#color(brown)("In this case the objective is to make all the denominators the same")#
Note that #(+1)xx(-3)=-3# and that #-3+1=-2# both of which are part of #x^2-2x-3#
So factorising #x^2-2x-3# we get #(x-3)(x+1)#
Thus we can write the given equation as:
#4/(x-3)+2/(x+1)=(-8)/((x-3)(x+1)#
Multiply by 1 and you do not change the value. However, 1 comes in many forms.
#color(white)("dd")color(green)([4/(x-3)color(red)(xx1)]color(white)("dd")+color(white)("dd")[2/(x+1)color(red)(xx1)]color(white)("dd")=(-8)/((x-3)(x+1))#
#color(green)([4/(x-3)color(red)(xx(x+1)/(x+1))]+[2/(x+1)color(red)(xx(x-3)/(x-3))]=(-8)/((x-3)(x+1))#
#color(green)( (4x+4)/((x-3)(x+1)) color(white)("dd")+color(white)("d") (2x-6)/((x-3)(x+1))color(white)("d") = (-8)/((x-3)(x+1))#
As all the denominators are the same we can forget about them and just solve for the numerators. Or, if you are a purist; multiply all of both sides by #(x-3)(x+1)# giving:
#color(white)("dd")color(green)(4x+4color(white)("dddddd")+color(white)("dddd")2x-6color(white)("dddddd")=color(white)("d")-8#
#color(white)("d")#
#color(white)("ddddddddddd")color(green)(6x-2=-8)#
#color(white)("dddddddddddddd")color(green)(6x=-6)#
#color(white)("dxxxxxxxxxxxxxx.") color(green)( x=-1)#
