if $X$ and $Y$ are Normal random variables with same parameters. is $(aX+bX)=(aX+bY)$? if $X$ and $Y$ are random variables and both are $\sim N(\mu,\sigma^2)$, ie share the same average and variance. Then we defined a new random variable as the linear combination of them as: $Z=(aX+bY)$ is that the same as saying $Z=(aX+bX)$? Is $Z=(aX+bX)$ the same as saying $Z=((a+b)X)$?
You have not said how $X$ and $Y$ are jointly distributed, but only how they are marginally distributed.

If they are both marginally distributed as $N(\mu,\sigma^2)$ then:


At one extreme you have $X=Y,$ in which case $X+Y= 2X \sim N(2\mu,4\sigma^2).$
At an opposite extremem you have $Y = 2\mu-X,$ in which case $X+Y\sim N(2\mu,0),$ i.e. the sum is constant.
Somewhere in between, $X$ and $Y$ are independent. In that case $X+Y\sim N(2\mu, 2\sigma^2).$
And there are cases where there is a nonzero correlation between $X$ and $Y$ but it's not $+1$ or $-1.$
There are also cases in which $X$ and $Y$ are not jointly normally distributed (but still have that same marginal distribution). One of those is where $Y = \pm X$ with plus and minus having equal probabilities and being chosen independently of $X$. In that case $X+Y$ is not normally distributed, since $\Pr(X+Y=0) = 1/2.$
I'm pretty sure the answer is no. An easy way to think about this is physically. AX+BX corresponds to the same event happening twice, whereas AX+BY corresponds to two separate, INDEPENDENT events occuring. With AX+BX, you pull the same point out of the distribution (causing an amplification), whereas with AX+BY, you pull different points (probably causing an increase in variance).

However, I'm not an expert, and there might be some weird mathematical symmetry I don't know about, but I would put it at 94.9% confidence that it is different.