<s>[INST] If $y=f(x)=\frac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$. Question: If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
We have $y=\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m} $
How can I find $m$? It is given than $m=3$.
 [/INST] If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
$$f(y) = \frac{3y -5}{2y-m} = x$$
$$2yx -mx = 3y-5$$
$$2yx - 3y +5 = mx$$
$$2\dfrac{3x-5}{2x-m}x-3\dfrac{3x-5}{2x-m} + 5 = mx$$
$$\dfrac{6x^2-10x-9x+15+10x -5m}{2x-m}=mx$$
$$ \dfrac{6x^2 -9x+15   -5m}{2x-m}=mx$$
$$ \dfrac{6x^2 -9x+15   -5m-2x^2m+ m^2x}{2x-m}= 0$$
$$  6x^2 -9x+15   -5m-2x^2m+ m^2x = 0$$
$$  3(2x^2 -3x+5)   =m(5+2x^2- mx)  $$
$$  3(2x^2 -3x+5)   =m(2x^2 -mx+5)  $$
You can easily see that
$$m =3$$
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