SOLUTION: Hi, I need help with a simplification that I need to do every time in a certain type of problem I basically need to go from equations in the form {{{ (1+x)/((1-x)(1-3x)) }}} to

Algebra ->  Functions -> SOLUTION: Hi, I need help with a simplification that I need to do every time in a certain type of problem I basically need to go from equations in the form {{{ (1+x)/((1-x)(1-3x)) }}} to       Log On


   



Question 698717: Hi, I need help with a simplification that I need to do every time in a certain type of problem
I basically need to go from equations in the form +%281%2Bx%29%2F%28%281-x%29%281-3x%29%29+ to equation in the form +%28-1%2F%281-x%29%29%2B%282%2F%281-3x%29%29+
I need a detailed way to go from the first to the second
Another one would be

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
To go from +%281%2Bx%29%2F%28%281-x%29%281-3x%29%29+ to +A%2F%281-x%29%2BB%2F%281-3x%29
you need to find the numbers A and B that make
+A%2F%281-x%29%2BB%2F%281-3x%29=%281%2Bx%29%2F%28%281-x%29%281-3x%29%29 true for all x values.
You could work from +A%2F%281-x%29%2BB%2F%281-3x%29 to %28%28A%2BB%29%2B%28-3A-B%29x%29%2F%28%281-x%29%281-3x%29%29,
showing how good you are at working with polynomials,
and then write the equations that make the coefficients in the numerator the same:
A%2BB=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%2F%281-x%29%2BB%2F%281-3x%29=%281%2Bx%29%2F%28%281-x%29%281-3x%29%29 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%2F%281%29%2BB%2F%281%29=1%2F%28%281%29%281%29%29 or A%2BB=1
For x=-1, I get

I do not even write that, confident that I can do that many calculations in my head without messing up, and just write
+A%2F2%2BB%2F4=0%2F%28%282%29%284%29%29 --> A%2F2%2BB%2F4=0 --> 2A%2BB=0

Next I solve system%28A%2BB=1%2C2A%2BB=0%29 to get highlight%28A=-1%29 and highlight%28B=2%29

For A%2F%281-8x%29%2BB%2F%281-10x%29=%281-9x%29%2F%28%281-8x%29%281-10x%29%29 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%2F%281%29%2BB%2F%281%29=+1%2F%28%281%29%281%29%29 --> A%2BB=1
For x=-1, the equation is
A%2F%281%2B8%29%2BB%2F%281%2B10%29=+%281%2B9%29%2F%28%281%2B8%29%281%2B10%29%29 --> A%2F9%2BB%2F11=10%2F%289%2A11%29 --> 11A%2B9B=10
Then I solve system%28A%2BB=1%2C11A%2B9B=10%29 to get highlight%28A=1%2F2%29 and highlight%28B=1%2F2%29