Question

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

Answer 1

`=> x=0" and "x=1`

Explanation:

Configuring the equation so that both sides end up with the same denominator.

Write `1" as " 1=x/x` giving

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

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

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

Consider just the left hand side

`2/(x-1)+x/x" " ->" "(2+x(x-1))/(x(x-1))`

Putting it all together

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

As the denominators are the same on both sides the equation is equally true if you just equate the numerators to each other and totally forget about the denominators.

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

`x(x-1)=0`

`=> x=0" and "x=1`

Answer 2

`x = -1, 2`

Explanation:

We have: `(2) / (x - 1) + 1 = (2) / (x^(2) - x)`

Let's simplify the left-hand side of the equation:

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

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

Then, let's cross-multiply:

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

We can expand the parentheses to get:

`=> x^(3) - 2 x^(2) + x = 2 x - 2`

Let's move all terms to the left-hand side to form a cubic equation:

`=> x^(3) - 2x^(2) - x + 2 = 0`

`=> (x^(3) - 2 x^(2)) + (- x + 2) = 0`

Now, let's factorise:

`=> x^(2) (x - 2) - (x - 2) = 0`

`=> (x - 2) (x^(2) - 1) = 0`

The difference of squares can be simplified to:

`=> (x - 2) (x + 1) (x - 1) = 0`

`=> x = -1, 1, 2`

However, using `x = 1` yields an undefined result.

Therefore, the solutions to the equation are `x = -1` and `x = 2`.