `x^2 + y^2 = 4` and `y^2 = 3x`?

Answer 1

`(x,y)=(1,-sqrt(3)),(1,sqrt(3)),(4,isqrt(12)),(4,-isqrt(12))`

Explanation:

Substitute the second equation into the first to obtain a quadratic equation for `x`:

`x^2+y^2=x^2+3x=4` => `x^2+3x-4=(x+4)(x-1)=0`

This has solutions `x=-4,1`, substituting this into the second equation we have `y=+-sqrt(3),+-isqrt(12)`.

Therefore we have:

`(x,y)=(1,-sqrt(3)),(1,sqrt(3)),(4,isqrt(12)),(4,-isqrt(12))`

Answer 2

Substitute the second equation into the first to get a quadratic in `x`, the positive root of which gives two possible Real values for `y` in the second equation.

`(x, y) = (1, +-sqrt(3))`

Explanation:

Substitute `y^2=3x` into the first equation to get:

`x^2+3x = 4`

Subtract `4` from both sides to get:

`0 = x^2+3x-4 = (x+4)(x-1)`

So `x = 1` or `x = -4`.

If `x = -4` then the second equation becomes `y^2 = -12`, which has no Real valued solutions.

If `x = 1` then the second equation becomes `y^2 = 3`, so `y = +-sqrt(3)`