ANSWER
The solution is
[tex](x=1,y=2),(x=2,y=3)[/tex]
EXPLANATION
We have
[tex]y=x+1---(1)[/tex]
and
[tex]y=x^2-1---(2)[/tex]
Let us substitute equation (1) in to equation (2). This gives us,
[tex]x+1=x^2-1(2)[/tex]
We rewrite this as a quadratic equation as the highest degree is 2.
[tex]x^2-x-1-1=0[/tex]
This implies that
[tex]x^2-x-2=0[/tex]
we factor to obtain,
[tex]x^2+x-2x-2=0[/tex]
[tex]x(x-1)-2(x-1)=0[/tex]
[tex](x-1)(x-2)=0[/tex]
This means,
[tex](x-1)=0\:\: or\:\:(x-2)=0[/tex]
[tex]x=1\:\: or\:\:x=2[/tex]
We substitute this values into any of the above equations, preferably equation (1)
When, [tex]x=1[/tex], [tex]y=1+1=2[/tex]
When, [tex]x=2[/tex], [tex]y=2+1=3[/tex]
The solution is
[tex](1,2),(2,3)[/tex]
