Question

How many real solutions are there in the system of equations `x=y^2-5` and `x=2y`?

Answer 1

I found:
`x_(1)=2(1+sqrt(6))`
`x_(2)=2(1-sqrt(6))`
`y_(1)=1+sqrt(6)`
`y_(2)=1-sqrt(6)`

Explanation:

We can substitute the second into the first to get:
`2y=y^2-5`
`y^2-2y-5=0`
Using the Quadratic Formula we get:
`y_(1,2)=(2+-sqrt(4+20))/2=(2+-sqrt(24))/2=(2+-sqrt(6*4))/2=`
`=(2+-2sqrt(6))/2=1+-sqrt(6)`
these two `y` values will give you two `x` values using the second equation:
`x_(1,2)=2(1+-sqrt(6))`