Question

a/1-2a + b/1-2b + c/1-2c=1/2
Find 1/1-2a

Answer 1

Answer:

$\dfrac{1}{1-2a}=\dfrac{8bc-2c-2b-1}{2a-4ab-4ac+8abc}$

Step-by-step explanation:

Given:

$=\dfrac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}=\frac{1}{2}$

To find the value of the following term:

$\dfrac{1}{1-2a}$

Solution:

Subtract the left term to the right side term :

$=\dfrac{a}{1-2a}=\frac{1}{2}-\dfrac{b}{1-2b}-\dfrac{c}{1-2c}$

By taking the LCM of the right hand side term:

$=\dfrac{a}{1-2a}=\dfrac{(1-2b)(1-2c)-2(1-2c)(b)-2(1-2b)}{2(1-2b)(1-2c)}$

In the solving process cancel the term which are cancel out the right hand side ,We get:

$=\dfrac{a}{1-2a}=\dfrac{1-2b-2c+4bc-2b+4bc-2+4b}{2(1-2b-2c+4bc)}$

$=\dfrac{a}{1-2a}=\dfrac{8bc-2c-2b-1}{2-4b-4c+8bc}$

$=\dfrac{1}{1-2a}=\dfrac{8bc-2c-2b-1}{2a-4ab-4ac+8abc}$

This is the final result.

Answer 2

Answer:

$\frac{1}{1-2a }  =  \frac{1}{a} \times \frac{[10bc -4b-4c+1]}{2 \times (1-2b) \times (1-2c)}$

Step-by-step explanation:

In these question the expression is given as :  a/1-2a + b/1-2b + c/1-2c=1/2

We have to find the value of 1/1-2a

Let us start evaluating the expression:

$\frac{a}{1-2a} = \frac{1}{2}  - \frac{b}{1-2b} - \frac{c}{1-2c}$

$\frac{1}{1-2a} =  \frac{1}{a}  \times [ \frac{1}{2}  - \frac{b}{1-2b} - \frac{c}{1-2c}]$

$= \frac{1}{a} \times [\frac{(1-2b) \times (1-2c) - 2b \times (1-2c) - 2c(1-2b)}{2 \times (1-2b) \times (1-2c)} ]$

$\frac{1}{a}  \times [ \frac{1 - 2c - 2b + 4bc - 2b + 4bc - 2c + 2bc}{2 \times (1-2b) \times (1-2c)} ]$

$\frac{1}{1-2a }  =  \frac{1}{a} \times \frac{[10bc -4b-4c+1]}{2 \times (1-2b) \times (1-2c)}$