Question

How do you solve `(x+1)/(x-1)=2/(2x-1)+2/(x-1)`?

Answer 1

`x=3/2" "` the root

`x=1" " "` the extraneous root

Explanation:

From the give equation, we multiply both sides by the
LCD`=(2x-1)(x-1)`

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

`(2x-1)(x-1)*(x+1)/(x-1)=(2x-1)(x-1)[2/(2x-1)+2/(x-1)]`

We simplify to obtain

`(2x-1)cancel(x-1)*(x+1)/cancel(x-1)=(2x-1)(x-1)[2/(2x-1)+2/(x-1)]`

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

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

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

We expand by multiplication

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

Transpose all terms to the left side of the equation

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

`2x^2-5x+3=0`

We can solve this now by Factoring method

`(2x-3)(x-1)=0`

Equate each factor to 0 and solve for the values of x

First factor
`2x-3=0`

`x=3/2`

Second Factor
`x-1=0`
`x=1`

By checking these values in the original equation, we will find that `x=1` is an extraneous root because it will have a division by zero

The value `x=3/2` is a Root.

God bless....I hope the explanation is useful.

Answer 2

`x=3/2`

Explanation:

Before starting, it is important to note the restrictions in this equation. When the denominator of each fraction is set to not equal `0`, the restrictions are:

`x-1!=0color(white)(XXXXXXXX)2x-1!=0`

`x!=1color(white)(XXXXXXXXXX)2x!=1`

`color(white)(XXXXXXXXXXXXX)x!=1/2`

Thus, the restrictions are `color(red)(|bar(ul(color(white)(a/a)x!=1,1/2color(white)(a/a)|)))`.

Solving the Equation
`1`. Start by finding the L.C.D. (lowest common denominator) for the right side of the equation. Then rewrite the fractions on the right side with the inclusion of the L.C.D.

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

`(x+1)/(x-1)=(2color(darkorange)((x-1)))/((2x-1)color(darkorange)((x-1)))+(2color(blue)((2x-1)))/((x-1)color(blue)((2x-1)))`

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

`2`. Simplify the right side of the equation.

`(x+1)/(x-1)=(2x-2+4x-2)/((2x-1)(x-1))`

`(x+1)/(x-1)=(6x-4)/((2x-1)(x-1))`

`3`. Multiply both sides of the equation by `color(teal)((x-1))` and `color(teal)((2x-1))` to get rid of the denominator.

`(x+1)/color(red)cancelcolor(black)((x-1))xxcolor(teal)((2x-1)color(red)cancelcolor(teal)((x-1)))=(6x-4)/(color(red)cancelcolor(black)((2x-1)(x-1)))xxcolor(red)cancelcolor(teal)((2x-1)(x-1))`

`(x+1)(2x-1)=6x-4`

`4`. Expand the brackets on the left side of the equation.

`2x^2+x-1=6x-4`

`5`. Bring all the terms to the left side of the equation.

`color(violet)2x^2` `color(brown)(-5)x` `color(turquoise)(+3)=0`

`6`. Use the quadratic formula to solve for the values of `x`.

`color(violet)(a=2)color(white)(XXXXXX)color(brown)(b=-5)color(white)(XXXXXX)color(turquoise)(c=3)`

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

`x=(-(color(brown)(-5))+-sqrt((color(brown)(-5))^2-4(color(violet)2)(color(turquoise)3)))/(2(color(violet)2))`

`x=(5+-sqrt(1))/4`

`x=6/4color(white)(X),color(white)(X)4/4`

`x=3/2color(white)(X),color(white)(X)1`

However, looking back at the restrictions, ( `color(red)(x!=1,1/2)` ), `x=1` is not a valid solution. Therefore:

`color(green)(|bar(ul(color(white)(a/a)x=3/2color(white)(a/a)|)))`