Question

Question #df22a

Answer 1

I got:
`x_1=3 and y_1=6`
`x_2=-1 and y_2=-2`

Explanation:

We can try substituting the first into the second equation for `y` to get:
`2x=x^2-3`
`x^3-2x-3=0`
solve using the Quadratic Formula:
`x_(1,2)=(2+-sqrt(4+12))/2=(2+-4)/2`
we get two solutions:
`x_1=6/2=3`
`x_2=-2/3=-1`

These two values, substituted into the first equation will give us:
`x_1=3` then `y_1=2*3=6`
`x_2=-1` then `y_2=2*(-1)=-2`

Graphically: