Под буквой б помогите

[latex] \frac{y+1}{y+2} + \frac{y+2}{y+1} = \frac{25}{12} \\ x^2-2xy+y^2=0 \\ \\ \frac{(y+1)^2+(y+2)^2}{(y+2)(y+1)} = \frac{25}{12} \\ (x-y)^2=0 \\ \\ \frac{y^2+2y+1+y^2+4y+4}{y^2+3y+2} = \frac{25}{12} \\ x=y \\ \\ 25(y^2+3y+2)=12(2y^2+6y+5) \\ y^2+3y-10=0 \\ D=49 \\ y_{1} =-5 \\ x_{1} =-5 \\ y_{2} =2 \\ x_{2} =2[/latex]