Question 698717
To go from {{{ (1+x)/((1-x)(1-3x)) }}} to {{{ A/(1-x)+B/(1-3x)}}}
you need to find the numbers {{{A}}} and {{{B}}} that make
{{{ A/(1-x)+B/(1-3x)=(1+x)/((1-x)(1-3x))}}} true for all {{{x}}} values.
You could work from {{{ A/(1-x)+B/(1-3x)}}} to {{{((A+B)+(-3A-B)x)/((1-x)(1-3x))}}},
showing how good you are at working with polynomials,
and then write the equations that make the coefficients in the numerator the same:
{{{A+B=1}}} and {{{-3A-B=-3}}}
That would give you a system of linear equations that you could solve for {{{A}}} and {{{B}}}.
 
I prefer a less risky approach, safer from my tendency to mess up work with polynomials.
Since {{{ A/(1-x)+B/(1-3x)=(1+x)/((1-x)(1-3x))}}} true for all {{{x}}} values,
I can use two cunningly-chosen values of {{{x}}} to generate two linear equations.
For {{{x=0}}}, the equation is
{{{ A/(1)+B/(1)=1/((1)(1))}}} or {{{A+B=1}}}
For {{{x=-1}}}, I get
{{{ A/(1-(-1))+B/(1-3(-1))=(1+(-1))/((1-(-1))(1-3(-1)))}}}
I do not even write that, confident that I can do that many calculations in my head without messing up, and just write
{{{ A/2+B/4=0/((2)(4))}}} --> {{{A/2+B/4=0}}} --> {{{2A+B=0}}}
 
Next I solve {{{system(A+B=1,2A+B=0)}}} to get {{{highlight(A=-1)}}} and {{{highlight(B=2)}}}
 
For {{{A/(1-8x)+B/(1-10x)=(1-9x)/((1-8x)(1-10x))}}} {{{x=0}}} and {{{x=-1}}} work well too.
If they did not, I would use some numbers that made the calculations easier, and  did not make the denominators zero.
For {{{x=0}}}, the equation is
{{{A/(1)+B/(1)= 1/((1)(1))}}} --> {{{A+B=1}}}
For {{{x=-1}}}, the equation is
{{{A/(1+8)+B/(1+10)= (1+9)/((1+8)(1+10))}}} --> {{{A/9+B/11=10/(9*11)}}} --> {{{11A+9B=10}}}
Then I solve {{{system(A+B=1,11A+9B=10)}}} to get {{{highlight(A=1/2)}}} and {{{highlight(B=1/2)}}}