Both equations have
as the left hand side.
Then, in a situation like
![y=a,\quad y=b](https://tex.z-dn.net/?f=y%3Da%2C%5Cquad%20y%3Db)
we can deduce
, since they both equal ![y](https://tex.z-dn.net/?f=y)
So, we can set up the equation
![4x^2-6x+4=x+1 \iff 4x^2-7x+3 = 0](https://tex.z-dn.net/?f=%204x%5E2-6x%2B4%3Dx%2B1%20%5Ciff%204x%5E2-7x%2B3%20%3D%200)
The solutions are
![x=\dfrac{3}{4},\quad x=1](https://tex.z-dn.net/?f=%20x%3D%5Cdfrac%7B3%7D%7B4%7D%2C%5Cquad%20x%3D1%20)
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](https://tex.z-dn.net/?f=%20x%3D%5Cdfrac%7B3%7D%7B4%7D%20%5Cimplies%20y%20%3D%20%5Cdfrac%7B7%7D%7B4%7D%2C%5Cquad%20x%3D1%20%5Cimplies%20y%3D2%20)
It would be the first one :)
7440. The pattern is the answer multiplied by 2. So following that pattern ends you at 7440.
#1=D #2=d #3=B hope i helped and please mark me as brainiest