Answer
$x=-9$
Work Step by Step
Given,
$y_{1}=\frac{1}{x}$
$ y_{2}=\frac{1}{x^{2}-x}=\frac{1}{x(x-1)} $
$ y_{3}=\frac{1}{x-1}$
The equation is
$6y_{1}-3y_{2}=7y_{3}$
Substitute $y_{1}, y_{2}$ and $y_{3}$ values in the equation.
$6y_{1}-3y_{2}=7y_{3}$
$6(\frac{1}{x})-3(\frac{1}{x(x-1)})=7(\frac{1}{x-1})$
$\frac{6}{x}-\frac{3}{x(x-1)}=\frac{7}{x-1}$
$x=0$ and $x=1$ makes the denominator zero. So,$0$ and $1$ are restricted values.
$\frac{6}{x}-\frac{3}{x(x-1)}=\frac{7}{x-1}; x\ne 0, x\ne 1;$
Multiply both sides by $x(x-1)$
$x(x-1)(\frac{6}{x}-\frac{3}{x(x-1)})=x(x-1)(\frac{7}{x-1})$
$x(x-1)(\frac{6}{x})-x(x-1)(\frac{3}{x(x-1)})=x(x-1)(\frac{7}{x-1})$
$6(x-1)-3=7x$
$6x-6-3=7x$
$6x-9=7x$
$-9=7x-6x$
$x=-9$
$x=-9$ is not the restricted value. The solution is $x=-9$
