Question

The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$. The roots of the latter equation are $4$ and $1$. Then the roots of the first equation are:

• A. $4$ and $1$
• B. only $1$
• C. only $4$
• D. neither $4$ nor $1$
• E. $4$ and some other root

Answer 1

The correct option is C only $4$

$\frac{{2x}^2}{x-1}+\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0\frac{{2x}^2+{2x}^2-2x+7x-7+4-6x+x-1}{3(x-1)}=0\frac{4}{3}\left[\frac{x^2-5x+4}{x-1}\right]=0\frac{4}{3}\left(\frac{(x-4)(x-1)}{(x-1)}\right)=0x-4=0x=4$