Question

I need help/ with the explanation/answer thanks :)

Answer 1

Both equations have $y$ as the left hand side.

Then, in a situation like

$y=a,\quad y=b$

we can deduce $a=b$, since they both equal $y$

So, we can set up the equation

$4x^2-6x+4=x+1 \iff 4x^2-7x+3 = 0$

The solutions are

$x=\dfrac{3}{4},\quad x=1$

From the second equation we know that y is one more than x, so we have

$x=\dfrac{3}{4} \implies y = \dfrac{7}{4},\quad x=1 \implies y=2$