Question

Hello, can somebody help me out with this problem?
(x)/(x-2)+(x-1)/(x+1)=1

Answer 1

`\frac x{x-2}+(x-1)/(x+1)=1`


`\frac x{x-2}\cdot(x+1)/(x+1)+(x-1)/(x+1)\cdot(x-2)/(x-2)=1`

So long as
`x\\eq-1` and
`x\\eq2`, we can carry out the manipulation above. Then


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


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


`(x^2+x+x^2-3x+2)/(x^2-x-2)=1`


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


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


`2x^2-2x+2=x^2-x-2`


`x^2-x+4=0`

We can complete the square to solve:



`x^2-x+\frac14+\frac{15}4=0`


`\left(x-\frac12\right)^2=-\frac{15}4`

However,
`y^2\ge0` for all (real) values of
`y`, which means there is no real solution to this equation.

If you're solving over the complex numbers, we can take the square root of both sides to get


`x-\frac12=\pm i\frac{√(15)}2`


`\implies x=\frac{1\pm i√(15)}2`