Question 1008226
 {{{10/(b^2 - 3b - 4) = 3b/(b -4) - 2/(b + 1 )}}}


 {{{10/(b^2 - 3b - 4) = 3b(b + 1 )/((b + 1 )(b -4)) - 2(b -4)/((b + 1 )(b -4))}}}


{{{10/(b^2+b - 4b - 4) = (3b(b + 1 )- 2(b -4))/((b + 1 )(b -4))}}}


{{{10/((b^2 +b)- (4b+ 4)) = (3b^2 + 3b - 2b +8)/((b + 1 )(b -4))}}}


{{{10/((b+1) (b-4)) = (3b^2 + b +8)/((b + 1 )(b -4))}}}


{{{10(cross((b + 1 ))cross((b -4)))/(cross((b+1))cross( (b-4))) = (3b^2 + b +8)}}}


{{{10 = 3b^2 + b +8}}}


{{{3b^2 + b +8-10=0}}}


{{{3b^2 + b -2=0}}}


{{{3b^2 + 3b-2b -2=0}}}


{{{(3b^2 + 3b)-(2b +2)=0


{{{3b(b + 1)-2(b +1)=0}}}


{{{(3b-2)(b +1)=0}}}


solutions:

if {{{3b-2=0}}} =>{{{3b=2}}} =>{{{b=2/3}}}

if {{{b +1=0}}} => {{{b= -1}}}-> since we have {{{b+1}}} in denominator at very beginning, we exclude this solution


so, the value of {{{b}}} that gives as real solution is  {{{b=2/3}}}