Find the value of $n$ that satisfies $2(n+1)!+6n!=3(n+1)!$, where $n! = n\cdot (n-1)\cdot (n-2) \cdots 2\cdot 1$.
I wonder if I can simplify the equation by dividing both sides by a common factor. I notice that $(n+1)! = (n+1)\cdot n!$, so I can rewrite the equation as $2(n+1)\cdot n!+6n!=3(n+1)\cdot n!$. Now I see that $n!$ is a common factor on both sides, so I can divide both sides by $n!$ and get $2(n+1)+6=3(n+1)$. This is a much simpler equation to solve for $n$. I can expand the brackets and collect like terms to get $2n+8=3n+3$. Then I can subtract $2n$ from both sides and get $8=n+3$. Finally, I can subtract $3$ from both sides and get $n=5$. 