In algebraic fractions the first approach is to factor wherever possible.
#(4c)/(c-5)-1/(c+1) = (3c^2+3)/(c^2-4c-5)#
#(4c)/(c-5)-1/(c+1) = (color(blue)((c-5)(c+1)xx)3(c^2+1))/((c-5)(c+1))#
If an equation has fractions, you can get rid of the denominators by multiplying by the LCD, in this case #color(blue)((c-5)(c+1))#
Working on the LHS first:
#(color(blue)(cancel(c-5)(c+1)xx)4c)/cancel(c-5)-color(blue)((c-5)cancel(c+1)xx1)/cancel(c+1) " "rarr# this simplifies to:
#rarr 4c(c+1) - (c-5)#
Working on the RHS:
#(color(blue)(cancel(c-5)cancel(c+1)xx)3(c^2+1))/(cancel(c-5)cancel(c+1))" "rarr = 3(c^2+1)#
The equation is now:
#4c(c+1) - (c-5) =3(c^2+1)" "larr# no fractions
#4c^2+4c -c+5 = 3c^2+3#
#c^2+3c+2 = 0" "larr# factorize
#(c+2)(c+1) = 0#
#c = -2 " or " c= -1#
