Impulse response of an LTI system given the input and output signals I have been given the input and output signals of an LTI system as: 

$x[n] = (\frac{1}{2})^nu[n] + 2^nu[-n-1]$

$y[n] = 6(\frac{1}{2})^nu[n] - 6(\frac{3}{4})^nu[n]$

I have found the system function $H(z)$ by using $H(z) = \frac{Y(z)}{X(z)}$ and using the standard transform tables to get: 

$X(z) = \frac{1}{1-\frac{1}{2} {z^{-1}}} - \frac{1}{1-2z^{-1}} = \frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}$

$Y(z) = 6*\frac{1}{1-\frac{1}{2} {z^{-1}}} - 6*\frac{1}{1-\frac{3}{4}z^{-1}} = \frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}$

$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}}{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}} = \frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$

Then this can be further simplified to: 

$H(z) = 1 - \frac{\frac{5}{4}z^{-1}}{1-\frac{3}{4}z^{-1}}$

However from here I am stuck on how to get the form of the second term back into the time domain as it does not match anything from the standard tables. 

I am just wondering if I have missed anything or made a mistake in my calculations, or if there is some way to get it into time domain form from here that I am not seeing, so that i can get the impule response $h[n]$.

Thank you for any help you can give me.
$H(z) = \frac{Y(z)}{X(z)} = \frac{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}}{\frac{-\frac{3}{2}z^{-1}}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}} = \frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$

$H(z)= \frac{1}{1-\frac{3}{4}z^{-1}}-\frac{2z^{-1}}{1-\frac{3}{4}z^{-1}}$

$h[n] = (\frac{3}{4})^nu[n] - 2*(\frac{3}{4})^{n-1}u[n-1]$

Since $z^{-1}$(the delay term) is multiplied to the z-transform of $(\frac{3}{4})^nu[n]$, the unit step function gets delayed by 1 i.e. $u[n-1]$