Answer
$x=5$
Work Step by Step
Given,
$y_{1}=\frac{1}{x}, y_{2}=\frac{1}{2x} , y_{3}=\frac{1}{x-1}$
The equation is
$3y_{1}+4y_{2}=4y_{3}$
Substitute $y_{1}, y_{2}$ and $y_{3}$ values in the equation.
$3y_{1}+4y_{2}=4y_{3}$
$3(\frac{1}{x})+4(\frac{1}{2x})=4(\frac{1}{x-1}) $
$x=0$ and $x=1$ makes the denominator zero. So,$0$ and $1$ are restricted values.
$\frac{3}{x}+\frac{4}{2x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$
$\frac{3}{x}+\frac{2}{x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$
$\frac{5}{x}=\frac{4}{x-1} ; x\ne 0, x\ne 1;$
$5(x-1) = 4x$
$5x-5=4x$
$5x-4x=5$
$x=5$
$x=5$ is not the restricted value. The solution is $x=5$