Fractional equations: two exercises

Text

Solve the following fractional equations:

• (a) $1+\dfrac{2}{3-2x}=0$;
• (b) $\dfrac{2}{x^2-2x+1}-\dfrac{1}{1-x}=0$.

Solution

(a) Existence condition: $3-2x\neq0$, that is $x\neq\dfrac32$. Reducing to a common denominator: $1+\frac{2}{3-2x}=\frac{(3-2x)+2}{3-2x}=\frac{5-2x}{3-2x}=0.$ A fraction is zero when its numerator is zero (and the denominator is not): $5-2x=0 \implies x=\frac52.$ Since $\dfrac52\neq\dfrac32$, the solution is acceptable.

(b) Notice that $x^2-2x+1=(x-1)^2$ and $1-x=-(x-1)$. Existence condition: $(x-1)^2\neq0$, that is $x\neq1$. The equation becomes: $\frac{2}{(x-1)^2}-\frac{1}{-(x-1)}=\frac{2}{(x-1)^2}+\frac{1}{x-1}=0.$ Common denominator $(x-1)^2$: $\frac{2+(x-1)}{(x-1)^2}=\frac{x+1}{(x-1)^2}=0 \implies x+1=0 \implies x=-1.$ Since $-1\neq1$, the solution is acceptable. $\boxed{x_a=\frac52,\qquad x_b=-1}$