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 $m=3$.
 A: 1-method. The hyperbola has assymptotes: $y=\frac32$ and $x=\frac m2$. The inverse function must be symmetric with respect to the line $y=x$. For the hyperbola to be self inverse, the assymptotes must intersect at the line $y=x$, consequently, $m=3$.
2-method. Find the inverse function and equate to the original:
$$y=\frac{3x-5}{2x-m}\Rightarrow x=\frac{mx-5}{2x-3}\Rightarrow \\
y^{-1}=\frac{mx-5}{2x-3}=\frac{3x-5}{2x-m}\Rightarrow m=3.$$
A: Now, $$\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m}=x$$ or
$$\frac{3(3x-5)-5(2x-m)}{2(3x-5)-m(2x-m)}=x$$ or
$$\frac{-x+5m-15}{x(6-2m)+m^2-10}=x$$ or $$-x+5m-15=x^2(6-2m)+x(m^2-10).$$
We need $$5m-15=0,$$ $$6-2m=0$$ and $$m^2-10=-1,$$ which gives $m=3.$
A: 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
$$m =3$$
