Fractional equation with integer roots

Text

Solve the following fractional equation: $\frac{2-x}{1-2x}-1 = \frac{x-3}{2-x}.$

Solution

Existence conditions: $x\neq \tfrac12$ and $x\neq 2$. Simplify the left-hand side: $\frac{2-x}{1-2x}-1 = \frac{(2-x)-(1-2x)}{1-2x} = \frac{x+1}{1-2x}.$ The equation is $\dfrac{x+1}{1-2x} = \dfrac{x-3}{2-x}$; cross-multiplying: $(x+1)(2-x) = (x-3)(1-2x) \;\Rightarrow\; -x^2+x+2 = -2x^2+7x-3.$ Rearranging: $x^2 - 6x + 5 = 0 \Rightarrow (x-1)(x-5)=0$, i.e. $x=1$ or $x=5$. Both acceptable. $\boxed{x=1 \ \vee\ x=5}$