Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$ I absolutely know I am not doing this right. :[ 
Could I get some input or point back in the right direction?
My work done so far is shown below. 

Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the Expected value and Variance of random variable $Y=\frac{1}{a} X-b$, where $a$ and $b$ are constant values. 

My work. 
$$
\begin{align*}
E(Y) &= aE(x)-b = \sum_x \Big(\frac{1}{a} x - b \Big) p_x(x) = \frac{1}{a} \sum_x x p_x(x) - b 
\\ &=  \frac{1}{a} \sum_x x p_x(x) - b \sum_x p_x(x) = a E(x) - b \cdot 1.
\end{align*}
$$
If $a = 0$, then $E(x-b) = E(x)$ and if $b = 0$, then $E(ax)= \frac{1}{a}E(x)$.
$$
\begin{align*}
\mu &= E(X) = \int_{-\infty}^{\infty} x f(x) dx = \int_{0}^{1} x \Big(\frac{1}{a} X - b \Big) dx = \frac{1}{a} \int_0^1 X^2 - xb ~dx 
\\ & = \frac{1}{a} \Big( \frac{x^3}{3} - \frac{bx^2}{2} \Big) \text{ from } 1 \text{ to } 0 
\\ & = \frac{1}{a} \Big( \frac{1}{3} - \frac{b}{2} \Big) .
\end{align*}
$$
$$
\text{RV } Y = (X - E(x)^2).
$$
$$
\sigma^2 = \operatorname{Var}(x) = E[(x) - E[x])^2]
$$
$$
\operatorname{Var} \Big( \frac{1}{a} X - b \Big) = a^2 \operatorname{Var}(x).
$$
$$
\int_{\Box}^{\Box}\Big( \frac{1}{a} X - b \Big) - \Big( \frac{1}{a} X - b \Big)^2 \ldots
$$

Since expectation is linear, you have that $\mathsf{E}(Y)=\mathsf{E}(1/a\:X-b)=1/a\:\mathsf{E}(X)-b$.  Furthermore,
$$
\begin{align}
\mathsf{Var}(Y)
&=\mathsf{E}(Y^{\;2})-\mathsf{E}(Y)^2\\
&=\mathsf{E}(1/a^2X^2-2b/aX+b^2)-(1/a\:\mathsf{E}(X)-b)^2\\
&=1/a^2\mathsf{E}(X^2)-2b/a\mathsf{E}(X)+b^2-1/a^2\mathsf{E}(X)^2+2b/a\mathsf{E}(X)-b^2\\
&=1/a^2\mathsf{E}(X^2)-1/a^2\mathsf{E}(X)^2\\
&=1/a^2\mathsf{Var}(X).
\end{align}
$$
