Question 834520
Let's simplify numerator and denominator first,
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=((3x)/((x+4)(x-3))-6/(x+4))}}}
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=(1/(x+4))*((3x)/(x-3)-(6(x-3))/(x-3))}}}
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=(1/((x+4)(x-3)))*(3x-6(x-3))}}}
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=(1/((x+4)(x-3)))*(3x-6x+18))}}}
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=(1/((x+4)(x-3)))*(-3x+18))}}}
{{{ ((3x)/(x^2+x-12)  -  (6)/(x+4))=(-3(x-6))/((x+4)(x-3)))}}}
Now the denominator,
 {{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (x+1)/(x-3)  +  (2x)/((x-3)(x-2)))}}}
 {{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/(x-3))*((x+1)  +  (2x)/((x-2)))}}}
 {{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/(x-3))*(((x+1)(x-2))/(x-2)  +  (2x)/((x-2)))}}}
 {{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/((x-3)(x-2))) *((x+1)(x-2)+2x)))}}}
{{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/((x-3)(x-2))) *(x^2-2x+x-2+2x))}}}
{{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/((x-3)(x-2))) *(x^2+x-2))}}}
{{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= (1/((x-3)(x-2))) *((x+2)(x-1)))}}}
{{{ (x+1)/(x-3)  +  (2x)/(x^2-5x+6)= ((x+2)(x-1))/((x-3)(x-2)))}}}
Dividing by a fraction is the same as multiplying by its reciprocal,
{{{ ((-3(x-6))/((x+4)(x-3))) /(((x+2)(x-1))/((x-3)(x-2)))=((-3(x-6))/((x+4)(x-3)))*(((x-3)(x-2))/((x+2)(x-1)))}}}
{{{ ((-3(x-6))/((x+4)(x-3))) /(((x+2)(x-1))/((x-3)(x-2)))=((-3(x-6))/((x+4)))*(((x-2))/((x+2)(x-1)))}}}