[tex]\dfrac{y+1}{y+2}+\dfrac{y+8}{y+9}=\dfrac{y+2}{y+3}+\dfrac{y+7}{y+8}[/tex]
Write all fractions in terms of a common denominator:
[tex]\dfrac{y+1}{y+2}=\dfrac{(y+1)(y+9)(y+3)(y+8)}{(y+2)(y+9)(y+3)(y+8)}[/tex]
[tex]\dfrac{y+8}{y+9}=\dfrac{(y+8)^2(y+2)(y+3)}{(y+2)(y+9)(y+3)(y+8)}[/tex]
[tex]\dfrac{y+2}{y+3}=\dfrac{(y+2)^2(y+9)(y+8)}{(y+2)(y+9)(y+3)(y+8)}[/tex]
[tex]\dfrac{y+7}{y+8}=\dfrac{(y+7)(y+2)(y+9)(y+3)}{(y+2)(y+9)(y+3)(y+8)}[/tex]
Then move all fractions to one side and simplify the numerator:
[tex]\dfrac{(y+1)(y+9)(y+3)(y+8)+(y+8)^2(y+2)(y+3)-(y+2)^2(y+9)(y+8)-(y+7)(y+2)(y+9)(y+3)}{(y+2)(y+9)(y+3)(y+8)}=0[/tex]
The numerator dictates when the fraction reduces to 0. The denominator can never be 0, so we know that y cannot take any of the values -2, -9, -3, nor -8.
So the equation reduces to
[tex](y+1)(y+9)(y+3)(y+8)+(y+8)^2(y+2)(y+3)-(y+2)^2(y+9)(y+8)-(y+7)(y+2)(y+9)(y+3)=0[/tex]
Expand the left side; you would end up with
[tex]-6(2y+11)=0[/tex]
[tex]2y+11=0[/tex]
[tex]2y=-11[/tex]
[tex]\implies\boxed{y=-\dfrac{11}2}[/tex]
