Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{y+1}{y-1} =\frac{y-5}{y-3} \\\\(y+1)\cdot(y-3)=(y-1)\cdot(y-5)\\\\y^{2} -3y+y-3=y^{2} -5y-y+5\\\\y^{2} -2y-3=y^{2} -6y+5\\\\y^{2} -2y-y^{2} +6y=5+3\\\\4y=8\\\\\boxed{y=2}\\\\\\2)\\\\\frac{x-8}{x+1} =-2\\\\x-8=-2\cdot(x+1)\\\\x-8=-2x-2\\\\x+2x=-2+8\\\\3x=6\\\\\boxed{x=2}[/tex]