Verified answer

[tex]\frac{5y+3}{2y+2}-\frac{7y+4}{3y+3}= \frac{5y+3}{2(y+1)}-\frac{7y+4}{3(y+1)}=\frac{3(5y+3)-2(7y+4)}{2*3(y+1)} =\frac{15y+9-14y-8}{6(y+1)}=\frac{y+1}{6(y+1)}=\frac{1}{6}[/tex]