Answer to Question #90760 in Algebra for Sayem

Question

x÷{(1-ax) (1-bx) } = {1÷(a-b)}÷(1-ax) +{1÷(b-a)}÷(1-bx)
How pls explain??

Answer 1

Let $\frac{x}{(1-ax)(1-bx)}=\frac{A}{1-ax}+\frac{B}{1-bx}$.

We have: $\frac{x}{(1-ax)(1-bx)}=\frac{A(1-bx)+B(1-ax)}{(1-ax)(1-bx)}$ .

So, $x=-(Ab+Ba)x+A+B$ or $Ab+Ba=-1, A+B=0.$

Solving this system for A and B we get: $A=\frac{1}{a-b}, \;\;B=\frac{1}{b-a}$.

Therefore, $\frac{x}{(1-ax)(1-bx)}=\frac{1}{(a-b)(1-ax)}+\frac{1}{(b-a)(1-bx)}$ .