Answer to simple algebraic formula manipulation I have to show that
$2y+(x + 1) = 3 \cdot 2^{x+1}− (x + 1) − 2$
is equal to
$y=3⋅2^x−x−2$
I can get this far:       
$2y+(x+1)=3⋅2^{x+1}−(x+1)−2$
$2y+(x+1)=3⋅2^{x+1}−x−1−2$
$2y+(x+1)=3⋅2^{x+1}−x−3$
$2y+x+1=3⋅2^{x+1}−x−3$
$2y=3⋅2^{x+1}−x−3−1−x$
$2y=3⋅2^{x+1}−2x−4$
Now I should divide by 2, but the 3 on the right side throws me off. Some help would be appreciated!

I think you missed one $x$ to be an exponent. Go ahead and divide by $2$ to get:
$$
y= 3\cdot \frac{2^{x+1}}2+x-2=3\cdot {2^{x}}+x-2
$$
