# Solve:(2/x-1)- (1/1-2x)=1Solve:(2/x-1)- (1/1-2x)=1 The 2/x-1 and the 1/1-2x are written as fractions.i need 2 possible solutions for x with steps on how to do it please that would be really helpful!

*print*Print*list*Cite

Solve:(2/x-1)- (1/1-2x)=1

`(2/(x-1)) - (1/(1-2x))=1`

Find your LCD (lowest common denominator).

In this case: (x-1)(1-2x)

Now multiply everything by your LCD-

the 1st term already has the (x-1) so only needs to be multiplied by the (1-2x)

the 2nd term already has the (1-2x) so only needs the (x-1)

the 1 needs both factors as it is effectively `1/1`

As this is an equation (as distinguished by the "equal" (=) sign) and not an algebraic fraction (when you keep the denominator) and you multiplied both side by all the factors, you can drop the denominator :

`(2/(x-1)) - (1/(1-2x) ) = 1`

becomes:

`2(1-2x) - 1(x-1) = 1(x-1)(1-2x)`

Simplify:

`2-4x -x +1` = `x-2x^2 -1 + 2x `

Note that the negative symbols changed to positive:(-1)(-1) and on the right side of the equation: (-1)(-2x)

`therefore 0 = -2x^2 +x+2x +4x +x -1 -2 -1`

`0 = - 2x^2 +8x - 4`

Multiply by (-1) to remove the negative symbol:

`0 = 2x^2 -8x + 4`

Reduce by the factor of 2:

`0 = x^2 -4x + 2`

We cannot factorize this without the formula:

`x= (-b +- sqrt(b^2-4ac))/(2a)`

a=1 b=-4 c=2 (ie the coefficients of `x^2` , -4x and 2)

`therefore x=(-(-4) +-(sqrt( (-4)^2 - 4 (1)(2))) )/ (2(1))`

`therefore x= (4 +sqrt(16-8))/(2)` or `x= (4-sqrt(16-8))/2`

`therefore x= 4/2 + sqrt8/2` or `x=4/2 - sqrt8/2`

Note that the denominator has been split between the two so that it can be simplified correctly.

`therefore x = 2+ sqrt(2.2.2)/2` or `x= 2 - sqrt(2.2.2)/2`

`therefore x= 2 + (2 sqrt2)/2` or `x= 2- (2sqrt2)/2`

Note the simplifed form of the square root (or surd form).

`therefore **x = 2 + `sqrt(2)` ` or `x= 2-`sqrt(2)` `**