There is NO question here!
We are given a fraction, $\frac{x}{y}$.
Apparently we are told that if we add 1 to both numerator and denominator we get $\frac{2}{3}$. That is, $\frac{x+1}{y+ 1}= \frac{2}{3}$. Multiplying both sides by 3(y+ 1) gives 3(x+ 1)= 2(y+ 1) so 3x+ 3= 2y+ 2. Subtract 2y+ 3 from both sides to get 3x- 2y= -1.
We are also told that if we subtract 1 from both numerator and denominator we get $\frac{1}{2}$. That is, $\frac{x- 1}{y- 1}= \frac{1}{2}$. Multiplying both sides by 2(y-1) gives 2(x- 1)= y- 1 so 2x- 2= y- 1. Subtract y and add 2 to both sides to get 2x- y= 1.
We now have the two equations 3x- 2y= -1 and 2x- y= 1 and, I presume, want to find values of x and y that satisfy both equations. From the second equation y= 2x- 1. Replace y in 3x- 2y= -1 by that to get an equation in x only and solve that equation for x. Then use y= 2x- 1 with that value of x to find y.