How is the linear combination of Normal Distribution also normal? X ~ N(μ, σ2)
z-square = (X-μ)/σ ~ N (0,1)

If Y = (X - α) / β
X = α + βY
Y ~ N((μ - α) / β, (1 / β^2) * σ2)


How can we derive from above two that linear combination of normal distribution is also normal ?
The $\text{pdf}$ of the sum $s$ of two r.v. $x,y$ can be obtained as the convolution product of the originating distributions, which is an integral, and
$$\text{pdf}(s)\propto\int_{-\infty}^\infty\exp\left(-\frac{(s-t-\mu_x)^2}{2\sigma_x^2}\right)\exp\left(-\frac{(t-\mu_y)^2}{2\sigma_y^2}\right)dt
\\=\int_{-\infty}^\infty\exp\left(-\frac{(s-t-\mu_x)^2}{2\sigma_x^2}-\frac{(t-\mu_y)^2}{2\sigma_y^2}\right)dt.$$
As the exponent is a quadratic polynomial in $t,s$, the value of the integral is the exponential of a quadratic polynomial in $s$, i.e. a Gaussian function.
The $\text{pdf}$ of the sum $s$ of two r.v. $x,y$ can be obtained as the convolution product of the originating distributions, which is an integral, and
$$\text{pdf}(s)\propto\int_{-\infty}^\infty\exp\left(-\frac{(s-t-\mu_x)^2}{2\sigma_x^2}\right)\exp\left(-\frac{(t-\mu_y)^2}{2\sigma_y^2}\right)dt
\\=\int_{-\infty}^\infty\exp\left(-\frac{(s-t-\mu_x)^2}{2\sigma_x^2}-\frac{(t-\mu_y)^2}{2\sigma_y^2}\right)dt.$$
As the exponent is a quadratic polynomial in $t,s$, the value of the integral is the exponential of a quadratic polynomial in $s$, i.e. a Gaussian function.