Both equations have [tex]y[/tex] as the left hand side.
Then, in a situation like
[tex]y=a,\quad y=b[/tex]
we can deduce [tex]a=b[/tex], since they both equal [tex]y[/tex]
So, we can set up the equation
[tex] 4x^2-6x+4=x+1 \iff 4x^2-7x+3 = 0[/tex]
The solutions are
[tex] x=\dfrac{3}{4},\quad x=1 [/tex]
From the second equation we know that y is one more than x, so we have
[tex] x=\dfrac{3}{4} \implies y = \dfrac{7}{4},\quad x=1 \implies y=2 [/tex]
