Question

Find the value of the variable y, for which:

Answer 1

Answer:  y = -11

Step-by-step explanation:

$\text{Sum:}\\\\.\quad \dfrac{y+1}{y-5}+\dfrac{10}{y+5}\\\\\\=\bigg(\dfrac{y+5}{y+5}\bigg)\dfrac{y+1}{y-5}+\dfrac{10}{y+5}\bigg(\dfrac{y-5}{y-5}\bigg)\\\\\\=\dfrac{y^2+6y+5}{(y+5)(y-5)}+\dfrac{10y-50}{(y+5)(y-5)}\\\\\\=\dfrac{y^2+6y+5+10y-50}{(y+5)(y-5)}\\\\\\=\dfrac{y^2+16y-45}{(y+5)(y-5)}$

$\text{Product:}\\\\.\quad \dfrac{(y+1)(10)}{(y-5)(y+5)}\\\\\\=\dfrac{10y+10}{(y-5)(y+5)}\\\\\\\text{Sum = Product:}\\\\\dfrac{y^2+16y-45}{(y+5)(y-5)}=\dfrac{10y+10}{(y+5)(y-5)}\qquad Restriction: y\neq -5, 5\\\\\\\rightarrow y^2+16y-45=10y+10\\\\\rightarrow y^2+6y-55=0\\\\\rightarrow (y+11)(y-5)=0\\\\\rightarrow y+11=0\quad and\quad y-5=0\\\\\rightarrow y=-11\qquad and\qquad y=5\\\\\\\text{Since y = 5 is a restricted value, it is not a valid solution}$