Question

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

Answer 1

$\dfrac x{x-2}+\dfrac{x-1}{x+1}=1$

$\dfrac x{x-2}\cdot\dfrac{x+1}{x+1}+\dfrac{x-1}{x+1}\cdot\dfrac{x-2}{x-2}=1$

So long as $x\neq-1$ and $x\neq2$, we can carry out the manipulation above. Then

$\dfrac{x(x+1)}{(x-2)(x+1)}+\dfrac{(x-1)(x-2)}{(x+1)(x-2)}=1$

$\dfrac{x(x+1)+(x-1)(x-2)}{(x-2)(x+1)}=1$

$\dfrac{x^2+x+x^2-3x+2}{x^2-x-2}=1$

$\dfrac{2x^2-2x+2}{x^2-x-2}=1$

$\dfrac{2x^2-2x+2}{x^2-x-2}\cdot(x^2-x-2)=1\cdot(x^2-x-2)$

$2x^2-2x+2=x^2-x-2$

$x^2-x+4=0$

We can complete the square to solve:


$x^2-x+\dfrac14+\dfrac{15}4=0$

$\left(x-\dfrac12\right)^2=-\dfrac{15}4$

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

$x-\dfrac12=\pm i\dfrac{\sqrt{15}}2$

$\implies x=\dfrac{1\pm i\sqrt{15}}2$